typeof(df)

## [1] "list"

attributes(df)

## $names
## [1] "x" "y"
## 
## $row.names
## [1] 1 2 3
## 
## $class
## [1] "data.frame"

df2 = list(x = 1:3, y = factor(c("a", "b", "c")))
attr(df2,"class") = "data.frame"
attr(df2,"row.names") = 1:3
str(df2)

## 'data.frame':    3 obs. of  2 variables:
##  $ x: int  1 2 3
##  $ y: Factor w/ 3 levels "a","b","c": 1 2 3

If an R vector is shorter than expected, R will increase the length by repeating elements of the short vector. If the lengths are evenly divisible this will occur without any output / warning.

For data frames if the lengths are not evenly divisible then there will be an error.

data.frame(x = 1:3, y = c("a"))

##   x y
## 1 1 a
## 2 2 a
## 3 3 a

data.frame(x = 1:3, y = c("a","b"))

## Error in data.frame(x = 1:3, y = c("a", "b")): arguments imply differing number of rows: 3, 2

We can add rows or columns to a data frame using rbind and cbind respectively.

df = data.frame(x = 1:3, y = c("a","b","c"))
str(rbind(df, c(TRUE,FALSE)))

## Warning in `[<-.factor`(`*tmp*`, ri, value = FALSE): invalid factor level,
## NA generated

## 'data.frame':    4 obs. of  2 variables:
##  $ x: int  1 2 3 1
##  $ y: Factor w/ 3 levels "a","b","c": 1 2 3 NA

str(cbind(df, z=TRUE))

## 'data.frame':    3 obs. of  3 variables:
##  $ x: int  1 2 3
##  $ y: Factor w/ 3 levels "a","b","c": 1 2 3
##  $ z: logi  TRUE TRUE TRUE

df1 = data.frame(x = 1:3, y = c("a","b","c"))
df2 = data.frame(m = 3:1, n = c(TRUE,TRUE,FALSE))
str(cbind(df1,df2))

## 'data.frame':    3 obs. of  4 variables:
##  $ x: int  1 2 3
##  $ y: Factor w/ 3 levels "a","b","c": 1 2 3
##  $ m: int  3 2 1
##  $ n: logi  TRUE TRUE FALSE

# Bad
str(rbind(cbind(df1,df2),c(1,"a",1,1)))

## 'data.frame':    4 obs. of  4 variables:
##  $ x: chr  "1" "2" "3" "1"
##  $ y: Factor w/ 3 levels "a","b","c": 1 2 3 1
##  $ m: chr  "3" "2" "1" "1"
##  $ n: chr  "TRUE" "TRUE" "FALSE" "1"

# Good
str(rbind(cbind(df1,df2),list(1,"a",1,1)))

## 'data.frame':    4 obs. of  4 variables:
##  $ x: num  1 2 3 1
##  $ y: Factor w/ 3 levels "a","b","c": 1 2 3 1
##  $ m: num  3 2 1 1
##  $ n: num  1 1 0 1

Construct a data frame that contains the following data (in as efficient a manner as possible). Hint - the rep function might prove useful.

  Patient    Gender          Treatment 1     Treatment 2     Treatment 3
---------- --------------- --------------- --------------- ---------------
  1          Male            Yes             Yes             Yes
  2          Male            Yes             Yes             No 
  3          Male            Yes             No              Yes
  4          Male            Yes             No              No
  5          Male            No              Yes             Yes
  6          Male            No              Yes             No
  7          Male            No              No              Yes
  8          Male            No              No              No
  9          Female          Yes             Yes             Yes 
  10         Female          Yes             Yes             No
  11         Female          Yes             No              Yes
  12         Female          Yes             No              No
  13         Female          No              Yes             Yes
  14         Female          No              Yes             No
  15         Female          No              No              Yes
  16         Female          No              No              No

Returns elements at the given location (note R uses a 1-based not a 0-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

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

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 in y%%2: non-numeric argument to binary operator

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

[[ subsets like [ except it only subsets a single value.

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

## [1] 1

x[["a"]]

## [1] 1

x[[1:2]]

## Error in x[[1:2]]: attempt to select more than one element in vectorIndex

Note that for lists the returned value may not be a list (more on this later).

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

## $b
## [1] 4

y[[2]]

## [1] 4

y[["b"]]

## [1] 4

y[[1:2]]

## Error in y[[1:2]]: subscript out of bounds

$ is equivalent to [[ but it only works for lists, by default it uses partial matching (exact=FALSE).

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

## Error in x$abc: $ operator is invalid for atomic vectors

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

## [1] 1

y$d

## [1] 5

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.

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

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

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

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

x[1,3]

## [1] 5

x[1:2, 1:2]

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

x[, 1:2]

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

x[-1,-3]

## M N 
## 2 4

By default R's [ subset operator is a preserving subset operator, in that the returned object will have the same type as the parent. Confusingly, when used with a matrix or array [ becomes a simplifying operator (does not preserve type) - this behavior can be controlled by the drop argument.

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

## M N O 
## 1 3 5

x[1, , drop=TRUE]

## M N O 
## 1 3 5

x[1, , drop=FALSE]

##   M N O
## A 1 3 5

If provided with a single value, data frames assume you want to subset a column or columns - multiple values then the data frame is treated as a matrix.

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

##   a
## 1 1
## 2 2

df[[1]]

## [1] 1 2

df[, "a"]

## [1] 1 2

df["a"]

##   a
## 1 1
## 2 2

df[, "a", drop = FALSE]

##   a
## 1 1
## 2 2

df[1,]

##   a b
## 1 1 3

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

##   a b a.1
## 1 1 3   1
## 2 2 4   2

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

grades = data.frame(
            student = c("Alice","Bob","Carol","Dan","Eve","Frank",
                        "Mallory","Oscar","Peggy","Sam","Wendy"),
            grade   = c(82, 78, 62, 98, 64, 53, 86, 73, 54, 57, 61),
            year    = c(3L, 2L, 2L, 1L, 3L, 3L, 4L, 3L, 2L, 2L, 1L),
            stringsAsFactors = FALSE
         )

For the above data frame use subsetting and subsetting assignment to add two new features (columns) to the data set:

• the student's letter grade (factor vector with labels A - F)
  • A (90-100), B (80-89), C (70-79), D (60-69), F (0-59)
• the student's passing status the class (logical vector)
  • TRUE for a grade of A, B, or C
  • FALSE for a grade of D or F

These changes should not be hard coded - if I gave you a new data frame your code should still produce the correct answer.

Hadley Wickham has a package that modifies data frames to be more modern, or as he calls them surly and lazy.

library(tibble)
class(iris)

## [1] "data.frame"

tbl_iris = as_tibble(iris)
class(tbl_iris)

## [1] "tbl_df"     "tbl"        "data.frame"

tbl_iris

## # A tibble: 150 x 5
##    Sepal.Length Sepal.Width Petal.Length Petal.Width Species
##           <dbl>       <dbl>        <dbl>       <dbl>  <fctr>
## 1           5.1         3.5          1.4         0.2  setosa
## 2           4.9         3.0          1.4         0.2  setosa
## 3           4.7         3.2          1.3         0.2  setosa
## 4           4.6         3.1          1.5         0.2  setosa
## 5           5.0         3.6          1.4         0.2  setosa
## 6           5.4         3.9          1.7         0.4  setosa
## 7           4.6         3.4          1.4         0.3  setosa
## 8           5.0         3.4          1.5         0.2  setosa
## 9           4.4         2.9          1.4         0.2  setosa
## 10          4.9         3.1          1.5         0.1  setosa
## # ... with 140 more rows

tbl_iris[1,]

## # A tibble: 1 x 5
##   Sepal.Length Sepal.Width Petal.Length Petal.Width Species
##          <dbl>       <dbl>        <dbl>       <dbl>  <fctr>
## 1          5.1         3.5          1.4         0.2  setosa

tbl_iris[,"Species"]

## # A tibble: 150 x 1
##    Species
##     <fctr>
## 1   setosa
## 2   setosa
## 3   setosa
## 4   setosa
## 5   setosa
## 6   setosa
## 7   setosa
## 8   setosa
## 9   setosa
## 10  setosa
## # ... with 140 more rows