Week 1 R Assignment

by Corey Chu

2.1 Suppose you keep track of your mileage each time you fill up. At your last 6 fill-ups the mileage was 65311 65624 65908 66219 66499 66821 67145 67447 Enter these numbers into R. Use the function diff on the data. What does it give? The difference between elements (2nd - 1st, 3rd - 2nd, etc.)

Note: even though the instructions use x as a variable I opted to name it “difference” to avoid confusion with the x variable used for 2.5.

miles = c(65311, 65624, 65908, 66219, 66499, 66821, 67145, 67447)
x = diff(miles)

You should see the number of miles between fill-ups. Use the max to find the maximum number of miles between fill-ups, the mean function to find the average number of miles and the min to get the minimum number of miles.

max(x)

## [1] 324

mean(miles)

## [1] 66371.75

min(miles)

## [1] 65311

2.2 Suppose you track your commute times for two weeks (10 days) and you find the following times in minutes 17 16 20 24 22 15 21 15 17 22 Enter this into R.

commute <- c(17, 16, 20, 24, 22, 15, 21, 15, 17, 22)

Use the function max to find the longest commute time, the function mean to find the average and the function min to find the minimum.

max(commute)

## [1] 24

mean(commute)

## [1] 18.9

min(commute)

## [1] 15

Oops, the 24 was a mistake. It should have been 18. How can you fix this? Do so, and then find the new average.

commute[4] <- 18
mean(commute)

## [1] 18.3

How many times was your commute 20 minutes or more? To answer this one can try (if you called your numbers commutes)

sum(commute>=20)

## [1] 4

What do you get? 4

What percent of your commutes are less than 17 minutes? 30. How can you answer this with R?

sum(commute<17)/length(commute)*100

## [1] 30

2.3 Your cell phone bill varies from month to month. Suppose your year has the following monthly amounts 46 33 39 37 46 30 48 32 49 35 30 48 Enter this data into a variable called bill.

bill <- c(46, 33, 39, 37, 46, 30, 48, 32, 49, 35, 30, 48)

Use the sum command to find the amount you spent this year on the cell phone.

sum(bill)

## [1] 473

What is the smallest amount you spent in a month? What is the largest?

min(bill)

## [1] 30

max(bill)

## [1] 49

How many months was the amount greater than $40? What percentage was this?

sum(bill>40)

## [1] 5

sum(bill>40)/length(bill)*100

## [1] 41.66667

2.4 You want to buy a used car and find that over 3 months of watching the classifieds you see the following prices (suppose the cars are all similar). 9000 9500 9400 9400 10000 9500 10300 10200 Use R to find the average value and compare it to Edmund’s (http://www.edmunds.com) estimate of $9500.

car <- c(9000, 9500, 9400, 9400, 10000, 9500, 10300, 10200)
mean(car)

## [1] 9662.5

mean(car) <= 9500

## [1] FALSE

Use R to find the minimum value and the maximum value. Which price would you like to pay?

min(car)

## [1] 9000

max(car)

## [1] 10300

I’d like to pay $9000.

2.5 Try to guess the results of these R commands. Remember, the way to access entries in a vector is with []. Suppose we assume

x = c(1,3,5,7,9)
y = c(2,3,5,7,11,13)

Note: These guesses are what I posted in the discussion forum.

x+1 Guess: each of the x vector’s values will be increased by 1 (2, 4, 6, 8, 10).

x+1

## [1]  2  4  6  8 10

y*2 Guess: each of the y vector’s values will be doubled (4, 6, 10, 14, 22, 26).

y*2

## [1]  4  6 10 14 22 26

length(x) and length(y) Guess: this will be the total number of values from both vectors (5+6 = 11)

length(x) + length(y)

## [1] 11

x+y Guess: the shorter vector has five values, so for the first five values the corresponding numbers will be added together. Then x will loop back to the beginning to have its first value added to y’s sixth value (ultimately resulting in 3, 6, 10, 14, 20, 14)

x+y

## Warning in x + y: longer object length is not a multiple of shorter object
## length

## [1]  3  6 10 14 20 14

sum(x>5) and sum(x[x>5]) Guess: x>5 is true when you only have the values 7 and 9, so the first sum is 7+9 = 16. I’m not too sure about the second one: is it (1+3+5+7+9)7 + (1+3+5+7+9)9, resulting in 400?

sum(x>5)

## [1] 2

sum(x[x>5])

## [1] 16

Note: I did 2.5 before I did any of the other problems in this set, which ask for similar queries. Response: I did not expect this. I figure sum(x>5) = 2 because it’s essentially asking to count how many elements fulfill the condition (two: 7 and 9). Is sum(x[x>5]) = 16 because it’s asking to sum all x elements for when x > 5?

sum(x>5 | x< 3) # read | as ’or’, & and ’and’ Guess: There are three elements that are either greater than 5 (7, 9) or less than 3 (1).

sum(x>5 | x< 3)

## [1] 3

y[3] Guess: This will select the third element of y (5)

y[3]

## [1] 5

y[-3] Guess: This will select all elements of y except the third, so 2, 3, 7, 11, 13.

y[-3]

## [1]  2  3  7 11 13

y[x] (What is NA?) Guess: this would select the first, third, fifth, seventh, and ninth elements of y; since y is only six elements long the seventh and ninth will be NA, or not available. The overall result should be 2, 5, 11, NA, NA.

y[x]

## [1]  2  5 11 NA NA

y[y>=7] Guess: assuming x[x>5] means to identify the x values greater than 5, then y[y>=7] manes to identify the y values >=7, namely 7, 11, and 13

y[y>=7]

## [1]  7 11 13

2.6 Let the data x be given by

x = c(1, 8, 2, 6, 3, 8, 5, 5, 5, 5)

Use R to compute the following functions. Note, we use X1 to denote the first element of x (which is 0) etc. Note: is the “0” a typo if x’s first element is 1?

(X1 + X2 + · · · + X10)/10 (use sum)

sum(x)/10

## [1] 4.8

Find log10(Xi) for each i. (Use the log function which by default is base e)

log10(x[1:10])

##  [1] 0.0000000 0.9030900 0.3010300 0.7781513 0.4771213 0.9030900 0.6989700
##  [8] 0.6989700 0.6989700 0.6989700

Find (Xi − 4.4)/2.875 for each i. (Do it all at once)

(x[1:10]-4.4)/2.875

##  [1] -1.1826087  1.2521739 -0.8347826  0.5565217 -0.4869565  1.2521739
##  [7]  0.2086957  0.2086957  0.2086957  0.2086957

Find the difference between the largest and smallest values of x. (This is the range. You can use max and min or guess a built in command.)

max(x) - min(x)

## [1] 7