R+dplyr: conditionally swap the elements of two columns - r

Question

Consider the dataframe df at the end of the post.
I simply would like to swap the elements of columns x and y whenever x>y.
There may be other columns in the dataframe which I do not want to touch.
In a sense, I would like to sort row wise the columns x and y.
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
df<-tibble(x=1:10, y=10:1, extra=LETTERS[1:10])
df
#> # A tibble: 10 × 3
#> # Rowwise:
#> x y extra
#> <int> <int> <chr>
#> 1 1 10 A
#> 2 2 9 B
#> 3 3 8 C
#> 4 4 7 D
#> 5 5 6 E
#> 6 6 5 F
#> 7 7 4 G
#> 8 8 3 H
#> 9 9 2 I
#> 10 10 1 J
Created on 2021-10-06 by the reprex package (v2.0.1)

Answer 1

base solution:
use which(df$x > df$y) to determine row numbers you want to change, then use rev to swap values for these:
df[which(df$x > df$y), c("x", "y")] <- rev(df[which(df$x > df$y), c("x", "y")])
df
# x y extra
# <int> <int> <chr>
# 1 1 10 A
# 2 2 9 B
# 3 3 8 C
# 4 4 7 D
# 5 5 6 E
# 6 5 6 F
# 7 4 7 G
# 8 3 8 H
# 9 2 9 I
# 10 1 10 J

Answer 2

Thanks everyone!
I wrote a small function which does what I need and generalizes to the case of multiple variables.
See the reprex
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
set.seed(1234)
set_colnames <- `colnames<-`
df<-tibble(x=1:10, y=10:1, z=rnorm(10), extra=LETTERS[1:10]) %>%
rowwise()
df
#> # A tibble: 10 × 4
#> # Rowwise:
#> x y z extra
#> <int> <int> <dbl> <chr>
#> 1 1 10 -1.21 A
#> 2 2 9 0.277 B
#> 3 3 8 1.08 C
#> 4 4 7 -2.35 D
#> 5 5 6 0.429 E
#> 6 6 5 0.506 F
#> 7 7 4 -0.575 G
#> 8 8 3 -0.547 H
#> 9 9 2 -0.564 I
#> 10 10 1 -0.890 J
sort_rows <- function(df, col_names, dec=F){
temp <- df %>%
select(all_of(col_names))
extra_names <- setdiff(colnames(df), col_names)
temp2 <- df %>%
select(all_of(extra_names))
res <- t(apply(temp, 1, sort, decreasing=dec)) %>%
as_tibble %>%
set_colnames(col_names) %>%
bind_cols(temp2)
return(res)
}
col_names <- c("x", "y", "z")
df_s <- df %>%
sort_rows(col_names, dec=F)
#> Warning: The `x` argument of `as_tibble.matrix()` must have unique column names if `.name_repair` is omitted as of tibble 2.0.0.
#> Using compatibility `.name_repair`.
df_s
#> # A tibble: 10 × 4
#> x y z extra
#> <dbl> <dbl> <dbl> <chr>
#> 1 -1.21 1 10 A
#> 2 0.277 2 9 B
#> 3 1.08 3 8 C
#> 4 -2.35 4 7 D
#> 5 0.429 5 6 E
#> 6 0.506 5 6 F
#> 7 -0.575 4 7 G
#> 8 -0.547 3 8 H
#> 9 -0.564 2 9 I
#> 10 -0.890 1 10 J
Created on 2021-10-06 by the reprex package (v2.0.1)

Answer 3

This looks like sorting for me:
library(tidyverse)
df <- tibble(x=1:10, y=10:1, extra=LETTERS[1:10])
df
#> # A tibble: 10 x 3
#> x y extra
#> <int> <int> <chr>
#> 1 1 10 A
#> 2 2 9 B
#> 3 3 8 C
#> 4 4 7 D
#> 5 5 6 E
#> 6 6 5 F
#> 7 7 4 G
#> 8 8 3 H
#> 9 9 2 I
#> 10 10 1 J
extra_cols <- df %>% colnames() %>% setdiff(c("x", "y"))
extra_cols
#> [1] "extra"
df %>%
mutate(row = row_number()) %>%
pivot_longer(-c(row, extra_cols)) %>%
group_by_at(c("row", extra_cols)) %>%
transmute(
value = value %>% sort(),
name = c("x", "y"),
) %>%
pivot_wider() %>%
ungroup() %>%
select(-row)
#> Note: Using an external vector in selections is ambiguous.
#> ℹ Use `all_of(extra_cols)` instead of `extra_cols` to silence this message.
#> ℹ See <https://tidyselect.r-lib.org/reference/faq-external-vector.html>.
#> This message is displayed once per session.
#> # A tibble: 10 x 3
#> extra x y
#> <chr> <int> <int>
#> 1 A 1 10
#> 2 B 2 9
#> 3 C 3 8
#> 4 D 4 7
#> 5 E 5 6
#> 6 F 5 6
#> 7 G 4 7
#> 8 H 3 8
#> 9 I 2 9
#> 10 J 1 10
Created on 2021-10-06 by the reprex package (v2.0.1)

Answer 4

Try using apply on axis 1 and transpose it with t, then use as_tibble to convert it to a tibble.
Then finally change the column names:
> df <- as_tibble(t(apply(df, 1, sort)))
> names(df) <- c('x', 'y')
> df
# A tibble: 10 x 2
x y
<int> <int>
1 1 10
2 2 9
3 3 8
4 4 7
5 5 6
6 5 6
7 4 7
8 3 8
9 2 9
10 1 10