Idiom #274 Remove all white space characters

Create the string t from the string s, removing all the spaces, newlines, tabulations, etc.

let t: String = s.chars().filter(|c| !c.is_whitespace()).collect();

using System;
using System.Linq;

var t = new string(s.Where(c => !Char.IsWhiteSpace(c)).ToArray());

[Ch || Ch <- "This is a\tstring with\nwhite\rspaces", Ch /= 8, Ch /= 9, Ch /= 10, Ch /= 13, Ch /= 32 ].

import "strings"
import "unicode"

t := strings.Map(func(r rune) rune {
	if unicode.IsSpace(r) {
		return -1
	}
	return r
}, s)

import Data.Char (isSpace)

t = filter (not . isSpace) s

let t = s.replace(/\s/g,'');

String t = s.replaceAll("\\s+", "");

import static java.lang.Character.isWhitespace;

String t = "";
for (char c : s.toCharArray())
    if (!isWhitespace(c)) t = t + c;

String t = "";
for (char c : s.toCharArray())
    switch (c) {
        case ' ', '\t', '\n', '\r' -> {}
        default -> t = t + c;
    }

import static java.lang.Character.isWhitespace;
import static java.lang.String.valueOf;
import static java.util.stream.Collectors.joining;

String t = s.chars()
    .filter(c -> !isWhitespace(c))
    .mapToObj(c -> valueOf((char) c))
    .collect(joining());

  SetLength(t, Length(s));
  i := 0;
  for Ch in S do
    if (Ch > #32) then
    begin
      Inc(i);
      t[i] := Ch;
    end;
  SetLength(t, i);

uses RegExpr;

t := ReplaceRegExpr('\s', s, '');

my $t = $s;
$t =~ s/\s*//g;

import re

t = re.sub('\\s', '', s)

t = ''.join(s.split())

from string import whitespace

f = lambda x: x not in whitespace
t = ''.join(filter(f, s))

t = s.gsub(/\s/, "")

t = s.gsub(/[[:space:]]/, "")

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Idiom created by programming-idioms.org

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