The most basic form of comparison is equality. Let’s briefly recap its syntax. The following statements all evaluate to TRUE (feel free to try them out in the console).
3 == (2 + 1)
"intermediate" != "r"
TRUE != FALSE
"Rchitect" != "rchitect"Notice from the last expression that R is case sensitive: “R” is not equal to “r”. Keep this in mind when solving the exercises in this chapter!
# Comparison of logicals
TRUE == FALSE
# Comparison of numerics
-6 * 14 != 17 - 101
# Comparison of character strings
"useR" =="user"
# Compare a logical with a numeric
TRUE==1Apart from equality operators, Filip also introduced the less than and greater than operators: < and >. You can also add an equal sign to express less than or equal to or greater than or equal to, respectively. Have a look at the following R expressions, that all evaluate to FALSE:
(1 + 2) > 4
"dog" < "Cats"
TRUE <= FALSERemember that for string comparison, R determines the greater than relationship based on alphabetical order. Also, keep in mind that TRUE corresponds to 1 in R, and FALSE coerces to 0 behind the scenes. Therefore, FALSE < TRUE is TRUE.
You are already aware that R is very good with vectors. Without having to change anything about the syntax, R’s relational operators also work on vectors.
Let’s go back to the example that was started in the video. You want to figure out whether your activity on social media platforms have paid off and decide to look at your results for LinkedIn and Facebook. The sample code in the editor initializes the vectors linkedin and facebook. Each of the vectors contains the number of profile views your LinkedIn and Facebook profiles had over the last seven days.
Using relational operators, find a logical answer, i.e. TRUE or FALSE, for the following questions:
# The linkedin and facebook vectors have already been created for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)
facebook <- c(17, 7, 5, 16, 8, 13, 14)
# Popular days
linkedin > 15
# Quiet days
linkedin <= 5
# LinkedIn more popular than Facebook
linkedin > facebookR’s ability to deal with different data structures for comparisons does not stop at vectors. Matrices and relational operators also work together seamlessly!
Instead of in vectors (as in the previous exercise), the LinkedIn and Facebook data is now stored in a matrix called views. The first row contains the LinkedIn information; the second row the Facebook information. The original vectors facebook and linkedin are still available as well.
Using the relational operators you’ve learned so far, try to discover the following:
# The social data has been created for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)
facebook <- c(17, 7, 5, 16, 8, 13, 14)
views <- matrix(c(linkedin, facebook), nrow = 2, byrow = TRUE)
# When does views equal 13?
views == 13
# When is views less than or equal to 14?
views <= 14Before you work your way through the next exercises, have a look at the following R expressions. All of them will evaluate to TRUE:
TRUE & TRUE FALSE | TRUE 5 <= 5 & 2 < 3 3 < 4 | 7 < 6
Watch out: 3 < x < 7 to check if x is between 3 and 7 will not work; you’ll need 3 < x & x < 7 for that.
In this exercise, you’ll be working with the last variable. This variable equals the last value of the linkedin vector that you’ve worked with previously. The linkedin vector represents the number of LinkedIn views your profile had in the last seven days, remember? Both the variables linkedin and last have already been defined in the editor.
Write R expressions to solve the following questions concerning the variable last:
Is last under 5 or above 10? Is last between 15 and 20, excluding 15 but including 20?
# The linkedin and last variable are already defined for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)
last <- tail(linkedin, 1)
# Is last under 5 or above 10?
last < 5 || last > 10
# Is last between 15 (exclusive) and 20 (inclusive)?
last > 15 && last <= 20Like relational operators, logical operators work perfectly fine with vectors and matrices.
Both the vectors linkedin and facebook are available again. Also a matrix - views - has been defined; its first and second row correspond to the linkedin and facebook vectors, respectively. Ready for some advanced queries to gain more insights into your social outreach?
# The social data (linkedin, facebook, views) has been created for you
# linkedin exceeds 10 but facebook below 10
linkedin > 10 & facebook < 10
# FALSE FALSE TRUE FALSE FALSE FALSE FALSE
# When were one or both visited at least 12 times?
linkedin >= 12 | facebook >= 12
# TRUE FALSE TRUE TRUE FALSE TRUE TRUE
# When is views between 11 (exclusive) and 14 (inclusive)?
views > 11 & views <= 14With the things you’ve learned by now, you’re able to solve pretty cool problems.
Instead of recording the number of views for your own LinkedIn profile, suppose you conducted a survey inside the company you’re working for. You’ve asked every employee with a LinkedIn profile how many visits their profile has had over the past seven days. You stored the results in a data frame called li_df. This data frame is available in the workspace; type li_df in the console to check it out.
second <- li_df$day2
# [1] 3 23 18 18 25 20 4 3 22 12 27 13 17 27 6 35 17 6 1 12 15 17 12 8 7
# [26] 25 15 32 29 1 22 11 5 17 12 26 13 10 37 33 19 29 8 22 10 19 27 18 15 28
# Build a logical vector, TRUE if value in second is extreme: extremes
extremes <- second > 25 | second < 5
# Count the number of TRUEs in extremes
sum(extremes)
# 16
# Variables related to your last day of recordings
medium <- "LinkedIn"
num_views <- 14Before diving into some exercises on the if statement, have another look at its syntax:
if (condition) {
expr
}Remember your vectors with social profile views? Let’s look at it from another angle. The medium variable gives information about the social website; the num_views variable denotes the actual number of views that particular medium had on the last day of your recordings. Both these variables have already been defined in the editor.
You can only use an else statement in combination with an if statement. The else statement does not require a condition; its corresponding code is simply run if all of the preceding conditions in the control structure are FALSE. Here’s a recipe for its usage:
if (condition) {
expr1
} else {
expr2
}It’s important that the else keyword comes on the same line as the closing bracket of the if part!
Both if statements that you coded in the previous exercises are already available in the editor. It’s now up to you to extend them with the appropriate else statements!
Add an else statement to both control structures, such that
# Variables related to your last day of recordings
medium <- "LinkedIn"
num_views <- 14
# Control structure for medium
if (medium == "LinkedIn") {
print("Showing LinkedIn information")
} else if (medium == "Facebook") {
# Add code to print correct string when condition is True
print("Showing Facebook information")
} else {
print("Unknown medium")
}
# Control structure for num_views
if (num_views > 15) {
print("You're popular!")
} else if (num_views <= 15 & num_views > 10) {
# Add code to print correct string when condition is TRUE
print("Your number of views is average")
} else {
print("Try to be more visible!")
}The else if statement allows you to further customize your control structure. You can add as many else if statements as you like. Keep in mind that R ignores the remainder of the control structure once a condition has been found that is TRUE and the corresponding expressions have been executed. Here’s an overview of the syntax to freshen your memory:
if (condition1) {
expr1
} else if (condition2) {
expr2
} else if (condition3) {
expr3
} else {
expr4
}Again, It’s important that the else if keywords comes on the same line as the closing bracket of the previous part of the control construct!
# Variables related to your last day of recordings
medium <- "LinkedIn"
num_views <- 14
# Control structure for medium
if (medium == "LinkedIn") {
print("Showing LinkedIn information")
} else if (medium == "Facebook") {
# Add code to print correct string when condition is True
print("Showing Facebook information")
} else {
print("Unknown medium")
}
# Control structure for num_views
if (num_views > 15) {
print("You're popular!")
} else if (num_views <= 15 & num_views > 10) {
# Add code to print correct string when condition is TRUE
print("Your number of views is average")
} else {
print("Try to be more visible!")
}In this exercise, you will combine everything that you’ve learned so far: relational operators, logical operators and control constructs. You’ll need it all!
In the editor, we’ve coded two values beforehand: li and fb, denoting the number of profile views your LinkedIn and Facebook profile had on the last day of recordings. Go through the instructions to create R code that generates a ‘social media score’, sms, based on the values of li and fb.
Finish the control-flow construct with the following behavior:
If both li and fb are strictly below 10, set sms equal to half the sum of li and fb.
Finally, print the resulting sms variable to the console.
# Variables related to your last day of recordings
li <- 15
fb <- 9
# Code the control-flow construct
if (li > 15 & fb > 15) {
sms <- c(sum(li) * 2 + sum(fb) * 2)
} else if (li < 10 & fb < 10) {
sms <- c(sum(li) / 2 + sum(fb) / 2)
} else {
sms <- c(sum(li) + sum(fb))
}
# Print the resulting sms to the console
sms## [1] 24Have another look at some useful math functions that R features:
abs(): Calculate the absolute value.sum(): Calculate the sum of all the values in a data structure.mean(): Calculate the arithmetic mean.round(): Round the values to 0 decimal places by default. Try out ?round in the console for variations of round() and ways to change the number of digits to round to.errors).Calculate the sum of the absolute rounded values of the training errors. You can work in parts, or with a single one-liner. There’s no need to store the result in a variable, just have R print it.
# The errors vector has already been defined for you
errors <- c(1.9, -2.6, 4.0, -9.5, -3.4, 7.3)
# Sum of absolute rounded values of errors
sum(abs(round(errors)))## [1] 29We went ahead and included some code on the right, but there’s still an error. Can you trace it and fix it?
In times of despair, help with functions such as sum() and rev() are a single command away; simply use ?sum and ?rev in the console.
Fix the error by including code on the last line. Remember: you want to call mean() only once!
# Don't edit these two lines
vec1 <- c(1.5, 2.5, 8.4, 3.7, 6.3)
vec2 <- rev(vec1)
# Fix the error
mean(c(abs(vec1), abs(vec2)))## [1] 4.48Let’s get you started with building a while loop from the ground up. Have another look at its recipe:
while (condition) {
expr
}Remember that the condition part of this recipe should become FALSE at some point during the execution. Otherwise, the while loop will go on indefinitely. In DataCamp’s learning interface, your session will be disconnected in this case.
Have a look at the code on the right; it initializes the speed variables and already provides a while loop template to get you started.
Instructions
Code a while loop with the following characteristics:
while loop should check if speed is higher than 30.while loop, print out "Slow down!".while loop, decrease the speed by 7 units. This step is crucial; otherwise your while loop will never stop.# Initialize the speed variable
speed <- 64
# Code the while loop
while (speed > 30) {
print("Slow down!")
speed <- speed - 7
}
# Print out the speed variable
speedR Console
> # Initialize the speed variable
> speed <- 64
>
> # Code the while loop
> while (speed > 30) {
print("Slow down!")
speed <- speed - 7
}
[1] "Slow down!"
[1] "Slow down!"
[1] "Slow down!"
[1] "Slow down!"
[1] "Slow down!"
>
> # Print out the speed variable
> speed
[1] 29In the previous exercise, you simulated the interaction between a driver and a driver’s assistant: When the speed was too high, “Slow down!” got printed out to the console, resulting in a decrease of your speed by 7 units.
There are several ways in which you could make your driver’s assistant more advanced. For example, the assistant could give you different messages based on your speed or provide you with a current speed at a given moment.
A while loop similar to the one you’ve coded in the previous exercise is already available in the editor. It prints out your current speed, but there’s no code that decreases the speed variable yet, which is pretty dangerous. Can you make the appropriate changes?
Instructions
If the speed is greater than 48, have R print out “Slow down big time!”, and decrease the speed by 11. Otherwise, have R simply print out “Slow down!”, and decrease the speed by 6.
# Initialize the speed variable
speed <- 64
# Extend/adapt the while loop
while (speed > 30) {
print(paste("Your speed is",speed))
if (speed > 48) {
print("Slow down big time!")
speed <- speed - 11
} else {
print("Slow down!")
speed <- speed - 6
}
}R Console
> # Initialize the speed variable
> speed <- 64
>
> # Extend/adapt the while loop
> while (speed > 30) {
print(paste("Your speed is",speed))
if (speed > 48) {
print("Slow down big time!")
speed <- speed - 11
} else {
print("Slow down!")
speed <- speed - 6
}
}
[1] "Your speed is 64"
[1] "Slow down big time!"
[1] "Your speed is 53"
[1] "Slow down big time!"
[1] "Your speed is 42"
[1] "Slow down!"
[1] "Your speed is 36"
[1] "Slow down!"There are some very rare situations in which severe speeding is necessary: what if a hurricane is approaching and you have to get away as quickly as possible? You don’t want the driver’s assistant sending you speeding notifications in that scenario, right?
This seems like a great opportunity to include the break statement in the while loop you’ve been working on. Remember that the break statement is a control statement. When R encounters it, the while loop is abandoned completely.
Instructions
Adapt the while loop such that it is abandoned when the speed of the vehicle is greater than 80. This time, the speed variable has been initialized to 88; keep it that way.
# Initialize the speed variable
speed <- 88
while (speed > 30) {
print(paste("Your speed is", speed))
# Break the while loop when speed exceeds 80
if (speed > 80) {
break
}
if (speed > 48) {
print("Slow down big time!")
speed <- speed - 11
} else {
print("Slow down!")
speed <- speed - 6
}
}R Console
> # Initialize the speed variable
> speed <- 88
>
> while (speed > 30) {
print(paste("Your speed is", speed))
# Break the while loop when speed exceeds 80
if (speed > 80) {
break
}
if (speed > 48) {
print("Slow down big time!")
speed <- speed - 11
} else {
print("Slow down!")
speed <- speed - 6
}
}
[1] "Your speed is 88"The previous exercises guided you through developing a pretty advanced while loop, containing a break statement and different messages and updates as determined by control flow constructs. If you manage to solve this comprehensive exercise using a while loop, you’re totally ready for the next topic: the for loop.
Instructions
Finish the while loop so that it:
i, so 3 * i, at each run.break if the triple of i is divisible by 8, but still prints out this triple before breaking.# Initialize i as 1
i <- 1
# Code the while loop
while (i <= 10) {
print(3 * i)
if ( (3 * i) %% 8 == 0 ) {
print(3 * i)
break
}
i <- i + 1
}R Console
> # Initialize i as 1
> i <- 1
>
> # Code the while loop
> while (i <= 10) {
print(3 * i)
if ( (3 * i) %% 8 == 0 ) {
print(3 * i)
break
}
i <- i + 1
}
[1] 3
[1] 6
[1] 9
[1] 12
[1] 15
[1] 18
[1] 21
[1] 24
[1] 24In the previous video, Filip told you about two different strategies for using the for loop. To refresh your memory, consider the following loops that are equivalent in R:
primes <- c(2, 3, 5, 7, 11, 13)
# loop version 1
for (p in primes) {
print(p)
}
# loop version 2
for (i in 1:length(primes)) {
print(primes[i])
}Remember our linkedin vector? It’s a vector that contains the number of views your LinkedIn profile had in the last seven days. The linkedin vector has already been defined in the editor on the right so that you can fully focus on the instructions!
Instructions
Write a for loop that iterates over all the elements of linkedin and prints out every element separately. Do this in two ways: using the loop version 1 and the loop version 2 in the example code above.
# The linkedin vector has already been defined for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)
# Loop version 1
for (p in linkedin) {
print(p)
}
# Loop version 2
for (i in 1:length(linkedin)) {
print(linkedin[i])
}R Console
> # The linkedin vector has already been defined for you
> linkedin <- c(16, 9, 13, 5, 2, 17, 14)
>
> # Loop version 1
> for (p in linkedin) {
print(p)
}
[1] 16
[1] 9
[1] 13
[1] 5
[1] 2
[1] 17
[1] 14
>
>
>
> # Loop version 2
> for (i in 1:length(linkedin)) {
print(linkedin[i])
}
[1] 16
[1] 9
[1] 13
[1] 5
[1] 2
[1] 17
[1] 14Looping over a list is just as easy and convenient as looping over a vector. There are again two different approaches here:
primes_list <- list(2, 3, 5, 7, 11, 13)
# loop version 1
for (p in primes_list) {
print(p)
}
# loop version 2
for (i in 1:length(primes_list)) {
print(primes_list[[i]])
}Notice that you need double square brackets - [[ ]] - to select the list elements in loop version 2.
Suppose you have a list of all sorts of information on New York City: its population size, the names of the boroughs, and whether it is the capital of the United States. We’ve already prepared a list nyc with all this information in the editor (source: Wikipedia).
Instructions
As in the previous exercise, loop over the nyc list in two different ways to print its elements:
nyc list (loop version 1).# The nyc list is already specified
nyc <- list(pop = 8405837,
boroughs = c("Manhattan", "Bronx", "Brooklyn", "Queens", "Staten Island"),
capital = FALSE)
# Loop version 1
for (p in nyc) {
print(p)
}
# Loop version 2
for (i in 1:length(nyc)) {
print(nyc[[i]])
}R Console
> # The nyc list is already specified
> nyc <- list(pop = 8405837,
boroughs = c("Manhattan", "Bronx", "Brooklyn", "Queens", "Staten Island"),
capital = FALSE)
>
> # Loop version 1
>
> for (p in nyc) {
print(p)
}
[1] 8405837
[1] "Manhattan" "Bronx" "Brooklyn" "Queens"
[5] "Staten Island"
[1] FALSE
>
>
>
> # Loop version 2
> for (i in 1:length(nyc)) {
print(nyc[[i]])
}
[1] 8405837
[1] "Manhattan" "Bronx" "Brooklyn" "Queens"
[5] "Staten Island"
[1] FALSEIn your workspace, there’s a matrix ttt, that represents the status of a tic-tac-toe game. It contains the values “X”, “O” and “NA”. Print out ttt in the console so you can have a closer look. On row 1 and column 1, there’s “O”, while on row 3 and column 2 there’s “NA”.
To solve this exercise, you’ll need a for loop inside a for loop, often called a nested loop. Doing this in R is a breeze! Simply use the following recipe:
for (var1 in seq1) {
for (var2 in seq2) {
expr
}
}Instructions
Finish the nested for loops to go over the elements in ttt:
i (use 1:nrow(ttt)).j (use 1:ncol(ttt)).print() and paste() to print out information in the following format: “On row i and column j the board contains x”, where x is the value on that position.# The tic-tac-toe matrix ttt has already been defined for you
# define the double for loop
for (i in 1:nrow(ttt)) {
for (j in 1:ncol(ttt)) {
print(paste("On row", i, "and column", j, "the board contains", ttt[i, j]))
}
}R Console
> # The tic-tac-toe matrix ttt has already been defined for you
>
> # define the double for loop
> for (i in 1:nrow(ttt)) {
for (j in 1:ncol(ttt)) {
print(paste("On row", i, "and column", j, "the board contains", ttt[i, j]))
}
}
[1] "On row 1 and column 1 the board contains O"
[1] "On row 1 and column 2 the board contains NA"
[1] "On row 1 and column 3 the board contains X"
[1] "On row 2 and column 1 the board contains NA"
[1] "On row 2 and column 2 the board contains O"
[1] "On row 2 and column 3 the board contains O"
[1] "On row 3 and column 1 the board contains X"
[1] "On row 3 and column 2 the board contains NA"
[1] "On row 3 and column 3 the board contains X"Let’s return to the LinkedIn profile views data, stored in a vector linkedin. In the first exercise on for loops you already did a simple printout of each element in this vector. A little more in-depth interpretation of this data wouldn’t hurt, right? Time to throw in some conditionals! As with the while loop, you can use the if and else statements inside the for loop.
Instructions
Add code to the for loop that loops over the elements of the linkedin vector:
# The linkedin vector has already been defined for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)
# Code the for loop with conditionals
for (li in linkedin) {
if (li > 10) {
print("You're popular!")
} else {
print("Be more visible!")
}
print(li)
}R Console
> # The linkedin vector has already been defined for you
> linkedin <- c(16, 9, 13, 5, 2, 17, 14)
>
> # Code the for loop with conditionals
> for (li in linkedin) {
if (li > 10) {
print("You're popular!")
} else {
print("Be more visible!")
}
print(li)
}
[1] "You're popular!"
[1] 16
[1] "Be more visible!"
[1] 9
[1] "You're popular!"
[1] 13
[1] "Be more visible!"
[1] 5
[1] "Be more visible!"
[1] 2
[1] "You're popular!"
[1] 17
[1] "You're popular!"
[1] 14In the editor on the right you’ll find a possible solution to the previous exercise. The code loops over the linkedin vector and prints out different messages depending on the values of li.
In this exercise, you will use the break and next statements:
The break statement abandons the active loop: the remaining code in the loop is skipped and the loop is not iterated over anymore. The next statement skips the remainder of the code in the loop, but continues the iteration.
Instructions
Extend the for loop with two new, separate if tests in the editor as follows:
for loop (break).next).# The linkedin vector has already been defined for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)
# Extend the for loop
for (li in linkedin) {
if (li > 10) {
print("You're popular!")
} else {
print("Be more visible!")
}
# Add if statement with break
if (li > 16) {
print("This is ridiculous, I'm outta here!")
break
}
# Add if statement with next
if (li < 5) {
print("This is too embarrassing!")
next
}
print(li)
}R Console
> # The linkedin vector has already been defined for you
> linkedin <- c(16, 9, 13, 5, 2, 17, 14)
>
> # Extend the for loop
> for (li in linkedin) {
if (li > 10) {
print("You're popular!")
} else {
print("Be more visible!")
}
# Add if statement with break
if (li > 16) {
print("This is ridiculous, I'm outta here!")
break
}
# Add if statement with next
if (li < 5) {
print("This is too embarrassing!")
next
}
print(li)
}
[1] "You're popular!"
[1] 16
[1] "Be more visible!"
[1] 9
[1] "You're popular!"
[1] 13
[1] "Be more visible!"
[1] 5
[1] "Be more visible!"
[1] "This is too embarrassing!"
[1] "You're popular!"
[1] "This is ridiculous, I'm outta here!"This exercise will not introduce any new concepts on for loops.
In the editor on the right, we already went ahead and defined a variable rquote. This variable has been split up into a vector that contains separate letters and has been stored in a vector chars with the strsplit() function.
Can you write code that counts the number of r’s that come before the first u in rquote?
Instructions
rcount, as 0.for loop:char equals "r", increase the value of rcount by 1.char equals "u", leave the for loop entirely with a break.rcount to the console to see if your code is correct.# Pre-defined variables
rquote <- "r's internals are irrefutably intriguing"
chars <- strsplit(rquote, split = "")[[1]]
# Initialize rcount
rcount <- 0
# Finish the for loop
for (char in chars) {
if (char == "r") rcount <- rcount + 1
if (char == "u") break
}
# Print out rcount
rcountR Console
> # Pre-defined variables
> rquote <- "r's internals are irrefutably intriguing"
> chars <- strsplit(rquote, split = "")[[1]]
>
> # Initialize rcount
> rcount <- 0
>
> # Finish the for loop
> for (char in chars) {
if (char == "r") rcount <- rcount + 1
if (char == "u") break
}
>
> # Print out rcount
> rcount
[1] 5Before even thinking of using an R function, you should clarify which arguments it expects. All the relevant details such as a description, usage, and arguments can be found in the documentation. To consult the documentation on the sample() function, for example, you can use one of following R commands:
help(sample)
?sampleIf you execute these commands in the console of the DataCamp interface, you’ll be redirected to www.rdocumentation.org.
A quick hack to see the arguments of the sample() function is the args() function. Try it out in the console:
args(sample)In the next exercises, you’ll be learning how to use the mean() function with increasing complexity. The first thing you’ll have to do is get acquainted with the mean() function.
The documentation on the mean() function gives us quite some information:
mean() function computes the arithmetic mean.x and ....x argument should be a vector containing numeric, logical or time-related information.Remember that R can match arguments both by position and by name. Can you still remember the difference? You’ll find out in this exercise!
Once more, you’ll be working with the view counts of your social network profiles for the past 7 days. These are stored in the linkedin and facebook vectors and have already been defined in the editor on the right.
linkedin and facebook and assign the result to avg_li and avg_fb, respectively. Experiment with different types of argument matching!avg_li and avg_fb.Check the documentation on the mean() function again:
?meanThe Usage section of the documentation includes two versions of the mean() function. The first usage,
mean(x, ...)is the most general usage of the mean function. The ‘Default S3 method’, however, is:
mean(x, trim = 0, na.rm = FALSE, ...)The ... is called the ellipsis. It is a way for R to pass arguments along without the function having to name them explicitly. The ellipsis will be treated in more detail in future courses.
For the remainder of this exercise, just work with the second usage of the mean function. Notice that both trim and na.rm have default values. This makes them optional arguments.
linkedin and facebook and store the result in a variable avg_sum.trim argument equal to 0.2 and assign the result to avg_sum_trimmed.avg_sum and avg_sum_trimmed; can you spot the difference?In the video, Filip guided you through the example of specifying arguments of the sd() function. The sd() function has an optional argument, na.rm that specified whether or not to remove missing values from the input vector before calculating the standard deviation.
If you’ve had a good look at the documentation, you’ll know by now that the mean() function also has this argument, na.rm, and it does the exact same thing. By default, it is set to FALSE, as the Usage of the default S3 method shows:
mean(x, trim = 0, na.rm = FALSE, ...)Let’s see what happens if your vectors linkedin and facebook contain missing values (NA).
You already know that R functions return objects that you can then use somewhere else. This makes it easy to use functions inside functions, as you’ve seen before:
speed <- 31
print(paste("Your speed is", speed))Notice that both the print() and paste() functions use the ellipsis - ... - as an argument. Can you figure out how they’re used?
Use abs() on linkedin - facebook to get the absolute differences between the daily Linkedin and Facebook profile views. Next, use this function call inside mean() to calculate the Mean Absolute Deviation. In the mean() call, make sure to specify na.rm to treat missing values correctly!
By now, you will probably have a good understanding of the difference between required and optional arguments. Let’s refresh this difference by having one last look at the mean() function:
mean(x, trim = 0, na.rm = FALSE, ...)x is required; if you do not specify it, R will throw an error. trim and na.rm are optional arguments: they have a default value which is used if the arguments are not explicitly specified.
Which of the following statements about the read.table() function are true?
header, sep and quote are all optional arguments.row.names and fileEncoding don’t have default values.read.table("myfile.txt", "-", TRUE) will throw an error.read.table("myfile.txt", sep = "-", header = TRUE) will throw an error.Wow, things are getting serious… you’re about to write your own function! Before you have a go at it, have a look at the following function template:
my_fun <- function(arg1, arg2) {
body
}Notice that this recipe uses the assignment operator (<-) just as if you were assigning a vector to a variable for example. This is not a coincidence. Creating a function in R basically is the assignment of a function object to a variable! In the recipe above, you’re creating a new R variable my_fun, that becomes available in the workspace as soon as you execute the definition. From then on, you can use the my_fun as a function.
pow_two(): it takes one argument and returns that number squared (that number times itself).12 as input.sum_abs(), that takes two arguments and returns the sum of the absolute values of both arguments.sum_abs() with arguments -2 and 3 afterwards.There are situations in which your function does not require an input. Let’s say you want to write a function that gives us the random outcome of throwing a fair die:
throw_die <- function() {
number <- sample(1:6, size = 1)
number
}
throw_die()Up to you to code a function that doesn’t take any arguments!
hello(). It prints out “Hi there!” and returns TRUE. It has no arguments.hello(), without specifying arguments of course.Do you still remember the difference between an argument with and without default values? Have another look at the sd() function by typing ?sd in the console. The usage section shows the following information:
sd(x, na.rm = FALSE)This tells us that x has to be defined for the sd() function to be called correctly, however, na.rm already has a default value. Not specifying this argument won’t cause an error.
You can define default argument values in your own R functions as well. You can use the following recipe to do so:
my_fun <- function(arg1, arg2 = val2) {
body
}The editor on the right already includes an extended version of the pow_two() function from before. Can you finish it?
print_info, that is TRUE by default.if construct around the print() function: this function should only be executed if print_info is TRUE.pow_two() function you’ve just coded.An issue that Filip did not discuss in the video is function scoping. It implies that variables that are defined inside a function are not accessible outside that function. Try running the following code and see if you understand the results:
pow_two <- function(x) {
y <- x ^ 2
return(y)
}
pow_two(4)
y
xy was defined inside the pow_two() function and therefore it is not accessible outside of that function. This is also true for the function’s arguments of course - x in this case.
Which statement is correct about the following chunk of code? The function two_dice() is already available in the workspace.
two_dice <- function() {
possibilities <- 1:6
dice1 <- sample(possibilities, size = 1)
dice2 <- sample(possibilities, size = 1)
dice1 + dice2
}two_dice() causes an error.res <- two_dice() makes the contents of dice1 and dice2 available outside the function.two_dice() function, R won’t have access to dice1 and dice2 outside the function.The title gives it away already: R passes arguments by value. What does this mean? Simply put, it means that an R function cannot change the variable that you input to that function. Let’s look at a simple example (try it in the console):
triple <- function(x) {
x <- 3*x
x
}
a <- 5
triple(a)
aInside the triple() function, the argument x gets overwritten with its value times three. Afterwards this new x is returned. If you call this function with a variable a set equal to 5, you obtain 15. But did the value of a change? If R were to pass a to triple() by reference, the override of the x inside the function would ripple through to the variable a, outside the function. However, R passes by value, so the R objects you pass to a function can never change unless you do an explicit assignment. a remains equal to 5, even after calling triple(a).
Can you tell which one of the following statements is false about the following piece of code?
increment <- function(x, inc = 1) {
x <- x + inc
x
}
count <- 5
a <- increment(count, 2)
b <- increment(count)
count <- increment(count, 2)a and b equal 7 and 6 respectively after executing this code block.increment(), where a is defined, a equals 7 and count equals 5.count will equal 10.count was actually changed because of the explicit assignment.Now that you’ve acquired some skills in defining functions with different types of arguments and return values, you should try to create more advanced functions. As you’ve noticed in the previous exercises, it’s perfectly possible to add control-flow constructs, loops and even other functions to your function body.
Remember our social media example? The vectors linkedin and facebook are already defined in the workspace so you can get your hands dirty straight away. As a first step, you will be writing a function that can interpret a single value of this vector. In the next exercise, you will write another function that can handle an entire vector at once.
interpret(), that interprets the number of profile views on a single day:num_views.num_views is greater than 15, the function prints out “You’re popular!” to the console and returns num_views.interpret() function twice: on the first value of the linkedin vector and on the second element of the facebook vector.A possible implementation of the interpret() function is already available in the editor. In this exercise you’ll be writing another function that will use the interpret() function to interpret all the data from your daily profile views inside a vector. Furthermore, your function will return the sum of views on popular days, if asked for. A for loop is ideal for iterating over all the vector elements. The ability to return the sum of views on popular days is something you can code through a function argument with a default value.
Finish the template for the interpret_all() function:
return_sum an optional argument, that is TRUE by default.for loop, iterate over all views: on every iteration, add the result of interpret(v) to count. Remember that interpret(v) returns v for popular days, and 0 otherwise. At the same time, interpret(v) will also do some printouts.if construct:return_sum is TRUE, return count.NULL.Call this newly defined function on both linkedin and facebook.
There are basically two extremely important functions when it comes down to R packages:
install.packages(), which as you can expect, installs a given package.library() which loads packages, i.e. attaches them to the search list on your R workspace.To install packages, you need administrator privileges. This means that install.packages() will thus not work in the DataCamp interface. However, almost all CRAN packages are installed on our servers. You can load them with library().
In this exercise, you’ll be learning how to load the ggplot2 package, a powerful package for data visualization. You’ll use it to create a plot of two variables of the mtcars data frame. The data has already been prepared for you in the workspace.
Before starting, execute the following commands in the console:
search(), to look at the currently attached packages andqplot(mtcars$wt, mtcars$hp), to build a plot of two variables of the mtcars data frame.An error should occur, because you haven’t loaded the ggplot2 package yet!
The library() and require() functions are not very picky when it comes down to argument types: both library(rjson) and library("rjson") work perfectly fine for loading a package.
Have a look at some more code chunks that (attempt to) load one or more packages:
# Chunk 1
library(data.table)
require(rjson)
# Chunk 2
library("data.table")
require(rjson)
# Chunk 3
library(data.table)
require(rjson, character.only = TRUE)
# Chunk 4
library(c("data.table", "rjson"))Select the option that lists all of the chunks that do not generate an error. The console on the right is yours to experiment in.
Before you go about solving the exercises below, have a look at the documentation of the lapply() function. The Usage section shows the following expression:
lapply(X, FUN, ...)To put it generally, lapply takes a vector or list X, and applies the function FUN to each of its members. If FUN requires additional arguments, you pass them after you’ve specified X and FUN (…). The output of lapply() is a list, the same length as X, where each element is the result of applying FUN on the corresponding element of X.
Now that you are truly brushing up on your data science skills, let’s revisit some of the most relevant figures in data science history. We’ve compiled a vector of famous mathematicians/statisticians and the year they were born. Up to you to extract some information!
Instructions
# The vector pioneers has already been created for you
pioneers <- c("GAUSS:1777", "BAYES:1702", "PASCAL:1623", "PEARSON:1857")
# Split names from birth year
split_math <- strsplit(pioneers, split = ":")
# Convert to lowercase strings: split_low
split_low <- lapply(split_math, tolower)
# Take a look at the structure of split_low
str(split_low)## List of 4
## $ : chr [1:2] "gauss" "1777"
## $ : chr [1:2] "bayes" "1702"
## $ : chr [1:2] "pascal" "1623"
## $ : chr [1:2] "pearson" "1857"As Filip explained in the instructional video, you can use lapply() on your own functions as well. You just need to code a new function and make sure it is available in the workspace. After that, you can use the function inside lapply() just as you did with base R functions.
In the previous exercise you already used lapply() once to convert the information about your favorite pioneering statisticians to a list of vectors composed of two character strings. Let’s write some code to select the names and the birth years separately.
The sample code already includes code that defined select_first(), that takes a vector as input and returns the first element of this vector.
Instructions
select_first() over the elements of split_low with lapply() and assign the result to a new variable namesselect_second() that does the exact same thing for the second element of an inputted vector.select_second() function over split_low and assign the output to the variable years.# Code from previous exercise:
pioneers <- c("GAUSS:1777", "BAYES:1702", "PASCAL:1623", "PEARSON:1857")
split <- strsplit(pioneers, split = ":")
split_low <- lapply(split, tolower)
# Write function select_first()
select_first <- function(x) {
x[1]
}
# Apply select_first() over split_low: names
names <- lapply(split_low, select_first)
# Write function select_second()
select_second <- function(x) {
x[2]
}
# Apply select_second() over split_low: years
years <- lapply(split_low, select_second)Writing your own functions and then using them inside lapply() is quite an accomplishment! But defining functions to use them only once is kind of overkill, isn’t it? That’s why you can use so-called anonymous functions in R.
Previously, you learned that functions in R are objects in their own right. This means that they aren’t automatically bound to a name. When you create a function, you can use the assignment operator to give the function a name. It’s perfectly possible, however, to not give the function a name. This is called an anonymous function:
# Named function
triple <- function(x) { 3 * x }
# Anonymous function with same implementation
function(x) { 3 * x }
# Use anonymous function inside lapply()
lapply(list(1,2,3), function(x) { 3 * x })Instructions
lapply() such that it uses an anonymous function that does the same thing.lapply to use an anonymous version of the select_second() function.select_first() and select_second(), as they are no longer useful.# Definition of split_low
pioneers <- c("GAUSS:1777", "BAYES:1702", "PASCAL:1623", "PEARSON:1857")
split <- strsplit(pioneers, split = ":")
split_low <- lapply(split, tolower)
# Transform: use anonymous function inside lapply
names <- lapply(split_low, function(x) {
x[1]
})
# Transform: use anonymous function inside lapply
years <- lapply(split_low, function(x) {
x[2]
})Now, there’s another way to solve the issue of using the select_*() functions only once: you can make a more generic function that can be used in more places.
In the video, the triple() function was transformed to the multiply() function to allow for a more generic approach. lapply() provides a way to handle functions that require more than one argument, such as the multiply() function:
multiply <- function(x, factor) {
x * factor
}
lapply(list(1,2,3), multiply, factor = 3)On the right we’ve included a generic version of the select functions that you’ve coded earlier: select_el(). It takes a vector as its first argument, and an index as its second argument. It returns the vector’s element at the specified index.
Instructions
# Definition of split_low
pioneers <- c("GAUSS:1777", "BAYES:1702", "PASCAL:1623", "PEARSON:1857")
split <- strsplit(pioneers, split = ":")
split_low <- lapply(split, tolower)
# Generic select function
select_el <- function(x, index) {
x[index]
}
# Use lapply() twice on split_low: names and years
names <- lapply(split_low, select_el, index = 1)
years <- lapply(split_low, select_el, index = 2)In all of the previous exercises, it was assumed that the functions that were applied over vectors and lists actually returned a meaningful result. For example, the tolower() function simply returns the strings with the characters in lowercase. This won’t always be the case. Suppose you want to display the structure of every element of a list. You could use the str() function for this, which returns NULL:
lapply(list(1, "a", TRUE), str)This call actually returns a list, the same size as the input list, containing all NULL values. On the other hand calling
str(TRUE)## logi TRUEon its own prints only the structure of the logical to the console, not NULL. That’s because str() uses invisible() behind the scenes, which returns an invisible copy of the return value, NULL in this case. This prevents it from being printed when the result of str() is not assigned.
What will the following code chunk return (split_low is already available in the workspace)? Try to reason about the result before simply executing it in the console!
lapply(split_low, function(x) {
if (nchar(x[1]) > 5) {
return(NULL)
} else {
return(x[2])
}
})(_) - list(NULL, NULL, “1623”, “1857”)
(_) - list(“gauss”, “bayes”, NULL, NULL)
(*) - list(“1777”, “1702”, NULL, NULL)
(_) - list(“1777”, “1702”)
You can use sapply() similar to how you used lapply(). The first argument of sapply() is the list or vector X over which you want to apply a function, FUN. Potential additional arguments to this function are specified afterwards (...):
sapply(X, FUN, ...)In the next couple of exercises, you’ll be working with the variable temp, that contains temperature measurements for 7 days. temp is a list of length 7, where each element is a vector of length 5, representing 5 measurements on a given day. This variable has already been defined in the workspace: type str(temp) to see its structure.
Instructions
lapply() to calculate the minimum (built-in function min()) of the temperature measurements for every day.sapply(). See how the output differs.lapply() to compute the the maximum (max()) temperature for each day.sapply() to solve the same question and see how lapply() and sapply() differ.# Use lapply() to find each day's minimum temperature
lapply(temp, min)## [[1]]
## [1] -1
##
## [[2]]
## [1] 5
##
## [[3]]
## [1] -3
##
## [[4]]
## [1] -2
##
## [[5]]
## [1] 2
##
## [[6]]
## [1] -3
##
## [[7]]
## [1] 1
##
## [[8]]
## [1] 7# Use sapply() to find each day's minimum temperature
sapply(temp, min)## [1] -1 5 -3 -2 2 -3 1 7# Use lapply() to find each day's maximum temperature
lapply(temp, max)## [[1]]
## [1] 9
##
## [[2]]
## [1] 13
##
## [[3]]
## [1] 8
##
## [[4]]
## [1] 7
##
## [[5]]
## [1] 9
##
## [[6]]
## [1] 9
##
## [[7]]
## [1] 9
##
## [[8]]
## [1] 7# Use sapply() to find each day's maximum temperature
sapply(temp, max)## [1] 9 13 8 7 9 9 9 7Can you tell the difference between the output of lapply() and sapply()? The former returns a list, while the latter returns a vector that is a simplified version of this list.
Like lapply(), apply() allows you to use self-defined functions and apply them over a vector or a list:
sapply(X, FUN, ...)Here, FUN can be one of R’s built-in functions, but it can also be a function you wrote. This self-written function can be defined before hand, or can be inserted directly as an anonymous function.
Instructions
extremes_avg(): it takes a vector of temperatures and calculates the average of the minimum and maximum temperatures of the vector.sapply() to apply it over the vectors inside temp.lapply() and see how the outputs differ.# Finish function definition of extremes_avg
extremes_avg <- function(x) {
( min(x) + max(x) ) / 2
}
# Apply extremes_avg() over temp using sapply()
sapply(temp, extremes_avg)## [1] 4.0 9.0 2.5 2.5 5.5 3.0 5.0 7.0# Apply extremes_avg() over temp using lapply()
lapply(temp, extremes_avg)## [[1]]
## [1] 4
##
## [[2]]
## [1] 9
##
## [[3]]
## [1] 2.5
##
## [[4]]
## [1] 2.5
##
## [[5]]
## [1] 5.5
##
## [[6]]
## [1] 3
##
## [[7]]
## [1] 5
##
## [[8]]
## [1] 7Of course, you could have solved this exercise using an anonymous function, but this would require you to use the code inside the definition of extremes_avg() twice. Duplicating code should be avoided as much as possible!
In the previous exercises, you’ve seen how sapply() simplifies the list that lapply() would return by turning it into a vector. But what if the function you’re applying over a list or a vector returns a vector of length greater than 1? If you don’t remember from the video, don’t waste more time in the valley of ignorance and head over to the instructions!
Instructions
extremes() function. It takes a vector of numerical values and returns a vector containing the minimum and maximum values of a given vector, with the names “min” and “max”, respectively.sapply().lapply() as well.# Create a function that returns min and max of a vector: extremes
extremes <- function(x) {
c(min = min(x), max = max(x))
}
# Apply extremes() over temp with sapply()
sapply(temp, extremes)## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
## min -1 5 -3 -2 2 -3 1 7
## max 9 13 8 7 9 9 9 7# Apply extremes() over temp with lapply()
lapply(temp, extremes)## [[1]]
## min max
## -1 9
##
## [[2]]
## min max
## 5 13
##
## [[3]]
## min max
## -3 8
##
## [[4]]
## min max
## -2 7
##
## [[5]]
## min max
## 2 9
##
## [[6]]
## min max
## -3 9
##
## [[7]]
## min max
## 1 9
##
## [[8]]
## min max
## 7 7Have a final look at the console and see how sapply() did a great job at simplifying the rather uninformative ‘list of vectors’ that lapply() returns. It actually returned a nicely formatted matrix!
It seems like we’ve hit the jackpot with sapply(). On all of the examples so far, sapply() was able to nicely simplify the rather bulky output of lapply(). But, as with life, there are things you can’t simplify. How does sapply() react?
We already created a function, below_zero(), that takes a vector of numerical values and returns a vector that only contains the values that are strictly below zero.
Instructions
# Definition of below_zero()
below_zero <- function(x) {
return(x[x < 0])
}
# Apply below_zero over temp using sapply(): freezing_s
freezing_s <- sapply(temp, below_zero)
# Apply below_zero over temp using lapply(): freezing_l
freezing_l <- lapply(temp, below_zero)
# Are freezing_s and freezing_l identical?
identical(freezing_l, freezing_s)## [1] TRUEGiven that the length of the output of below_zero() changes for different input vectors, sapply() is not able to nicely convert the output of lapply() to a nicely formatted matrix. Instead, the output values of sapply() and lapply() are exactly the same, as shown by the TRUE output of identical().
You already have some apply tricks under your sleeve, but you’re surely hungry for some more, aren’t you? In this exercise, you’ll see how sapply() reacts when it is used to apply a function that returns NULL over a vector or a list.
A function print_info(), that takes a vector and prints the average of this vector, has already been created for you. It uses the cat() function.
Instructions
# Definition of print_info()
print_info <- function(x) {
cat("The average temperature is", mean(x), "\n")
}
# Apply print_info() over temp using sapply()
sapply(temp, print_info)## The average temperature is 5.6
## The average temperature is 9
## The average temperature is 2.2
## The average temperature is 2.4
## The average temperature is 5.4
## The average temperature is 4.6
## The average temperature is 4.6
## The average temperature is 7## [[1]]
## NULL
##
## [[2]]
## NULL
##
## [[3]]
## NULL
##
## [[4]]
## NULL
##
## [[5]]
## NULL
##
## [[6]]
## NULL
##
## [[7]]
## NULL
##
## [[8]]
## NULL# Apply print_info() over temp using lapply()
lapply(temp, print_info)## The average temperature is 5.6
## The average temperature is 9
## The average temperature is 2.2
## The average temperature is 2.4
## The average temperature is 5.4
## The average temperature is 4.6
## The average temperature is 4.6
## The average temperature is 7## [[1]]
## NULL
##
## [[2]]
## NULL
##
## [[3]]
## NULL
##
## [[4]]
## NULL
##
## [[5]]
## NULL
##
## [[6]]
## NULL
##
## [[7]]
## NULL
##
## [[8]]
## NULLNotice here that, quite surprisingly, sapply() does not simplify the list of NULL's. That’s because the ‘vector-version’ of a list of NULL's would simply be a NULL, which is no longer a vector with the same length as the input. Proceed to the next exercise.
sapply(list(runif (10), runif (10)),
function(x) c(min = min(x), mean = mean(x), max = max(x)))Without going straight to the console to run the code, try to reason through which of the following statements are correct and why.
Select the option that lists all correct statements. (_) - (1) and (3) (*) - (2) and (3) () - (1) and (4) () - (2), (3) and (4)
Before you get your hands dirty with the third and last apply function that you’ll learn about in this intermediate R course, let’s take a look at its syntax. The function is called vapply(), and it has the following syntax:
vapply(X, FUN, FUN.VALUE, ..., USE.NAMES = TRUE)Over the elements inside X, the function FUN is applied. The FUN.VALUE argument expects a template for the return argument of this function FUN. SE.NAMES is TRUE by default; in this case vapply() tries to generate a named array, if possible.
For the next set of exercises, you’ll be working on the temp list again, that contains 7 numerical vectors of length 5. We also coded a function basics() that takes a vector, and returns a named vector of length 3, containing the minimum, mean and maximum value of the vector respectively.
Instructions
# Definition of basics()
basics <- function(x) {
c(min = min(x), mean = mean(x), max = max(x))
}
# Apply basics() over temp using vapply()
vapply(temp, basics, numeric(3))## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
## min -1.0 5 -3.0 -2.0 2.0 -3.0 1.0 7
## mean 5.6 9 2.2 2.4 5.4 4.6 4.6 7
## max 9.0 13 8.0 7.0 9.0 9.0 9.0 7Notice how, just as with sapply(), vapply() neatly transfers the names that you specify in the basics() function to the row names of the matrix that it returns.
So far you’ve seen that vapply() mimics the behavior of sapply() if everything goes according to plan. But what if it doesn’t?
In the video, Filip showed you that there are cases where the structure of the output of the function you want to apply, FUN, does not correspond to the template you specify in FUN.VALUE. In that case, vapply() will throw an error that informs you about the misalignment between expected and actual output.
Instructions
# Definition of the basics() function
basics <- function(x) {
c(min = min(x), mean = mean(x), median = median(x), max = max(x))
}
# Fix the error:
vapply(temp, basics, numeric(4))## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
## min -1.0 5 -3.0 -2.0 2.0 -3.0 1.0 7
## mean 5.6 9 2.2 2.4 5.4 4.6 4.6 7
## median 7.0 9 3.0 2.0 5.0 5.0 4.0 7
## max 9.0 13 8.0 7.0 9.0 9.0 9.0 7As highlighted before, vapply() can be considered a more robust version of sapply(), because you explicitly restrict the output of the function you want to apply. Converting your sapply() expressions in your own R scripts to vapply() expressions is therefore a good practice (and also a breeze!).
Instructions
sapply() expressions on the right to their vapply() counterparts. Their results should be exactly the same; you’re only adding robustness. You’ll need the templates numeric(1) and logical(1).# Convert to vapply() expression
vapply(temp, max, numeric(1))## [1] 9 13 8 7 9 9 9 7# Convert to vapply() expression
vapply(temp, function(x, y) { mean(x) > y }, y = 5, logical(1))## [1] TRUE TRUE FALSE FALSE TRUE FALSE FALSE TRUER features a bunch of functions to juggle around with data structures:: * seq(): Generate sequences, by specifying the from, to, and by arguments.
* rep(): Replicate elements of vectors and lists.
* sort(): Sort a vector in ascending order. Works on numerics, but also on character strings and logicals.
* rev(): Reverse the elements in a data structures for which reversal is defined.
* str(): Display the structure of any R object.
* append(): Merge vectors or lists.
* is.*(): Check for the class of an R object.
* as.*(): Convert an R object from one class to another.
* unlist(): Flatten (possibly embedded) lists to produce a vector.
linkedin and facebook lists to a vector, and store them as li_vec and fb_vec respectively.fb_vec to the li_vec (Facebook data comes last). Save the result as social_vec.social_vec from high to low. Print the resulting vector.# The linkedin and facebook lists have already been created for you
linkedin <- list(16, 9, 13, 5, 2, 17, 14)
facebook <- list(17, 7, 5, 16, 8, 13, 14)
# Convert linkedin and facebook to a vector: li_vec and fb_vec
li_vec <- unlist(linkedin)
fb_vec <- unlist(facebook)
# Append fb_vec to li_vec: social_vec
social_vec <- append(li_vec, fb_vec)
# Sort social_vec
sort(social_vec, decreasing = TRUE)## [1] 17 17 16 16 14 14 13 13 9 8 7 5 5 2Just as before, let’s switch roles. It’s up to you to see what unforgivable mistakes we’ve made. Go fix them!
Correct the expression. Make sure that your fix still uses the functions rep() and seq().
# Fix me
rep(seq(1, 7, by = 2), times = 7)## [1] 1 3 5 7 1 3 5 7 1 3 5 7 1 3 5 7 1 3 5 7 1 3 5 7 1 3 5 7There is a popular story about young Gauss. As a pupil, he had a lazy teacher who wanted to keep the classroom busy by having them add up the numbers 1 to 100. Gauss came up with an answer almost instantaneously, 5050. On the spot, he had developed a formula for calculating the sum of an arithmetic series. There are more general formulas for calculating the sum of an arithmetic series with different starting values and increments. Instead of deriving such a formula, why not use R to calculate the sum of a sequence?
seq(), create a sequence that ranges from 1 to 500 in increments of 3. Assign the resulting vector to a variable seq1.seq(), create a sequence that ranges from 1200 to 900 in increments of -7. Assign it to a variable seq2.sum() function twice and adding the two results, or by first concatenating the sequences and then using the sum() function once. Print the result to the console.# Create first sequence: seq1
seq1 <- seq(1, 500, by = 3)
# Create second sequence: seq2
seq2 <- seq(1200, 900, by = -7)
# Calculate total sum of the sequences
sum(seq1) + sum(seq2)## [1] 87029In their most basic form, regular expressions can be used to see whether a pattern exists inside a character string or a vector of character strings. For this purpose, you can use:
grepl(), which returns TRUE when a pattern is found in the corresponding character string.grep(), which returns a vector of indices of the character strings that contains the pattern.Both functions need a pattern and an x argument, where pattern is the regular expression you want to match for, and the x argument is the character vector from which matches should be sought.
In this and the following exercises, you’ll be querying and manipulating a character vector of email addresses! The vector emails has already been defined in the editor on the right so you can begin with the instructions straight away!
grepl() to generate a vector of logicals that indicates whether these email addressess contain "edu". Print the result to the output.grep(), but this time save the resulting indexes in a variable hits.hits to select from the emails vector only the emails that contain "edu".# The emails vector has already been defined for you
emails <- c("john.doe@ivyleague.edu", "education@world.gov", "dalai.lama@peace.org",
"invalid.edu", "quant@bigdatacollege.edu", "cookie.monster@sesame.tv")
# Use grepl() to match for "edu"
grepl("edu", emails)## [1] TRUE TRUE FALSE TRUE TRUE FALSE# Use grep() to match for "edu", save result to hits
hits <- grep("edu", emails)
# Subset emails using hits
emails[hits]## [1] "john.doe@ivyleague.edu" "education@world.gov"
## [3] "invalid.edu" "quant@bigdatacollege.edu"You can use the caret, ^, and the dollar sign, $ to match the content located in the start and end of a string, respectively. This could take us one step closer to a correct pattern for matching only the “.edu” email addresses from our list of emails. But there’s more that can be added to make the pattern more robust:
@, because a valid email must contain an at-sign..*, which matches any character (.) zero or more times (*). Both the dot and the asterisk are metacharacters. You can use them to match any character between the at-sign and the “.edu” portion of an email address.\\.edu$, to match the “.edu” part of the email at the end of the string. The \\ part escapes the dot: it tells R that you want to use the . as an actual character.grepl() with the more advanced regular expression to return a logical vector. Simply print the result.grep() to create a vector of indices. Store the result in the variable hits.emails[hits] again to subset the emails vector.# The emails vector has already been defined for you
emails <- c("john.doe@ivyleague.edu", "education@world.gov", "dalai.lama@peace.org",
"invalid.edu", "quant@bigdatacollege.edu", "cookie.monster@sesame.tv")
# Use grepl() to match for .edu addresses more robustly
grepl("@.*\\.edu$", emails)## [1] TRUE FALSE FALSE FALSE TRUE FALSE# Use grep() to match for .edu addresses more robustly, save result to hits
hits <- grep("@.*\\.edu$", emails)
# Subset emails using hits
emails[hits]## [1] "john.doe@ivyleague.edu" "quant@bigdatacollege.edu"While grep() and grepl() were used to simply check whether a regular expression could be matched with a character vector, sub() and gsub() take it one step further: you can specify a replacement argument. If inside the character vector x, the regular expression pattern is found, the matching element(s) will be replaced with replacement.sub() only replaces the first match, whereas gsub() replaces all matches.
Suppose that emails vector you’ve been working with is an excerpt of DataCamp’s email database. Why not offer the owners of the .edu email addresses a new email address on the datacamp.edu domain? This could be quite a powerful marketing stunt: Online education is taking over traditional learning institutions! Convert your email and be a part of the new generation!
With the advanced regular expression "@.*\\.edu$", use sub() to replace the match with "@datacamp.edu". Since there will only be one match per character string, gsub() is not necessary here. Inspect the resulting output.
# The emails vector has already been defined for you
emails <- c("john.doe@ivyleague.edu", "education@world.gov", "global@peace.org",
"invalid.edu", "quant@bigdatacollege.edu", "cookie.monster@sesame.tv")
# Use sub() to convert the email domains to datacamp.edu
sub("@.*\\.edu$", "@datacamp.edu", emails)## [1] "john.doe@datacamp.edu" "education@world.gov"
## [3] "global@peace.org" "invalid.edu"
## [5] "quant@datacamp.edu" "cookie.monster@sesame.tv"Regular expressions are a typical concept that you’ll learn by doing and by seeing other examples. Before you rack your brains over the regular expression in this exercise, have a look at the new things that will be used:
.*: A usual suspect! It can be read as “any character that is matched zero or more times”.\\s: Match a space. The “s” is normally a character, escaping it (\\) makes it a metacharacter.[0-9]+: Match the numbers 0 to 9, at least once (+).([0-9]+): The parentheses are used to make parts of the matching string available to define the replacement. The \\1 in the replacement argument of sub() gets set to the string that is captured by the regular expression [0-9]+1.awards <- c("Won 1 Oscar.",
"Won 1 Oscar. Another 9 wins & 24 nominations.",
"1 win and 2 nominations.",
"2 wins & 3 nominations.",
"Nominated for 2 Golden Globes. 1 more win & 2 nominations.",
"4 wins & 1 nomination.")
sub(".*\\s([0-9]+)\\snomination.*$", "\\1", awards)What does this code chunk return? awards is already defined in the workspace so you can start playing in the console straight away.
awards gets returned as there isn’t a single element in awards that matches the regular expression.Correct answer is: A vector of character strings containing “Won 1 Oscar.”, “24”, “2”, “3”, “2”, “1”.
In R, dates are represented by Date objects, while times are represented by POSIXct objects. Under the hood, however, these dates and times are simple numerical values. Date objects store the number of days since the 1st of January in 1970. POSIXct objects on the other hand, store the number of seconds since the 1st of January in 1970.
The 1st of January in 1970 is the common origin for representing times and dates in a wide range of programming languages. There is no particular reason for this; it is a simple convention. Of course, it’s also possible to create dates and times before 1970; the corresponding numerical values are simply negative in this case.
# Get the current date: today
today <- Sys.Date()
# See what today looks like under the hood
unclass(today)## [1] 17476# Get the current time: now
now <- Sys.time()
# See what now looks like under the hood
unclass(now)## [1] 1509989320To create a Date object from a simple character string in R, you can use the as.Date() function. The character string has to obey a format that can be defined using a set of symbols (the examples correspond to 13 January, 1982):
%Y: 4-digit year (1982)%y: 2-digit year (82)%m: 2-digit month (01)%d: 2-digit day of the month (13)%A: weekday (Wednesday)%a: abbreviated weekday (Wed)%B: month (January)%b: abbreviated month (Jan)The following R commands will all create the same Date object for the 13th day in January of 1982:
as.Date("1982-01-13")
as.Date("Jan-13-82", format = "%b-%d-%y")
as.Date("13 January, 1982", format = "%d %B, %Y")Notice that the first line here did not need a format argument, because by default R matches your character string to the formats "%Y-%m-%d" or "%Y/%m/%d".
In addition to creating dates, you can also convert dates to character strings that use a different date notation. For this, you use the format() function. Try the following lines of code:
today <- Sys.Date()
format(Sys.Date(), format = "%d %B, %Y")
format(Sys.Date(), format = "Today is a %A!")as.Date(), and assign them to date1, date2, and date3 respectively. The code for date1 is already included.format(). From the first date, select the weekday. From the second date, select the day of the month. From the third date, you should select the abbreviated month and the 4-digit year, separated by a space.# Definition of character strings representing dates
str1 <- "May 23, '96"
str2 <- "2012-03-15"
str3 <- "30/January/2006"
# Convert the strings to dates: date1, date2, date3
date1 <- as.Date(str1, format = "%b %d, '%y")
date2 <- as.Date(str2)
date3 <- as.Date(str3, format = "%d/%B/%Y")
# Convert dates to formatted strings
format(date1, "%A")## [1] "Thursday"format(date2, "%d")## [1] "15"format(date3, "%b %Y")## [1] "Jan 2006"Similar to working with dates, you can use as.POSIXct() to convert from a character string to a POSIXct object, and format() to convert from a POSIXct object to a character string. Again, you have a wide variety of symbols:
%H: hours as a decimal number (00-23) %I: hours as a decimal number (01-12) %M: minutes as a decimal number %S: seconds as a decimal number %T: shorthand notation for the typical format %H:%M:%S %p: AM/PM indicator
For a full list of conversion symbols, consult the strptime documentation in the console:
?strptimeAgain, as.POSIXct() uses a default format to match character strings. In this case, it’s %Y-%m-%d %H:%M:%S. In this exercise, abstraction is made of different time zones.
str1 and str2, to POSIXct objects called time1 and time2.format(), create a string from time1 containing only the minutes.time2, extract the hours and minutes as “hours:minutes AM/PM”. Refer to the assignment text above to find the correct conversion symbols!# Definition of character strings representing times
str1 <- "May 23, '96 hours:23 minutes:01 seconds:45"
str2 <- "2012-3-12 14:23:08"
# Convert the strings to POSIXct objects: time1, time2
time1 <- as.POSIXct(str1, format = "%B %d, '%y hours:%H minutes:%M seconds:%S")
time2 <- as.POSIXct(str2)
# Convert times to formatted strings
format(time1, "%M")## [1] "01"format(time2, "%I:%M %p")## [1] "02:23 PM"Both Date and POSIXct R objects are represented by simple numerical values under the hood. This makes calculation with time and date objects very straightforward: R performs the calculations using the underlying numerical values, and then converts the result back to human-readable time information again.
You can increment and decrement Date objects, or do actual calculations with them (try it out in the console!):
today <- Sys.Date()
today + 1
today - 1
as.Date("2015-03-12") - as.Date("2015-02-27")To control your eating habits, you decided to write down the dates of the last five days that you ate pizza. In the workspace, these dates are defined as five Date objects, day1 to day5. The code on the right also contains a vector pizza with these 5 Date objects.
diff() on pizza to calculate the differences between consecutive pizza days. Store the result in a new variable day_diff.# day1, day2, day3, day4 and day5 are already available in the workspace
# Difference between last and first pizza day
day5 - day1
# Create vector pizza
pizza <- c(day1, day2, day3, day4, day5)
# Create differences between consecutive pizza days: day_diff
day_diff <- diff(pizza)
# Average period between two consecutive pizza days
mean(day_diff)## [1] "Time difference of 18 days"## [1] "Time difference of 4.5 days"Calculations using POSIXct objects are completely analogous to those using Date objects. Try to experiment with this code to increase or decrease POSIXct objects:
now <- Sys.time()
now + 3600 # add an hour
now - 3600 * 24 # subtract a dayAdding or substracting time objects is also straightforward:
birth <- as.POSIXct("1879-03-14 14:37:23")
death <- as.POSIXct("1955-04-18 03:47:12")
einstein <- death - birth
einsteinYou’re developing a website that requires users to log in and out. You want to know what is the total and average amount of time a particular user spends on your website. This user has logged in 5 times and logged out 5 times as well. These times are gathered in the vectors login and logout, which are already defined in the workspace.
logout and login, i.e. the time the user was online in each independent session. Store the result in a variable time_online.time_online by printing it.# login and logout are already defined in the workspace
# Calculate the difference between login and logout: time_online
time_online <- logout - login
# Inspect the variable time_online
time_online
# Calculate the total time online
sum(time_online)
# Calculate the average time online
mean(time_online)## [1] "Difference in seconds: 2305.11818 34.18472 837.18182 2397.90153 1851.30411"## [1] "Time difference of 7425.69 secs"## [1] "Time difference of 1485.138 secs"The dates when a season begins and ends can vary depending on who you ask. People in Australia will tell you that spring starts on September 1st. The Irish people in the Northern hemisphere will swear that spring starts on February 1st, with the celebration of St. Brigid’s Day. Then there’s also the difference between astronomical and meteorological seasons: while astronomers are used to equinoxes and solstices, meteorologists divide the year into 4 fixed seasons that are each three months long. (source: www.timeanddate.com)
A vector astro, which contains character strings representing the dates on which the 4 astronomical seasons start, has been defined on your workspace. Similarly, a vector meteo has already been created for you, with the meteorological beginnings of a season.
as.Date() to convert the astro vector to a vector containing Date objects. You will need the %d, %b and %Y symbols to specify the format. Store the resulting vector as astro_dates.as.Date() to convert the meteo vector to a vector with Date objects. This time, you will need the %B, %d and %y symbols for the format argument. Store the resulting vector as meteo_dates.max(), abs() and -, calculate the maximum absolute difference between the astronomical and the meteorological beginnings of a season, i.e. astro_dates and meteo_dates. Simply print this maximum difference to the console output.# Convert astro to vector of Date objects: astro_dates
astro_dates <- as.Date(astro, format = "%d-%b-%Y")
# Convert meteo to vector of Date objects: meteo_dates
meteo_dates <- as.Date(meteo, format = "%B %d, %y")
# Calculate the maximum absolute difference between astro_dates and meteo_dates
max(abs(meteo_dates - astro_dates))## [1] "Time difference of 24 days"