Stat 133 Lab 8

⇒ /Users/sambamamba/Desktop/Stat_133/Stat_133 Stuff/Lab_8.Rmd

1. The following loop can be done as a vectorized calculation. What is the equivalent vectorized calculation?

vec = 1:50
i = 1

while(i < 50) {
  vec[i] <- i^2
  i <- i + 1
}

Equivalent vectorized calculation:

vec <- c(1:50)
vec[1:49] <- vec[1:49]^2
vec

##  [1]    1    4    9   16   25   36   49   64   81  100  121  144  169  196
## [15]  225  256  289  324  361  400  441  484  529  576  625  676  729  784
## [29]  841  900  961 1024 1089 1156 1225 1296 1369 1444 1521 1600 1681 1764
## [43] 1849 1936 2025 2116 2209 2304 2401   50

2. Write down what the vector x will contain after each line of R code, if they are executed sequentially.

x <- seq(from = 0 , to = 8, by = 2)

x will contain

## [1] 0 2 4 6 8

x[x < 4] <- NA

if x is the same as in part (a), then x will contain

y <- c("NA", "NA", 4, 6, 8)
y

## [1] "NA" "NA" "4"  "6"  "8"

x[] <- 0

x will return

z <- c(0,0,0,0,0)
z

## [1] 0 0 0 0 0

x <- 0

x will return

0

## [1] 0

**3. Fill in the appropriate types of R objects in the sentence: The lapply function in R operates on ____ and returns ___. There may be more than one correct answer.**

lapply in R operates on vectors, lists, matrices, data frames and returns a list.

4. Suppose we have a 5000 by 3 matrix m

set.seed(1337)
m <- matrix(runif(15000, -3, 3), ncol = 3)
head(m)

##            [,1]       [,2]       [,3]
## [1,]  0.4579293 -1.5301484  2.7691707
## [2,]  0.3884528  1.1452789 -1.8301459
## [3,] -2.5560586  1.3678571 -1.1991257
## [4,] -0.2768063 -2.5663416 -0.7060263
## [5,] -0.7603244 -0.5228916 -2.2818951
## [6,] -1.0120953 -2.1568870  2.9346426

1. We need to create a vector containing the sum of the squared entries in each row of m. Write R code to do this in two different ways:

• Using a for loop and saving the result in an object called m.ssq.loop:

m.ssq.loop <- c()
for (i in 1:nrow(m)) {
  b <- sum(m[i,]^2)
  m.ssq.loop <- c(m.ssq.loop, b)
}
head(m.ssq.loop)

## [1] 10.219360  4.811994  9.842371  7.161204  6.058554 14.288626

• Using the apply function and saving the result in an object called m.ssq.apply (See ?apply):

m.ssq.apply <- apply(m, 1, function(x) {sum(x^2)})
head(m.ssq.apply)

## [1] 10.219360  4.811994  9.842371  7.161204  6.058554 14.288626

2. Write a script that will check if your answers in part (a) are the same. It should return TRUE if the two vectors are exactly the same and FALSE otherwise. This is called a sanity check, and you should do them often when coding.

identical(m.ssq.loop, m.ssq.apply)

## [1] TRUE

5. table1 is given below:

Year <- c(2000, 2001)
Algeria <- c(7, 9)
Brazil <- c(12, "NA")
Columbia <- c(16, 18)

table1 <- data.frame(Year, Algeria, Brazil, Columbia)
table1

##   Year Algeria Brazil Columbia
## 1 2000       7     12       16
## 2 2001       9     NA       18

1. Wrangle table1 into the object table2.

table2 <- table1 %>%
  gather(key = Country, value = Value, Algeria, Brazil, Columbia)

## Warning: attributes are not identical across measure variables; they will
## be dropped

table2

##   Year  Country Value
## 1 2000  Algeria     7
## 2 2001  Algeria     9
## 3 2000   Brazil    12
## 4 2001   Brazil    NA
## 5 2000 Columbia    16
## 6 2001 Columbia    18

2. Use table2 to produce table3.

table3 <- table2 %>%
  group_by(Year) %>%
  mutate(Average = mean(as.numeric(Value), na.rm = TRUE)) %>%
  spread(key = Country, value = Value) 

## Warning: NAs introduced by coercion

## Warning: NAs introduced by coercion

table3

## Source: local data frame [2 x 5]
## Groups: Year [2]
## 
##    Year  Average Algeria Brazil Columbia
##   (dbl)    (dbl)   (chr)  (chr)    (chr)
## 1  2000 11.66667       7     12       16
## 2  2001 13.50000       9     NA       18

6. Consider the graphic in the lab prompt.

1. What are the glyphs in this graph?

Answer: The glyphs in this graph are the points on the graph and the smooth lines approximating the points.

2. What are the aesthetics of each glyph?

Answer:

• Points:
  • location (x and y axes), color
• Smooth lines:
  • color, x and y axes

3. Which variables are being mapped to each aesthetic?

Answer:

• X-axis: educ variable

• Y-axis: wage variable

• color: sex variable, with layers “F” and “M”

4. Are the variables qualitative or quantitative?

Answer:

• wage: quantitative

• educ: quantitative

• sex: qualitative

5. What are the guides on this graph?

Answer:

Guide is the display of the scale, and the scale is the relationship between a varibale and the aesthetic to which it is mapped to. As such, a guide is the display of the relationship between variable and aesthetic.

• The guides for the variable educ are the years of education mapped as tick marks on the x-axis. We are shown only the tick marks 5,10, and 15, corresponding to 5, 10, and 15 years of education.

• The guides for wage are the tick marks on the y-axis, with each tick mark denoting dollars per hour, or hourly wage, of each individual case.

• The sex variable guides are displayed by the coloring of the glyphs. For both the lines and the points, the green color represents “M” level of the sex variable, or represents the males, and the red colors on the lines and points represent the “F” layer, or the female sex.

6. Reconstruct this graphic using ggplot and the CPS85 data table. (Hint: the font size of the plot title is 20)

Bernie <- CPS85 %>%
  ggplot(aes(x = educ, y = wage, col = sex)) + geom_point() + geom_smooth(method = "lm", se = TRUE) + theme(plot.title = element_text(size = 20), axis.title = element_text(size = 13)) + labs(title = "Wage vs. Education in CPS85") + ylim(0,15)
Bernie

## Warning: Removed 55 rows containing non-finite values (stat_smooth).

## Warning: Removed 55 rows containing missing values (geom_point).

7. Consider the string my.string given below:

my.string <- "ggplot2 is a data visualization package f or the statistical programming language R"

1. Write a function SplitChars taht takes a character string as input and splits it intoa vector of single character elements. Hint: the function strsplit() may be useful. The result of calling the function on my.string is shown in the prompt.

SplitChars <- function(x) {
  if (!(is.character(x))) {
    stop("x must be a character vector")
  }
  strsplit(x, "")
}
SplitChars(my.string)

## [[1]]
##  [1] "g" "g" "p" "l" "o" "t" "2" " " "i" "s" " " "a" " " "d" "a" "t" "a"
## [18] " " "v" "i" "s" "u" "a" "l" "i" "z" "a" "t" "i" "o" "n" " " "p" "a"
## [35] "c" "k" "a" "g" "e" " " "f" " " "o" "r" " " "t" "h" "e" " " "s" "t"
## [52] "a" "t" "i" "s" "t" "i" "c" "a" "l" " " "p" "r" "o" "g" "r" "a" "m"
## [69] "m" "i" "n" "g" " " "l" "a" "n" "g" "u" "a" "g" "e" " " "R"

2. Write a for loop that counts the number of times that a, s, R, r appears in my.string and saves it in an object called count.

count <- 0
for (i in SplitChars(my.string)[[1]]) {
  if (i == "a" | i == "s" | i == "R" | i == "r") {
    count <- count + 1
  }
}
print(count)

## [1] 20

3. Write a function Reverse that reverses the order of single characters. The result of using it on SplitChars(mystring) is shown in the prompt.

sep <- SplitChars(my.string)
unlist(lapply(sep, rev))

##  [1] "R" " " "e" "g" "a" "u" "g" "n" "a" "l" " " "g" "n" "i" "m" "m" "a"
## [18] "r" "g" "o" "r" "p" " " "l" "a" "c" "i" "t" "s" "i" "t" "a" "t" "s"
## [35] " " "e" "h" "t" " " "r" "o" " " "f" " " "e" "g" "a" "k" "c" "a" "p"
## [52] " " "n" "o" "i" "t" "a" "z" "i" "l" "a" "u" "s" "i" "v" " " "a" "t"
## [69] "a" "d" " " "a" " " "s" "i" " " "2" "t" "o" "l" "p" "g" "g"