在Unix中用sed反转四个字母的长度 [英] Reverse four length of letters with sed in unix

问题描述

如何用 sed 反转四个字母的长度?

How can I reverse a four length of letters with sed?

例如:

the year was 1815.

反向:

the raey was 5181.

这是我的尝试:

cat filename | sed's/\([a-z]*\) *\([a-z]*\)/\2, \1/'

但是它没有按我的预期工作.

But it does not work as I intended.

推荐答案

不确定在所有情况下都可以使用GNU sed来实现.如果 _ 不在四个字母单词前后立即出现，则可以使用

not sure it is possible to do it with GNU sed for all cases. If _ doesn't occur immediately before/after four letter words, you can use

sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'

\ b 是单词边界，单词定义是任何字母，数字或下划线字符.因此 \ b 将确保仅匹配整个单词而不是单词的一部分

\b is word boundary, word definition being any alphabet or digit or underscore character. So \b will ensure to match only whole words not part of words

$ echo 'the year was 1815.' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
the raey was 5181.
$ echo 'two time five three six good' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
two emit evif three six doog

$ # but won't work if there are underscores around the words
$ echo '_good food' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
_good doof

具有环视支持的工具适用于所有情况

tool with lookaround support would work for all cases

$ echo '_good food' | perl -pe 's/(?<![a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])(?!=[a-z0-9])/$4$3$2$1/gi'
_doog doof

(?<！[a-z0-9])和(?！= [a-z0-9])分别是负向后看和负向后看

(?<![a-z0-9]) and (?!=[a-z0-9]) are negative lookbehind and negative lookahead respectively

可以缩写为

perl -pe 's/(?<![a-z0-9])[a-z0-9]{4}(?!=[a-z0-9])/reverse $&/gie'

使用 e 修饰符将Perl代码放在替换部分中.这种形式适合于轻松更改要反转的单词的长度

which uses the e modifier to place Perl code in substitution section. This form is suitable to easily change length of words to be reversed

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