如果一个单词也位于下一个单词(sed)中，该如何从一个单词中删除呢? [英] How to remove characters from a word if they are also in the next word (sed)?

问题描述

我正在尝试找到一种方法来删除第一个单词中的所有字符，前提是该字符位于第二个单词中.输入看起来像这样:

I'm trying to find a way to delete all characters in the first word IF that character is in the second word. The input looks like this:

计算机费用

结果应为:"mpuer"，因为删除了c，o和t.像这样的多行之间用回车符分隔，两个单词之间用空格分隔.

And the result should be: "mpuer" because the c, o and t were deleted. There are multiple lines like this separated by a return, the 2 words are separated by a space.

我一直在寻找解决方案的时间，但我真的被困住了.感谢所有帮助.

I've been searching quite some time for the solution, but I'm really stuck. All help is appreciated.

推荐答案

这可能对您有用:

echo "computer cost" |
sed ':a;s/\(.\)\(.* .*\1.*\)/\2/;ta;s/ .*//'
mpuer

说明:

• 为将来的分支命令:a;
添加标签
• 删除第一个单词中与第二个单词中相同字符匹配的字符 s/\(.\)\(.*.* \ 1.* \)/\ 2/
• 如果发生替换，则分支到标签 ta
• 如果没有其他替换，请删除第二个单词. s/.*//

替换正则表达式可以进一步解释:

The substitution regexp can be further explained:

• \(.\)与单词word中的任何字符匹配(后来称为 \ 1 )
• \(.*.* \ 1.* \)匹配单词.* 的其余部分中的所有字符，后跟空格 ，然后是第二个.* 中的所有字符，然后是来自第一个 \ 1 的匹配字符，然后是来自第二个的其余字符.*此分组以后称为 \ 2 .
• 如果以上匹配项替换为 \ 2 ，从而有效删除了匹配字符 \ 1

• \(.\) matches any character in word one (later refered to as \1)
• \(.* .*\1.*\) matches any characters in the remainder of a word one .* followed by a space followed by some on none characters in word two .* followed by a matching character from word one \1 followed by the remaining characters from word two .* this grouping will later be known as \2.
• If the above matches replace it by \2 thus effectively deleting the matching character \1

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