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How to remove all whitespace from a string?

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

" xx yy 11 22 33 "

"xxyy112233"

7 Answers
7

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

whitespace <- " tnrvf" # 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] " tnrvfx tnrvfy tnrvf"
## [4] NA


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

fixed = TRUE

fixed = FALSE

gsub(" ", "", x, fixed = TRUE)
## [1] "xy" "←→"
## [3] "tnrvfxtnrvfytnrvf" 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.

fixed = TRUE

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.

"[:space:]"

s

stringr

str_replace_all

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:

stringr

stringi

str_replace_all][3]

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.

stringr

str_trim

str_trim(x)
## [1] "x y" "← →" "x tnrvfy" NA
str_trim(x, "left")
## [1] "x y " "← → "
## [3] "x tnrvfy tnrvf" NA
str_trim(x, "right")
## [1] " x y" " ← →"
## [3] " tnrvfx tnrvfy" NA


stringi

stri_replace_all_charclass

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:

stringi

library(stringi)
stri_replace_all_fixed(x, " ", "")
stri_replace_all_charclass(x, "\pWHITE_SPACE", "")


Here "\pWHITE_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.

"\pWHITE_SPACE"

"[[:space:]]"

"\s"

space()

stri_replace_all_regex

stringi also has trim functions.

stringi

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


"[[:space:]]"

"\s"

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"


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.

stri_replace_all_charclass

stringi

library(stringi)
stri_replace_all_charclass(" ala t ma n kota ", "\pWHITE_SPACE", "")
## [1] "alamakota"


stringi

stringi

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

[[:blank:]]

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


x = "xx yy 11 22 33"

gsub(" ", "", x)

[1] "xxyy112233"

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

str_squish()

stringr

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


str_squish

Try this:

remove fill blank.

2. 1.
| |
V V


display subinstr(stritrim(" xx yy 11 22 33 "), " ", "",.)

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