Subsetting in R

• Vectors

x <- c(2, 4, 6, 8)
x[c(3, 1)]

[1] 6 2

x[c(2:4, 1, 3)]

[1] 4 6 8 2 6

• Vectors can be named

x <- c(first = 2, second = 4,
       third = 6, forth = 8)
x[c(3, 1)]

third first 
    6     2 

x["second"]

second 
     4 

• Negative indices to exclude elements

x

 first second  third  forth 
     2      4      6      8 

x[-c(3, 2)]

first forth 
    2     8 

x[c(-1, -2)]

third forth 
    6     8 

x[c(-1, 2)]

Error in x[c(-1, 2)] : only 0's may be mixed with negative subscripts

x <- matrix(c(2, 4, 6, 8, 10, 12), ncol = 3)
x

     [,1] [,2] [,3]
[1,]    2    6   10
[2,]    4    8   12

x[2, 1]  # row, column

[1] 4

• Additional Arguments for []

x[1, ]

[1]  2  6 10

x[1, , drop = FALSE]

     [,1] [,2] [,3]
[1,]    2    6   10

• For more, search in help: ?"["

• Structure

str(x)

 num [1:2, 1:3] 2 4 6 8 10 12

• Dimension

dim(x)

[1] 2 3

dat <- data.frame(x = c(1, 2),
                  y = c("A", "B"),
                  z = c("a", "b"))
dat

  x y z
1 1 A a
2 2 B b

dat[1, 2]  # access first row, second column

[1] A
Levels: A B

These are all the same!

dat[1, 2]
dat[1, "y"]  # Recommended!
dat[1, 'y']
dat[1, 2:2]

dat$y[1]
dat[["y"]][1]

• dat[1, 2:2] is redundant, but just so you know sequences are do-able in subsetting, too.
• We recommend you to always use double quotes (" vs ') in R, and always use names to access columns, if possible.

• Subsetting by row names, just as column names

dat

  x y z
1 1 A a
2 2 B b

row.names(dat) <- c("first", "second")
dat["first", "y", drop = FALSE]

      y
first A

• when omitting “,”, it picks columns

dat["y"]

       y
first  A
second B

dat[c(2, 3)]

       y z
first  A a
second B b

• This is much better.

dat[, c(2, 3)]  # better

       y z
first  A a
second B b

dat[, c("y", "z")]  # even better

       y z
first  A a
second B b

• Because you know immediately the code will return a new data frame.

• Data Frames Are Special Lists

x <- list(x = c(1, 2, 3, 3, 4), y = c(4, 8),
          z = c("burp", "blah"))
str(x)

List of 3
 $ x: num [1:5] 1 2 3 3 4
 $ y: num [1:2] 4 8
 $ z: chr [1:2] "burp" "blah"

x[["z"]][1]

[1] "burp"

What is behind dat[dat$x > 5] ?

• Boolean data type: TRUE, FALSE
• Logical Expressions Returns a Logical Vector (Booleans)

# constructing a list of random numbers
x <- round(rnorm(n = 6, mean = 5, sd = 3))
x

[1]  9 -2  8  3  7  7

x < 5  # a logical expression

[1] FALSE  TRUE FALSE  TRUE FALSE FALSE

• Remember our x is:

[1]  9 -2  8  3  7  7

• Booleans can be used to subset vectors

# if we just pass in a vector of booleans
x[c(FALSE, TRUE)]

[1] -2  3  7

• The (short) vector repeats itself.

[1]  9 -2  8  3  7  7

y <- x < 5
y

[1] FALSE  TRUE FALSE  TRUE FALSE FALSE

x[y] == x[x < 5]

[1] TRUE TRUE

• Make sure you understand why!

[1]  9 -2  8  3  7  7

# X & Y <-> intersect(x, y)
x[x < 5 & x > 2]

[1] 3

# X | Y <-> union(x, y)
x[x < 5 | x > 7]

[1]  9 -2  8  3

[1]  9 -2  8  3  7  7

# X & !Y <-> setdiff(x, y)
y3 <- x < 5 & !(x < 3)   # !(x < 3) == x >= 3
y3

[1] FALSE FALSE FALSE  TRUE FALSE FALSE

x[y3]

[1] 3

dat <- data.frame(x = c(1, NA, 3),
                  y = c("A", "B", "B"),
                  z = c("a", "b", "c"))
dat

   x y z
1  1 A a
2 NA B b
3  3 B c

dat[dat$x < 2, ]

    x    y    z
1   1    A    a
NA NA <NA> <NA>

• Why is this? Try split steps.

dat$x < 2

[1]  TRUE    NA FALSE

dat[c(TRUE, NA, FALSE), ]

    x    y    z
1   1    A    a
NA NA <NA> <NA>

• Solution: use which()

dat[which(dat$x < 2), ]

  x y z
1 1 A a

   x y z
1  1 A a
2 NA B b
3  3 B c

subset(dat, x < 2)

  x y z
1 1 A a

• But, as you remember…

dat[dat$x < 2, ]

    x    y    z
1   1    A    a
NA NA <NA> <NA>

• So subset function can eliminate NAs for you. But it is still not recommended for production code because of some other reasons.

That's all you need to know about subsetting in R.

Questions? Email me: yang.jianc@husky.neu.edu