关于r：一次将多列强制转换为因子

Coerce multiple columns to factors at once

我有一个示例数据框，如下所示：

1	data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))

我想知道如何选择多个列并将它们一起转换为因子。我通常以data$A = as.factor(data$A)的方式进行操作。但是，当数据帧很大并且包含许多列时，这种方式将非常耗时。有谁知道更好的方法吗？

相关讨论

这是使用dplyr的选项。 magrittr中的%<>%运算符使用结果值更新lhs对象。

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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

或者，如果我们使用的是data.table，请对set使用for循环

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setDT(data)
for(j in cols){
set(data, i=NULL, j=j, value=factor(data[[j]]))
}

或者我们可以在.SDcols中指定'cols'并将rhs分配(:=)到'cols'

1	setDT(data)[, (cols):= lapply(.SD, factor), .SDcols=cols]

最新的tidyverse方法是使用mutate_at函数：

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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

相关讨论

您可以使用mutate_if(dplyr)：

例如，在factor中强制integer：

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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

使用功能：

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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

并且，为了完整起见，关于这个仅询问更改字符串列的问题，有一个mutate_if：

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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(.)))

这是一个data.table示例。在本示例中，我使用了grep，因为那是我经常通过对名称使用部分匹配来选择许多列。

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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]

似乎在data.frame上使用SAPPLY将变量立即转换为因子不起作用，因为它会产生矩阵/数组。我的方法是改为使用LAPPLY，如下所示。

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## 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.

这是使用purrr包中的modify_at()函数的另一种方法。

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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

如果您还有另一个从表中获取值然后使用它们进行转换的目的，则可以尝试以下方法

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### 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]

这将选择专门基于字符的列，然后将其转换为因数。