How to remove all whitespace from a string? - r

Question

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

Answer 1

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

Answer 2

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"

Answer 3

x = "xx yy 11 22 33"
gsub(" ", "", x)
> [1] "xxyy112233"

Answer 4

Use [[:blank:]] to match any kind of horizontal white_space characters.
gsub("[[:blank:]]", "", " xx yy 11 22 33 ")
# [1] "xxyy112233"

Answer 5

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"

Answer 6

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

Answer 7

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

Answer 8

income<-c("$98,000.00 ", "$90,000.00 ", "$18,000.00 ", "")
To remove space after .00 use the trimws() function.
income<-trimws(income)

Answer 9

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 "), " ", "")