1

How could I replace whitespace of varying length with a single space in R?

R has plenty of methods for replacing characters, e.g. gsub, tidyr, or the stringr package. However, I cannot figure out how to best replace whitespace (with varying lengths) with a single \n space.

Here is a data.frame in R.

df = data.frame(column1= c("A B", "C  D", "E   F", "G    H", "I     J", "K      L", "M       N"))

Note that the number of spaces varies:

print(df)
#     column1
# 1       A B
# 2      C  D
# 3     E   F
# 4    G    H
# 5   I     J
# 6  K      L
# 7 M       N

My naïve approach would be to use gsub()

df$column1 <- gsub("[[:space:]]", " ", df$column1)

However, this doesn't change the varying length of spaces for replacement with a single space.

Wiktor Stribiżew
  • 607,720
  • 39
  • 448
  • 563
EB2127
  • 1,788
  • 3
  • 22
  • 43
  • Possible duplicate https://stackoverflow.com/questions/25707647/merge-multiple-spaces-to-single-space-remove-trailing-leading-spaces – Ronak Shah Sep 20 '21 at 04:09

2 Answers2

2

You could also use str_squish:

library(tidyverse)

df %>%
   mutate(column1 = str_squish(column1))

  column1
1     A B
2     C D
3     E F
4     G H
5     I J
6     K L
7     M N
Onyambu
  • 67,392
  • 3
  • 24
  • 53
1

We may need to use + i.e. one or more instead of just [[:space:]]

df$column1 <- gsub("[[:space:]]+", "\n", df$column1) 
df$column1
[1] "A\nB" "C\nD" "E\nF" "G\nH" "I\nJ" "K\nL" "M\nN"

If it is a single space, change the \n to " "

df$column1 <- gsub("[[:space:]]+", " ", df$column1) 
df$column1
[1] "A B" "C D" "E F" "G H" "I J" "K L" "M N"

The OP's code with [[:space:]] result in replacing the space with either space or \n

> gsub("[[:space:]]", "\n", df$column1)
[1] "A\nB"             "C\n\nD"           "E\n\n\nF"         "G\n\n\n\nH"       "I\n\n\n\n\nJ"     "K\n\n\n\n\n\nL"   "M\n\n\n\n\n\n\nN"
akrun
  • 874,273
  • 37
  • 540
  • 662