How to remove all whitespace from a string?

r, r-faq, regex, string

Solution

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

Problem

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

Original source