SED 替换了一些首次出现(和范围)的模式

Question

是否可以使用 SED 更改此场景中出现的前 4 个(或更多)字符串(与 sed -r 's/[^[:space:]]*/TEST/4g' 相反)):

TEST TEST TEST TEST five six seven

我使用 AWK 两次反转单词顺序，但这很长，很复杂，我想只用 SED 来实现:

echo one two three four five six seven | awk '{for(i=NF;i>=1;i--) printf "%s ", $i;print ""}'  | sed -r 's/[^ ]*/TEST/4g' |  awk '{for(i=NF;i>=1;i--) printf "%s ", $i;print ""}'

也许还可以选择更改出现范围，例如 3-5、6-12，...？

输入示例为:

one two three four five six seven

eight nine ten eleven twelve thirteen fourteen

fifteen sixteen seventeen eighteen nineteen twenty twenty-one

Answer 1

单个AWK怎么样:

awk '{for(i=1;i<=NF;i++) if(i<5){$i="TEST"}; print}'

测试运行:

$ echo one two three four five six seven | awk '{for(i=1;i<=NF;i++) if(i<5){$i="TEST"}; print}'
TEST TEST TEST TEST five six seven

这个解决方案简短、可读且可维护。如果它不能满足您的要求，请添加一些有关您的具体问题的详细信息。

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Perl等效解决方案:

perl -pe 's/\S+/$i++<4?"TEST":$&/ge'

测试运行:

$ echo one two three four five six seven | perl -pe 's/\S+/$i++<4?"TEST":$&/ge'
TEST TEST TEST TEST five six seven

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maybe there is option to change ranges of occurence like 3-5, 6-12

AWK:

awk '{for(i=3;i<6;i++)$i="TEST";print}'

在新提供的输入文件上测试运行:

$ awk '{for(i=3;i<6;i++)$i="TEST";print}' input
one two TEST TEST TEST six seven
eight nine TEST TEST TEST thirteen fourteen
fifteen sixteen TEST TEST TEST twenty twenty-one

Perl:

perl -pe 's/\S+/++$c~~[3..5]?"TEST":$&/ge'

在新提供的输入文件上测试运行:

$ perl -pe '$c=0;s/\S+/++$c~~[3..5]?"TEST":$&/ge' input
Smartmatch is experimental at -e line 1. <== This is a warning that goes to STDERR
one two TEST TEST TEST six seven
eight nine TEST TEST TEST thirteen fourteen
fifteen sixteen TEST TEST TEST twenty twenty-one