当列数可以更改时,如何移动数据帧的最后x列

这里最简单的解决方案是使用dplyr包:

install.packages("dplyr")
library(dplyr)
n <- length(mtcars)
mtcars2 <- select(mtcars, 1:3, (n-2):n, everything())

通过查看名称,可以看到已将最后3列移动到前三列的正后方:

names(mtcars)
# [1] "mpg"  "cyl"  "disp" "hp"   "drat" "wt"   "qsec" "vs"   "am"   "gear" "carb"
names(mtcars2)
# [1] "mpg"  "cyl"  "disp" "am"   "gear" "carb" "hp"   "drat" "wt"   "qsec" "vs"  

您也可以在base R中执行此操作,但它需要一个额外的计算(现在我想起来了,这很简单)。

mtcars3 <- data.frame(mtcars[1:3], mtcars[(n-2):n], mtcars[4:(n-3)])
names(mtcars3)
# [1] "mpg"  "cyl"  "disp" "am"   "gear" "carb" "hp"   "drat" "wt"   "qsec" "vs"  

df <- as.data.frame(matrix(0, ncol = 21, nrow = 5))


df[,c(1:10, (length(df)-9):length(df), 11:(length(df)-10))]

收益率

  V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21 V11
1  0  0  0  0  0  0  0  0  0   0   0   0   0   0   0   0   0   0   0   0   0
2  0  0  0  0  0  0  0  0  0   0   0   0   0   0   0   0   0   0   0   0   0
3  0  0  0  0  0  0  0  0  0   0   0   0   0   0   0   0   0   0   0   0   0
4  0  0  0  0  0  0  0  0  0   0   0   0   0   0   0   0   0   0   0   0   0
5  0  0  0  0  0  0  0  0  0   0   0   0   0   0   0   0   0   0   0   0   0

这适用于base R

#
 test_frame <- as.data.frame(matrix(runif(100 * 100, 0, 100), 
                                     nrow = 100,
                                     ncol = 100))
#
# define the columns on left and right to bring together
#  
 left_same_cols <- 10
 right_same_cols <- 10
#  
# populate data so we can visually verify result of move
#  
 test_frame[, 1:10] <- 1
 test_frame[, 90:100] <- 2
#
# define new index sequence
#
 cols <- c(1:left_same_cols, 
            (ncol(test_frame) - right_same_cols):ncol(test_frame), 
            (left_same_cols + right_same_cols):(ncol(test_frame) - 
 right_same_cols))
#
# create new data.frame with desired column order
#  
 test_frame2 <- test_frame[, cols] 
#