我有一个如下所示的示例数据框:
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。我通常像data$A = as.factor(data$A)
那样做。但是当数据框非常大并且包含大量列时,这种方式将非常耗时。有谁知道更好的方法吗?
选择一些列来强制使用因子:
cols <- c("A", "C", "D", "H")
使用lapply()
强制替换所选列:
data[cols] <- lapply(data[cols], factor) ## as.factor() could also be used
检查结果:
sapply(data, class)
# A B C D E F G
# "factor" "integer" "factor" "factor" "integer" "integer" "integer"
# H I J
# "factor" "integer" "integer"
这是一个使用dplyr
的选项。来自%<>%
的magrittr
运算符用结果值更新lhs对象。
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data.table
,使用for
循环与set
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指定为(colsxswpoi)'cols'
:=
最近的setDT(data)[, (cols):= lapply(.SD, factor), .SDcols=cols]
方式是使用tidyverse
函数:
mutate_at
而且,为了完整性和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
,有this question asking about changing string columns only:
mutate_if
你可以使用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(.)))
(mutate_if
):
例如,在dplyr
强制integer
:
factor
使用功能:
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
如果您有另一个目标是从表中获取值然后使用它们进行转换,您可以尝试以下方式
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
这将选择特定于字符的列,然后将它们转换为因子。
这是一个### 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]
的例子。我在这个例子中使用了data.table
,因为我经常通过对其名称使用部分匹配来选择许多列。
grep