How to remove all whitespace from a string?

Question

So " xx yy 11 22 33 " will become "xxyy112233". How can I achieve this?

Richie Cotton · Accepted Answer · 2014-12-31 10:57:53Z

320

In general, we want a solution that is vectorised, so here's a better test example:

whitespace <- " \t\n\r\v\f" # space, tab, newline, 
                            # carriage return, vertical tab, form feed
x <- c(
  " x y ",           # spaces before, after and in between
  " \u2190 \u2192 ", # contains unicode chars
  paste0(            # varied whitespace     
    whitespace, 
    "x", 
    whitespace, 
    "y", 
    whitespace, 
    collapse = ""
  ),   
  NA                 # missing
)
## [1] " x y "                           
## [2] " ← → "                           
## [3] " \t\n\r\v\fx \t\n\r\v\fy \t\n\r\v\f"
## [4] NA

The base R approach: `gsub`

gsub replaces all instances of a string (fixed = TRUE) or regular expression (fixed = FALSE, the default) with another string. To remove all spaces, use:

gsub(" ", "", x, fixed = TRUE)
## [1] "xy"                            "←→"             
## [3] "\t\n\r\v\fx\t\n\r\v\fy\t\n\r\v\f" NA

As DWin noted, in this case fixed = TRUE isn't necessary but provides slightly better performance since matching a fixed string is faster than matching a regular expression.

If you want to remove all types of whitespace, use:

gsub("[[:space:]]", "", x) # note the double square brackets
## [1] "xy" "←→" "xy" NA 

gsub("\\s", "", x)         # same; note the double backslash

library(regex)
gsub(space(), "", x)       # same

"[:space:]" is an R-specific regular expression group matching all space characters. \s is a language-independent regular-expression that does the same thing.

The `stringr` approach: `str_replace_all` and `str_trim`

stringr provides more human-readable wrappers around the base R functions (though as of Dec 2014, the development version has a branch built on top of stringi, mentioned below). The equivalents of the above commands, using [str_replace_all][3], are:

library(stringr)
str_replace_all(x, fixed(" "), "")
str_replace_all(x, space(), "")

stringr also has a str_trim function which removes only leading and trailing whitespace.

str_trim(x) 
## [1] "x y"          "← →"          "x \t\n\r\v\fy" NA    
str_trim(x, "left")    
## [1] "x y "                   "← → "    
## [3] "x \t\n\r\v\fy \t\n\r\v\f" NA     
str_trim(x, "right")    
## [1] " x y"                   " ← →"    
## [3] " \t\n\r\v\fx \t\n\r\v\fy" NA

The `stringi` approach: `stri_replace_all_charclass` and `stri_trim`

stringi is built upon the platform-independent ICU library, and has an extensive set of string manipulation functions. The equivalents of the above are:

library(stringi)
stri_replace_all_fixed(x, " ", "")
stri_replace_all_charclass(x, "\\p{WHITE_SPACE}", "")

Here "\\p{WHITE_SPACE}" is an alternate syntax for the set of Unicode code points considered to be whitespace, equivalent to "[[:space:]]", "\\s" and space(). For more complex regular expression replacements, there is also stri_replace_all_regex.

stringi also has trim functions.

stri_trim(x)
stri_trim_both(x)    # same
stri_trim(x, "left")
stri_trim_left(x)    # same
stri_trim(x, "right")  
stri_trim_right(x)   # same

edited Dec 31, 2014 at 10:57

Richie Cotton

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answered May 13, 2011 at 12:55

Aniko

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8 Comments

IRTFM Over a year ago

@Aniko. Is there a reason you used fixed=TRUE?

Aniko Over a year ago

@DWin Supposedly it is faster if R knows that it does not have to invoke the regular expression stuff. In this case it does not really make any difference, I am just in the habit of doing so.

Sacha Epskamp Over a year ago

Is there a difference between "[[:space:]]" and "\\s"?

Sir Ksilem Over a year ago

if you check on flyordie.sin.khk.be/2011/05/04/day-35-replacing-characters or just type in ?regex then you see that [:space:] is used for "Space characters: tab, newline, vertical tab, form feed, carriage return, and space." That's a lot more than space alone

Richie Cotton Over a year ago

@Aniko Hope you don't mind about the big edit. Since this question is highly popular, it looked like the answer needed to be more thorough.

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JollyRoger · Accepted Answer · 2013-06-26 13:02:40Z

33

I just learned about the "stringr" package to remove white space from the beginning and end of a string with str_trim( , side="both") but it also has a replacement function so that:

a <- " xx yy 11 22 33 " 
str_replace_all(string=a, pattern=" ", repl="")

[1] "xxyy112233"

answered Jun 26, 2013 at 13:02

JollyRoger

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1 Comment

bartektartanus Over a year ago

stringr package doesn't work well with every encoding. stringi package is better solution, for more info check github.com/Rexamine/stringi

Bear · Accepted Answer · 2019-02-06 00:37:59Z

23

x = "xx yy 11 22 33"

gsub(" ", "", x)

> [1] "xxyy112233"

edited Feb 6, 2019 at 0:37

Bear

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answered Sep 26, 2017 at 8:49

ZWL

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Comments

Avinash Raj · Accepted Answer · 2015-10-01 10:16:39Z

22

Use [[:blank:]] to match any kind of horizontal white_space characters.

gsub("[[:blank:]]", "", " xx yy 11 22  33 ")
# [1] "xxyy112233"

answered Oct 1, 2015 at 10:16

Avinash Raj

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Comments

gagolews · Accepted Answer · 2014-12-28 22:47:31Z

10

Please note that soultions written above removes only space. If you want also to remove tab or new line use stri_replace_all_charclass from stringi package.

library(stringi)
stri_replace_all_charclass("   ala \t  ma \n kota  ", "\\p{WHITE_SPACE}", "")
## [1] "alamakota"

edited Dec 28, 2014 at 22:47

gagolews

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answered Jul 16, 2013 at 11:20

bartektartanus

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3 Comments

bartektartanus Over a year ago

stringi package is on CRAN now, enjoy! :)

Lucas Fortini Over a year ago

This command above is incorrect. The right way is stri_replace_all_charclass(" ala \t ma \n kota ", "\\p{WHITE_SPACE}", "")

Rich Scriven Over a year ago

After using stringi for a few months now and seen/learned how powerful and efficient it is, it has become my go-to package for string operations. You guys did an awesome job with it.

divibisan · Accepted Answer · 2018-08-07 20:22:28Z

10

The function str_squish() from package stringr of tidyverse does the magic!

library(dplyr)
library(stringr)

df <- data.frame(a = c("  aZe  aze s", "wxc  s     aze   "), 
                 b = c("  12    12 ", "34e e4  "), 
                 stringsAsFactors = FALSE)
df <- df %>%
  rowwise() %>%
  mutate_all(funs(str_squish(.))) %>%
  ungroup()
df

# A tibble: 2 x 2
  a         b     
  <chr>     <chr> 
1 aZe aze s 12 12 
2 wxc s aze 34e e4

edited Aug 7, 2018 at 20:22

divibisan

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answered Aug 7, 2018 at 13:43

damianooldoni

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3 Comments

R Balasubramanian Over a year ago

Please do not link to code. Add it in the text body of your answer and explain it here, to give your answer more longterm value.

damianooldoni Over a year ago

Thanks @RBalasubramanian for reminding me of this guideline. I will follow it in the future.

Nettle Over a year ago

I don't see how this answers the question. str_squish doesn't remove all spaces. It just trims and substitutes multiple spaces for one.

Mike V · Accepted Answer · 2020-08-09 20:35:16Z

6

Another approach can be taken into account

library(stringr)
str_replace_all(" xx yy 11 22  33 ", regex("\\s*"), "")

#[1] "xxyy112233"

\\s: Matches Space, tab, vertical tab, newline, form feed, carriage return

*: Matches at least 0 times

answered Aug 9, 2020 at 20:35

Mike V

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1 Comment

Sky Scraper Over a year ago

Should be noted that the regex() command also comes from the stringr package.

E_net4 · Accepted Answer · 2021-07-22 12:41:47Z

5

income<-c("$98,000.00 ", "$90,000.00 ", "$18,000.00 ", "")

To remove space after .00 use the trimws() function.

income<-trimws(income)

edited Jul 22, 2021 at 12:41

E_net4

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answered Jul 22, 2021 at 12:40

Onkar Kokane

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Comments

Data Jules · Accepted Answer · 2019-08-22 17:07:25Z

0

From stringr library you could try this:

Remove consecutive fill blanks

Remove fill blank

library(stringr)

            2.         1.
            |          |
            V          V

    str_replace_all(str_trim(" xx yy 11 22  33 "), " ", "")

edited Aug 22, 2019 at 17:07

answered Aug 29, 2018 at 18:29

Data Jules

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1 Comment

Sky Scraper Over a year ago

only removes spaces and maybe tabs, not new lines

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How to remove all whitespace from a string?

9 Answers 9

The base R approach: `gsub`

The `stringr` approach: `str_replace_all` and `str_trim`

The `stringi` approach: `stri_replace_all_charclass` and `stri_trim`

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Collectives™ on Stack Overflow

9 Answers 9

The base R approach: gsub

The stringr approach: str_replace_all and str_trim

The stringi approach: stri_replace_all_charclass and stri_trim

8 Comments

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The base R approach: `gsub`

The `stringr` approach: `str_replace_all` and `str_trim`

The `stringi` approach: `stri_replace_all_charclass` and `stri_trim`