我有一个数据框架,我想在列名前面放以下文字:"high_"one_answers";low_"。应该将X2-X4中的列的名称重命名为eg。high_X2和X5-X7 eg。low_X6 .
请参见下面的示例。
X1 X2 X3 X4 X5 X6 X7
a 1 0 1 1 1 1 0
b 2 2 1 1 1 1 0
结果
X1 high_X2 high_X3 high_X4 low_X5 low_X6 low_X7
a 1 0 1 1 1 1 0
b 2 2 1 1 1 1 0
您可以使用rep
和paste
-
names(df)[-1] <- paste(rep(c('high', 'low'), each = 3), names(df)[-1], sep = '_')
df
# X1 high_X2 high_X3 high_X4 low_X5 low_X6 low_X7
#a 1 0 1 1 1 1 0
#b 2 2 1 1 1 1 0
如果你想依赖列的范围,那么dplyr
代码将更容易。
library(dplyr)
df %>%
rename_with(~paste('high', ., sep = '_'), X2:X4) %>%
rename_with(~paste('low', ., sep = '_'), X5:X7)
基本解决方案(对于这类事情来说更直接)
df <- data.frame(X1=c(a=1L,b=2L),
X2=c(a=0L,b=2L),
X3=c(a=1L,b=1L),
X4=c(a=1L,b=1L),
X5=c(a=1L,b=1L),
X6=c(a=1L,b=1L),
X7=c(a=1L,b=1L))
cn <- colnames(df)
cond <- as.integer(substr(cn,2L,nchar(cn))) %% 2L == 0L
colnames(df)[cond] <- paste0(cn[cond],"_is_pair")
一个整洁的解决方案(由于整洁而更尴尬)
library(dplyr)
library(stringr)
library(tidyselect)
df <- data.frame(X1=c(a=1L,b=2L),
X2=c(a=0L,b=2L),
X3=c(a=1L,b=1L),
X4=c(a=1L,b=1L),
X5=c(a=1L,b=1L),
X6=c(a=1L,b=1L),
X7=c(a=1L,b=1L))
is_pair <- function(vars = peek_vars(fn = "is_pair")) {
vars[as.integer(str_sub(vars,2L,nchar(vars))) %% 2L == 0L]
}
df %>% rename_with(~paste0(.x,"_is_pair"),
is_pair())