I have a sample data frame like below:
data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
I want to know how can I select multiple columns and convert them together to factors. I usually do it in the way like data$A = as.factor(data$A). But when the data frame is very large and contains lots of columns, this way will be very time consuming. Does anyone know of a better way to do it?
Choose some columns to coerce to factors:
cols <- c("A", "C", "D", "H")
Use lapply() to coerce and replace the chosen columns:
data[cols] <- lapply(data[cols], factor) ## as.factor() could also be used
Check the result:
sapply(data, class)
# A B C D E F G
# "factor" "integer" "factor" "factor" "integer" "integer" "integer"
# H I J
# "factor" "integer" "integer"
Here is an option using dplyr. The %<>% operator from magrittr update the lhs object with the resulting value.
library(magrittr)
library(dplyr)
cols <- c("A", "C", "D", "H")
data %<>%
mutate_each_(funs(factor(.)),cols)
str(data)
#'data.frame': 4 obs. of 10 variables:
# $ A: Factor w/ 4 levels "23","24","26",..: 1 2 3 4
# $ B: int 15 13 39 16
# $ C: Factor w/ 4 levels "3","5","18","37": 2 1 3 4
# $ D: Factor w/ 4 levels "2","6","28","38": 3 1 4 2
# $ E: int 14 4 22 20
# $ F: int 7 19 36 27
# $ G: int 35 40 21 10
# $ H: Factor w/ 4 levels "11","29","32",..: 1 4 3 2
# $ I: int 17 1 9 25
# $ J: int 12 30 8 33
Or if we are using data.table, either use a for loop with set
setDT(data)
for(j in cols){
set(data, i=NULL, j=j, value=factor(data[[j]]))
}
Or we can specify the 'cols' in .SDcols and assign (:=) the rhs to 'cols'
setDT(data)[, (cols):= lapply(.SD, factor), .SDcols=cols]
The more recent tidyverse way is to use the mutate_at function:
library(tidyverse)
library(magrittr)
set.seed(88)
data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
cols <- c("A", "C", "D", "H")
data %<>% mutate_at(cols, funs(factor(.)))
str(data)
$ A: Factor w/ 4 levels "5","17","18",..: 2 1 4 3
$ B: int 36 35 2 26
$ C: Factor w/ 4 levels "22","31","32",..: 1 2 4 3
$ D: Factor w/ 4 levels "1","9","16","39": 3 4 1 2
$ E: int 3 14 30 38
$ F: int 27 15 28 37
$ G: int 19 11 6 21
$ H: Factor w/ 4 levels "7","12","20",..: 1 3 4 2
$ I: int 23 24 13 8
$ J: int 10 25 4 33
You can use mutate_if (dplyr):
For example, coerce integer in factor:
mydata=structure(list(a = 1:10, b = 1:10, c = c("a", "a", "b", "b",
"c", "c", "c", "c", "c", "c")), row.names = c(NA, -10L), class = c("tbl_df",
"tbl", "data.frame"))
# A tibble: 10 x 3
a b c
<int> <int> <chr>
1 1 1 a
2 2 2 a
3 3 3 b
4 4 4 b
5 5 5 c
6 6 6 c
7 7 7 c
8 8 8 c
9 9 9 c
10 10 10 c
Use the function:
library(dplyr)
mydata%>%
mutate_if(is.integer,as.factor)
# A tibble: 10 x 3
a b c
<fct> <fct> <chr>
1 1 1 a
2 2 2 a
3 3 3 b
4 4 4 b
5 5 5 c
6 6 6 c
7 7 7 c
8 8 8 c
9 9 9 c
10 10 10 c
and, for completeness and with regards to this question asking about changing string columns only, there's mutate_if:
data <- cbind(stringVar = sample(c("foo","bar"),10,replace=TRUE),
data.frame(matrix(sample(1:40), 10, 10, dimnames = list(1:10, LETTERS[1:10]))),stringsAsFactors=FALSE)
factoredData = data %>% mutate_if(is.character,funs(factor(.)))
Here is a data.table example. I used grep in this example because that's how I often select many columns by using partial matches to their names.
library(data.table)
data <- data.table(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
factorCols <- grep(pattern = "A|C|D|H", x = names(data), value = TRUE)
data[, (factorCols) := lapply(.SD, as.factor), .SDcols = factorCols]
If you have another objective of getting in values from the table then using them to be converted, you can try the following way
### pre processing
ind <- bigm.train[,lapply(.SD,is.character)]
ind <- names(ind[,.SD[T]])
### Convert multiple columns to factor
bigm.train[,(ind):=lapply(.SD,factor),.SDcols=ind]
This selects columns which are specifically character based and then converts them to factor.
Here is another tidyverse approach using the modify_at() function from the purrr package.
library(purrr)
# Data frame with only integer columns
data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
# Modify specified columns to a factor class
data_with_factors <- data %>%
purrr::modify_at(c("A", "C", "E"), factor)
# Check the results:
str(data_with_factors)
# 'data.frame': 4 obs. of 10 variables:
# $ A: Factor w/ 4 levels "8","12","33",..: 1 3 4 2
# $ B: int 25 32 2 19
# $ C: Factor w/ 4 levels "5","15","35",..: 1 3 4 2
# $ D: int 11 7 27 6
# $ E: Factor w/ 4 levels "1","4","16","20": 2 3 1 4
# $ F: int 21 23 39 18
# $ G: int 31 14 38 26
# $ H: int 17 24 34 10
# $ I: int 13 28 30 29
# $ J: int 3 22 37 9
It appears that the use of SAPPLY on a data.frame to convert variables to factors at once does not work as it produces a matrix/ array. My approach is to use LAPPLY instead, as follows.
## let us create a data.frame here
class <- c("7", "6", "5", "3")
cash <- c(100, 200, 300, 150)
height <- c(170, 180, 150, 165)
people <- data.frame(class, cash, height)
class(people) ## This is a dataframe
## We now apply lapply to the data.frame as follows.
bb <- lapply(people, as.factor) %>% data.frame()
## The lapply part returns a list which we coerce back to a data.frame
class(bb) ## A data.frame
##Now let us check the classes of the variables
class(bb$class)
class(bb$height)
class(bb$cash) ## as expected, are all factors.
A simple and updated solution
data <- data %>%
mutate_at(cols, list(~factor(.)))
