Subsetting in General

R has several different subsetting operators ([, [[, and $).

The behavior of these operators will depend on the object they are being used with.

In general there are 6 different data types that can be used to subset:

Positive integers
Negative integers
Logical values
Empty
Zero
Character values (names)

Subsetting Vectors

Vectors - Positive Ints

Returns elements at the given location - note R uses a 1-based indexing scheme.

x = c(1,4,7)
x[c(1,3)]

## [1] 1 7

x[c(1,1)]

## [1] 1 1

x[c(1.9,2.1)]

## [1] 1 4

y = list(1,4,7)
str( y[c(1,3)] )

## List of 2
##  $ : num 1
##  $ : num 7

str( y[c(1,1)] )

## List of 2
##  $ : num 1
##  $ : num 1

str( y[c(1.9,2.1)] )

## List of 2
##  $ : num 1
##  $ : num 4

Vectors - Negative Ints

Excludes elements at the given location

x = c(1,4,7)
x[-1]

## [1] 4 7

x[-c(1,3)]

## [1] 4

x[c(-1,2)]

## Error: only 0's may be mixed with negative subscripts

y = list(1,4,7)
str( y[-1] )

## List of 2
##  $ : num 4
##  $ : num 7

str( y[-c(1,3)] )

## List of 1
##  $ : num 4

y[c(-1,2)]

## Error: only 0's may be mixed with negative subscripts

Vectors - Logical Values

Returns elements that correspond to TRUE in the logical vector. Length of the logical vector is expected to be the same of the vector being subsetted.

x = c(1,4,7,12)
x[c(TRUE,TRUE,FALSE,TRUE)]

## [1]  1  4 12

x[c(TRUE,FALSE)]

## [1] 1 7

x[x %% 2 == 0]

## [1]  4 12

y = list(1,4,7,12)
str( y[c(TRUE,TRUE,FALSE,TRUE)] )

## List of 3
##  $ : num 1
##  $ : num 4
##  $ : num 12

str( y[c(TRUE,FALSE)] )

## List of 2
##  $ : num 1
##  $ : num 7

str( y[y %% 2 == 0] )

## Error: non-numeric argument to binary operator

Vectors - Empty

Returns the original vector.

x = c(1,4,7)
x[]

## [1] 1 4 7

y = list(1,4,7)
str(y[])

## List of 3
##  $ : num 1
##  $ : num 4
##  $ : num 7

Vectors - Zero

Returns an empty vector

x = c(1,4,7)
x[0]

## numeric(0)

y = list(1,4,7)
str(y[0])

##  list()

Vectors - Character Values

If the vector has names, select elements whose names correspond to the character vector.

x = c(a=1,b=4,c=7)
x["a"]

## a 
## 1

x[c("b","c")]

## b c 
## 4 7

y = list(a=1,b=4,c=7)
str(y["a"])

## List of 1
##  $ a: num 1

str(y[c("b","c")])

## List of 2
##  $ b: num 4
##  $ c: num 7

Vectors - Out of bound subsetting

x = c(1,4,7)
x[4]

## [1] NA

x["a"]

## [1] NA

y = list(1,4,7)
str(y[4])

## List of 1
##  $ : NULL

str(y["a"])

## List of 1
##  $ : NULL

Vectors - Missing and NULL

x = c(1,4,7)
x[NA]

## [1] NA NA NA

x[NULL]

## numeric(0)

y = list(1,4,7)
str(y[NA])

## List of 3
##  $ : NULL
##  $ : NULL
##  $ : NULL

str(y[NULL])

##  list()

Vectors - [ vs. [[

[[ subsets like [ except it only subsets a single value. Note that for lists the returned value may not be a list (more on this later).

x = c(1,4,7)
x[[1]]

## [1] 1

y = list(1,4,7)
y[2]

## [[1]]
## [1] 4

y[[2]]

## [1] 4

Vectors - [[ vs. $

$ is equivalent to [[ for character subsetting of lists, by default it uses partial matching (exact=FALSE).

x = c("abc"=1, "def"=5)
x$abc

## Error: $ operator is invalid for atomic vectors

y = list("abc"=1, "def"=5)
y$abc

## [1] 1

y$d

## [1] 5

Logical operators and comparisons

op	Elementwise	Comp	Elementwise
x \| y	True	x < y	True
x & y	True	x > y	True
!x	True	x <= y	True
x \|\| y	False	x >= y	True
x && y	False	x != y	True
xor(x,y)	True	x == y	True
		x %in% y	True (for x)

Exercise 1

Below are 100 values,

x = c(56, 3, 17, 2, 4, 9, 6, 5, 19, 5, 2, 3, 5, 0, 13, 12, 6, 31, 10, 21, 8, 4, 1, 1, 2, 5, 16, 1, 3, 8, 1, 
      3, 4, 8, 5, 2, 8, 6, 18, 40, 10, 20, 1, 27, 2, 11, 14, 5, 7, 0, 3, 0, 7, 0, 8, 10, 10, 12, 8, 82, 
      21, 3, 34, 55, 18, 2, 9, 29, 1, 4, 7, 14, 7, 1, 2, 7, 4, 74, 5, 0, 3, 13, 2, 8, 1, 6, 13, 7, 1, 10, 
      5, 2, 4, 4, 14, 15, 4, 17, 1, 9)

write down how you would create a subset to accomplish each of the following:

Select every third value starting at position 2 in x.
Remove all values with an odd index (e.g. 1, 3, etc.)
Select only the values that are primes. (You may assume all values are less than 100)
Remove every 4th value, but only if it is odd.

Matrices, Data Frames, and Arrays

Matrices and Arrays

Atomic vectors can be treated as multidimensional (2 or more) objects by adding a dim attribute.

x = 1:8
dim(x) = c(2,4)
x

##      [,1] [,2] [,3] [,4]
## [1,]    1    3    5    7
## [2,]    2    4    6    8

matrix(1:8, nrow=2, ncol=4)

##      [,1] [,2] [,3] [,4]
## [1,]    1    3    5    7
## [2,]    2    4    6    8

x = 1:8
attr(x,"dim") = c(2,2,2)
x

## , , 1
## 
##      [,1] [,2]
## [1,]    1    3
## [2,]    2    4
## 
## , , 2
## 
##      [,1] [,2]
## [1,]    5    7
## [2,]    6    8

x = array(1:8,c(2,2,2))
x

## , , 1
## 
##      [,1] [,2]
## [1,]    1    3
## [2,]    2    4
## 
## , , 2
## 
##      [,1] [,2]
## [1,]    5    7
## [2,]    6    8

Naming dimensions

x = array(1:8,c(2,2,2))
colnames(x) = LETTERS[1:2]
rownames(x) = LETTERS[3:4]
dimnames(x)[[3]] = LETTERS[5:6]
x

## , , E
## 
##   A B
## C 1 3
## D 2 4
## 
## , , F
## 
##   A B
## C 5 7
## D 6 8

str(x)

##  int [1:2, 1:2, 1:2] 1 2 3 4 5 6 7 8
##  - attr(*, "dimnames")=List of 3
##   ..$ : chr [1:2] "C" "D"
##   ..$ : chr [1:2] "A" "B"
##   ..$ : chr [1:2] "E" "F"

Subsetting Matrices

x = matrix(1:6, nrow=2, ncol=3, dimnames=list(c("A","B"),c("M","N","O")))

x[1,3]

## [1] 5

x[1:2, 1:2]

##   M N
## A 1 3
## B 2 4

x[1:2,]

##   M N O
## A 1 3 5
## B 2 4 6

x[, 1:2]

##   M N
## A 1 3
## B 2 4

x[-1,-3]

## M N 
## 2 4

x[2,-1]

## N O 
## 4 6

x["A","M"]

## [1] 1

x["A", c("M","O")]

## M O 
## 1 5

x["B",]

## M N O 
## 2 4 6

x[, "C"]

## Error: subscript out of bounds

x[1,"M"]

## [1] 1

x["B"]

## [1] NA

x[1]

## [1] 1

x[-1]

## [1] 2 3 4 5 6

Preserving Subsetting

x = matrix(1:6, nrow=2, ncol=3, dimnames=list(c("A","B"),c("M","N","O")))
x[1, , drop=FALSE]

##   M N O
## A 1 3 5

x[, 2, drop=FALSE]

##   N
## A 3
## B 4

Preserving vs Simplifying Subsets

	Simplifying	Preserving
Vector	`x[[1]]`	`x[1]`
List	`x[[1]]`	`x[1]`
Array	`x[1, ]` `x[, 1]`	`x[1, , drop = FALSE]` `x[, 1, drop = FALSE]`
Factor	`x[1:4, drop = TRUE]`	`x[1:4]`
Data frame	`x[, 1]` `x[[1]]`	`x[, 1, drop = FALSE]` `x[1]`

Factor Subsetting

(x = factor(c("BS", "MS", "PhD", "MS")))

## [1] BS  MS  PhD MS 
## Levels: BS MS PhD

x[1:2]

## [1] BS MS
## Levels: BS MS PhD

x[1:2, drop=TRUE]

## [1] BS MS
## Levels: BS MS

Data Frame Subsetting

df = data.frame(a = 1:2, b = 3:4)
str(df[1])

## 'data.frame':    2 obs. of  1 variable:
##  $ a: int  1 2

str(df[[1]])

##  int [1:2] 1 2

str(df[, "a", drop = FALSE])

## 'data.frame':    2 obs. of  1 variable:
##  $ a: int  1 2

str(df[, "a"])

##  int [1:2] 1 2

str(df["a"])

## 'data.frame':    2 obs. of  1 variable:
##  $ a: int  1 2

str(df[c("a","b","a")])

## 'data.frame':    2 obs. of  3 variables:
##  $ a  : int  1 2
##  $ b  : int  3 4
##  $ a.1: int  1 2

str(df[c(FALSE,TRUE)])

## 'data.frame':    2 obs. of  1 variable:
##  $ b: int  3 4

Subsetting and assignment

Subsets can also be used with assignment to update specific values within an object.

x = c(1, 4, 7)

x[2] = 2
x

## [1] 1 2 7

x[x %% 2 != 0] = x[x %% 2 != 0] + 1
x

## [1] 2 2 8

x[c(1,1)] = c(2,3)
x

## [1] 3 2 8

x = 1:6

x[c(2,NA)] = 1
x

## [1] 1 1 3 4 5 6

x[c(TRUE,NA)] = 1
x

## [1] 1 1 1 4 1 6

x[c(-1,-3)] = 3
x

## [1] 1 3 1 3 3 3

x[] = 6:1
x

## [1] 6 5 4 3 2 1

Deleting list (df) elements

df = data.frame(a = 1:2, b = TRUE, c = c("A", "B"))

df[["b"]] = NULL
str(df)

## 'data.frame':    2 obs. of  2 variables:
##  $ a: int  1 2
##  $ c: Factor w/ 2 levels "A","B": 1 2

df[,"c"] = NULL
str(df)

## 'data.frame':    2 obs. of  1 variable:
##  $ a: int  1 2

Subsets of Subsets

df = data.frame(a = c(5,1,NA,3))

df$a[df$a == 5] = 0
df[["a"]][df[["a"]] == 1] = 0
df[1][df[1] == 3] = 0

str(df)

## 'data.frame':    4 obs. of  1 variable:
##  $ a: num  0 0 NA 0

Exercise 2

Load the course eval data set using the following command:

d = read.csv("~cr173/Sta523/data/evals.csv")

This data frame contains the following variables (columns):

cls_val - students’ average course value rating
prof_val - students’ average professor value rating
rank - professor’s rank (0 - teaching, 1 - tenure track, 2 - tenured)
gender - professor’s gender (0 - male, 1 - female)
cls_level - class level (0 - lower division, 1 - upper division)

Some of the values in data frame are missing. They have been coded using the value -999, make sure that they are properly treated as NAs.

Use subsetting to replace the values of the categorical variables with the appropriate character strings (do not use factors).

Acknowledgments

Above materials are derived in part from the following sources:

Hadley Wickham - Advanced R
R Language Definition