For this assignment you will practice creating, examining, and combining different data structures in R. This assignment is different from others in that it takes a worksheet format with built-in error checking. Each time you complete an answer, if you knit the document it should check your answer (but don’t remove “NULL” on an answer before you want it checked or you’ll get an error!) For this reason, you should re-knit every time you answer a question, so that if something goes wrong, you know the last edit caused the issue!
You will need to do the following:
#
); remove and replace the comments with
your code:# DO NOT EDIT
”;
these are used to test your code.>
) in designated locations to answer questions.
Example for Code Questions:
# Ex 1) Create a vector of the numbers 1, 5, 3, 2, 4.
example <- NULL
# You would write this:
example <- c(1,5,3,2,4)
# Ex 2) Add the number 10 to the sixth position of example2
example2 <- example
example2 <- NULL
# You would write:
example2[6] <- 10
Example for Text Questions: 1) “How do vectors differ from lists?”
A vector is a one dimensional object where every element is the same type of data. Lists are one dimensional but elements can be different data types.
# 1) Use "seq()" to create a vector of numbers from 0 to 45 in increments of 5
vec_num <- seq(0,45,by=5)
# 2) Use ":" to create an integer vector of the numbers 11 through 20.
vec_int <- 11:20
# 3) "LETTERS" contains the 26 capital letters in order. Use "LETTERS" and "[ ]" to create a vector of the last 10 capital letters.
vec_cha <- LETTERS[17:26]
# 4) "letters" contains the 26 lowercase letters in order. Use "factor", "letters", and "[ ]" to create a factor variable using the first ten lower case letters.
vec_fac <- factor(letters[1:10])
# 5) Use "c()" to combine "vec_cha" and "vec_fac" into "vec_let". Do not convert it to a factor!
vec_let <- c(vec_cha,vec_fac)
# 6) Use "c()" and "[ ]" to combine the first 4 elements of "vec_num" with the last
# 4 elements of "vec_int" to create "vec_ni".
vec_ni <- c(vec_num[1:4],vec_int[7:10])
# 7) Use "rev()" to reverse the order of "vec_fac".
fac_vec <- rev(vec_fac)
How’d you do?
Good job on vec_num!
Good job on vec_int!
Good job on vec_cha!
Good job on vec_fac!
Good job on vec_let!
Good job on vec_ni!
Good job on fac_vec!
c()
to combine vec_int
with
vec_fac
, what class of vector would you get?
Why?Answer: “integer”, because vectors can only be of one type and factors have an underlying numeric representation (as integers!)
new_vec <- c(TRUE, FALSE, TRUE, TRUE)
Answer:(1) “logical”; (2) you get 1s and 0s, you get “TRUE”s and “FALSE”s, you get “0”s, and “1”s; (3) you get an error because R doesn’t know what to do!
# 1) Use matrix() to create a matrix with 10 rows and four columns filled with NA
mat_empty <- matrix(NA,nrow=10,ncol=4)
# 2) Assign "vec_num" to the first column of "mat_1" below.
mat_1 <- mat_empty # DO NOT EDIT THIS LINE; add code below it.
mat_1[,1] <- vec_num
# 3) Assign "vec_int" to the second column of "mat_2" below
mat_2 <- mat_1 # DO NOT EDIT THIS LINE; add code below it.
mat_2[,2] <- vec_int
# 4) Assign "vec_cha" and "vec_fac" to the third and fourth columns of "mat_3" using one assignment operator.
mat_3 <- mat_2 # DO NOT EDIT THIS LINE; add code below it.
mat_3[,3:4] <- c(vec_cha,vec_fac)
# 5) Select the fourth row from "mat_3" and assign it to the object "row_4" as a vector.
row_4 <- mat_3[4,]
# 6) Assign the element in the 6th row and 2nd column of "mat_3" to "val_6_2" as a numeric value (using as.numeric).
val_6_2 <- as.numeric(mat_3[6,2])
# 7) Use "cbind()" to combine "vec_num", "vec_int", "vec_cha", and "vec_fac" into "mat_4".
mat_4 <- cbind(vec_num,vec_int,vec_cha,vec_fac)
# 8) Next, first transpose mat_4, then select only the first four columns and assign to mat_t
mat_t <- t(mat_4)[,1:4]
# 9) Then use rbind() to add the rows from mat_3 to mat_t (mat_t first, mat_3 second) and assign this combination to mat_big.
mat_big <- rbind(mat_t,mat_3)
How’d you do?
Good job on mat_empty!
Good job on mat_1!
Good job on mat_2!
Good job on mat_3!
Good job on row_4!
Good job on val_6_2!
Good job on mat_4!
Good job on mat_t!
Good job on mat_big!
names()
from mat_4
? What about
colnames()
? What about rownames()
? Can you
guess why you get all these results?Answer: You get NULL for “names” and “rownames”, but the underlying vector names with “colnames”. That’s because matrices stores “column names” but nothing else! “names” is a command reserved for data.frames and tibbles.
mat_letters <- matrix(letters, ncol=2)
"a"
to "m"
in the
first column and "n"
to "z"
in the second.
What would be an easy way to make the matrix go in alphabetical order
left to right, top to bottom?Answer: Add the argument “nrow=TRUE”.
math_mat <- matrix(1:5, nrow=5, ncol=5)
math_vec <- 1:5
math_mat
and math_vec
. When you add
math_mat + math_vec
, what happens?math_mat %*% math_vec
and from
math_mat * math_vec
. Can you tell what is happening?Answer: (1) the vector mat_vec is added to each columh of math_mat; (2) the first does matrix multiplication, while the second does element-wise multiplication.
# 1) Use "list()" to create a list that contains "vec_num" and "row_4", and assign the names
# "vec_num" and "row_4" to these two elements of "list_1".
list_1 <- list("vec_num"=vec_num,"row_4"=row_4)
# 2) Using "$", extract "row_4" from "list_1" and assign it to the object "row_4_2".
row_4_2 <- list_1$row_4
# 3) Create another list that contains "val_6_2" and "mat_big".
list_2 <- list("val_6_2"=val_6_2,"mat_big"=mat_big)
# 4) Combine list_1 and list_2 together using "c()" and assign them to "list_3"
list_3 <- c(list_1,list_2)
# 5) Use "unlist()" to turn "list_3" into a vector and assign it to "vector_3"
vector_3 <- unlist(list_3)
# 6) Use "as.list()" to convert "vector_3" into a list and assign it to "list_big"
list_big <- as.list(vector_3)
# 7) Now copy "list_3" as "list_4" (one line of code). Then use "[[ ]]" to assign "list_3" as the last (fifth) element of "list_4";
# that is, you should have a list object with five elements named "list_4" that contains the same four
# elements as "list_3" plus a fifth element that is -all- four elements of "list_3" as one object.
list_4 <- list_3
list_4[[5]] <- list_3
# 8) Select the third element (that is, the sub-element) of the the fifth element of "list_4" and assign it
# to element_5_3 using "[[ ]]".
element_5_3 <- list_4[[5]][[3]]
# 9) Lastly, repeat the previous assignment of the third element of the fifth element, but
# extract the element as a list rather than scalar using "[ ]" and assign it to "list_5_3".
list_5_3 <- list_4[[5]][3]
How’d you do?
Good job on list_1!
Good job on row_4_2!
Good job on list_2!
Good job on list_3!
Good job on vector_3!
Good job on list_big!
Good job on list_4!
Good job on element_5_3!
Good job on list_5_3!
Many functions in R produce lists as output because they produce
objects with different types of data and of different lengths. For
instance, consider the linear regression saved to lm.output
below. Don’t worry if you are not familiar with regression, we’re just
concerned with what the function produces!
lm.output <- lm(mpg ~ wt, data=mtcars)
lm.output
##
## Call:
## lm(formula = mpg ~ wt, data = mtcars)
##
## Coefficients:
## (Intercept) wt
## 37.285 -5.344
lm()
produce? Hint: use the function
length()
.Answer: 12 (length(lm.output))
model
element?Answer: 32 rows and 2 columns (dim(lm.output$model))
coefficients
for the intercept
and
wt
? Remember to call on them from lm.output
object for your answer!Answer: 37.285126 and -5.344472 (lm.output$coefficients)
# 1) Use "data.frame()" to combine "vec_num" (first column) and "vec_int" (second column) into "df_1".
df_1 <- data.frame(vec_num,vec_int)
# 2) Use "$" to extract "vec_num" from "df_1", reverse it with "rev()", and assign it as the vector "vec_num_2".
vec_num_2 <- rev(df_1$vec_num)
# 3) Use "$" to add "vec_num_2" to "df_2" as a new column with the name "number_vector".
df_2 <- df_1 # DO NOT EDIT THIS LINE; add code below it.
df_2$number_vector <- vec_num_2
# 4) Combine "df_2" with itself using "rbind()" to create "df_3"
df_3 <- rbind(df_2,df_2)
How’d you do?
Good job on df_1!
Good job on vec_num_2!
Good job on df_2!
Good job on df_3!
names()
, colnames()
, and
rownames()
on df_1. How does this compare to the behavior
of these functions on lists and matrices?Answer: All these functions “work” now! names and colnames tell us the vector names, while rownames is just the numbers between 1 and 10.
length()
and
dim()
differ between data frames, lists, matrices, and
vectors?Answer: length gives number of columns in a dataframe, number of elents in a list, number of total numbers in a matrix, and length of a vector. dim gives number of rows/columsn for data frames and matrices, but doesn’t work for lists and vectors.