Suppose I have a data.frame like:
set.seed(123)
df <- data.frame(a=rnorm(10, 0,1), b=rnorm(10,1,2), c=rnorm(10, 2, 1),
x=rnorm(10, 1,2), y=rnorm(10,2,3), z=rnorm(10, 3, 4))
# a b c x y z
#1 -0.56047565 3.44816359 0.9321763 1.8529284 -0.08412094 4.013274
#2 -0.23017749 1.71962765 1.7820251 0.4098570 1.37624817 2.885813
#3 1.55870831 1.80154290 0.9739956 2.7902513 -1.79618905 2.828518
#4 0.07050839 1.22136543 1.2711088 2.7562670 8.50686790 8.474409
#5 0.12928774 -0.11168227 1.3749607 2.6431622 5.62388599 2.096916
#6 1.71506499 4.57382627 0.3133067 2.3772805 -1.36932575 9.065882
#7 0.46091621 1.99570096 2.8377870 2.1078353 0.79134549 -3.195011
#8 -1.26506123 -2.93323431 2.1533731 0.8761766 0.60003394 5.338455
#9 -0.68685285 2.40271180 0.8618631 0.3880747 4.33989536 3.495417
#10 -0.44566197 0.05441718 3.2538149 0.2390580 1.74989280 3.863766
How can I reorder the columns to get the following?
# a x b y c z
#1 -0.56047565 1.8529284 3.44816359 -0.08412094 0.9321763 4.013274
#2 -0.23017749 0.4098570 1.71962765 1.37624817 1.7820251 2.885813
#3 1.55870831 2.7902513 1.80154290 -1.79618905 0.9739956 2.828518
#4 0.07050839 2.7562670 1.22136543 8.50686790 1.2711088 8.474409
#5 0.12928774 2.6431622 -0.11168227 5.62388599 1.3749607 2.096916
#6 1.71506499 2.3772805 4.57382627 -1.36932575 0.3133067 9.065882
#7 0.46091621 2.1078353 1.99570096 0.79134549 2.8377870 -3.195011
#8 -1.26506123 0.8761766 -2.93323431 0.60003394 2.1533731 5.338455
#9 -0.68685285 0.3880747 2.40271180 4.33989536 0.8618631 3.495417
#10 -0.44566197 0.2390580 0.05441718 1.74989280 3.2538149 3.863766
CodePudding user response：
Using modulo (%%)
d2 = df[ , order((seq_along(df) - 1) %% (ncol(df) / 2))]
names(d2)
# [1] "a" "x" "b" "y" "c" "z"
To make it work with both even and odd number of columns, use ceiling in the divisor:
df_odd = df[-6]
d2 = df_odd[ , order((seq_along(df_odd) - 1) %% ceiling(ncol(df) / 2))]
names(d2)
# [1] "a" "x" "b" "y" "c"
Sidenote: Because OP mentioned that they have over 100 columns it may be relevant to considerdata.table::setcolorder, which reorders the columns without copying the data:
library(data.table)
setDT(df)
setcolorder(df, order((seq_along(df) - 1) %% ceiling(ncol(df) / 2)))
CodePudding user response：
We can generate the key shuffle index as follows. Note how it handles odd/even n at the same time without an if().
ShufInd <- function (n) {
m <- ceiling(n / 2)
sequence(rep(2, each = m), seq_len(m), m)[1:n]
}
ShufInd(6)
#[1] 1 4 2 5 3 6
ShufInd(5)
#[1] 1 4 2 5 3
To shuffle a vector (atomic or list) or a data frame of length n:
## OP's data frame
df[ShufInd(length(df))]
## drop the last column they try again
df <- df[-6]
df[ShufInd(length(df))]
## an atomic vector
x <- letters[1:5]
x[ShufInd(length(x))]
## a list
x <- as.list(x)
x[ShufInd(length(x))]
To shuffle columns of a matrix:
mat <- matrix(1:10, 2, 5)
mat[, ShufInd(ncol(mat))]
This supersedes my initial answer that treats odd and even n separately.
Henrik's answer is also a unified approach that can be written as:
Henrik <- function (n) order(seq(0, n - 1) %% ceiling(n / 2))
This is impressively concise!
CodePudding user response：
A possible solution in base R:
df <- df[c(t(matrix(names(df), ncol = 2)))]
df
#> a x b y c z
#> 1 -0.56047565 1.8529284 3.44816359 -0.08412094 0.9321763 4.013274
#> 2 -0.23017749 0.4098570 1.71962765 1.37624817 1.7820251 2.885813
#> 3 1.55870831 2.7902513 1.80154290 -1.79618905 0.9739956 2.828518
#> 4 0.07050839 2.7562670 1.22136543 8.50686790 1.2711088 8.474409
#> 5 0.12928774 2.6431622 -0.11168227 5.62388599 1.3749607 2.096916
#> 6 1.71506499 2.3772805 4.57382627 -1.36932575 0.3133067 9.065882
#> 7 0.46091621 2.1078353 1.99570096 0.79134549 2.8377870 -3.195011
#> 8 -1.26506123 0.8761766 -2.93323431 0.60003394 2.1533731 5.338455
#> 9 -0.68685285 0.3880747 2.40271180 4.33989536 0.8618631 3.495417
#> 10 -0.44566197 0.2390580 0.05441718 1.74989280 3.2538149 3.863766
Using dplyr and the same idea:
library(dplyr)
df %>%
relocate(as.vector(t(matrix(names(df), ncol = 2))))
#> a x b y c z
#> 1 -0.56047565 1.8529284 3.44816359 -0.08412094 0.9321763 4.013274
#> 2 -0.23017749 0.4098570 1.71962765 1.37624817 1.7820251 2.885813
#> 3 1.55870831 2.7902513 1.80154290 -1.79618905 0.9739956 2.828518
#> 4 0.07050839 2.7562670 1.22136543 8.50686790 1.2711088 8.474409
#> 5 0.12928774 2.6431622 -0.11168227 5.62388599 1.3749607 2.096916
#> 6 1.71506499 2.3772805 4.57382627 -1.36932575 0.3133067 9.065882
#> 7 0.46091621 2.1078353 1.99570096 0.79134549 2.8377870 -3.195011
#> 8 -1.26506123 0.8761766 -2.93323431 0.60003394 2.1533731 5.338455
#> 9 -0.68685285 0.3880747 2.40271180 4.33989536 0.8618631 3.495417
#> 10 -0.44566197 0.2390580 0.05441718 1.74989280 3.2538149 3.863766
CodePudding user response：
You can use dplyr for this purpose as given below,
> library(dplyr)
> df
player position points rebounds
1 a G 12 5
2 b F 15 7
3 c F 19 7
4 d G 22 12
5 e G 32 11
> df %>% select(rebounds, position, points, player)
rebounds position points player
1 5 G 12 a
2 7 F 15 b
3 7 F 19 c
4 12 G 22 d
5 11 G 32 e