删除单个字母之间的空格 [英] Removing spaces between single letters

问题描述

我有一个字符串，它可能包含任意数量的由空格分隔的单字母.我正在寻找一个正则表达式(在 Perl 中)，它将删除所有(未知数量)单个字母之间的空格.

I have a string that may contain an arbitrary number of single-letters separated by spaces. I am looking for a regex (in Perl) that will remove spaces between all (unknown number) of single letters.

例如:

ab c d 应该变成 ab cd

a bcd e f gh 应该变成 a bcd ef gh

a b c 应该变成 abc

和

abc d 应该保持不变(因为没有单个字母后跟或前面有单个空格).

abc d should be unchanged (because there are no single letters followed by or preceded by a single space).

感谢您的任何想法.

推荐答案

您的描述与示例不符.在我看来，您想删除以下任何空格:(1) 前面不是一个字母，并且 (2) 后面是一个字母，而这个字母本身不是一个字母.这些条件可以精确地表示为嵌套环视:

Your description doesn't really match your examples. It looks to me like you want to remove any space that is (1) preceded by a letter which is not itself preceded by a letter, and (2) followed by a letter which is not itself followed by a letter. Those conditions can be expressed precisely as nested lookarounds:

/(?<=(?<!\pL)\pL) (?=\pL(?!\pL))/

已测试:

use strict;
use warnings;

use Test::Simple tests => 4;

sub clean {
  (my $x = shift) =~ s/(?<=(?<!\pL)\pL) (?=\pL(?!\pL))//g;
  $x;
}

ok(clean('ab c d')        eq 'ab cd');
ok(clean('a bcd e f gh')  eq 'a bcd ef gh');
ok(clean('a b c')         eq 'abc');
ok(clean('ab c d')        eq 'ab cd');

输出:

1..4
ok 1
ok 2
ok 3
ok 4

我假设您的意思是一个空格字符 (U+0020)；如果你想匹配任何空格，你可能想用 \s+ 替换空格.

I'm assuming you really meant one space character (U+0020); if you want to match any whitespace, you might want to replace the space with \s+.

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