Question

我有一个像这样的文本文件：

AAAAAA this is some content.
This is AAAAAA some more content AAAAAA. AAAAAA
This is yet AAAAAA some more [AAAAAA] content.

我需要用递增的数字替换所有出现的AAAAAA，例如，输出如下所示：

x1 this is some content.
This is x2 some more content x3. x4
This is yet x5 some more [x6] content.

如何将所有匹配项替换为递增的数字和字符串前缀？

我的问题与 How to replace all matches with an incrementing number in BASH?但是我尝试修改的给定解决方案似乎无法正常工作。

awk '{for(x=1;x<=NF;x++)if($x~/AAAAAA/){sub(/AAAAAA/,"x"++i)}}1' file

谢谢。

Answer 1

另一个awk

$ awk -v w='AAAAAA' '{while($0~w) sub(w,"x"++c)}1' file

x1 this is some content.
This is x2 some more content x3. x4
This is yet x5 some more [x6] content.

本质上是效率低下的gsub。

此人使用记录分隔符作为搜索词

$ awk -v RS='AAAAAA' -v ORS='' 'NR>1 && $0="x"++c $0' file

x1 this is some content.
This is x2 some more content x3. x4
This is yet x5 some more [x6] content.

抑制最后一个比较困难，而是推迟替换一个并抑制第一个。

Answer 2

awk保存在此处：

awk '{for(i=1;i<=NF;i++){if($i~/A+/){val="x"++count;sub(/A+/,val,$i)}}} 1' Input_file

OR

awk '{for(i=1;i<=NF;i++){if($i~/A+/){sub(/A+/,"x"++count,$i)}}} 1' Input_file

输出如下。

x1 this is some content.
This is x2 some more content x3. x4
This is yet x5 some more [x6] content.

Answer 3

最短的perl技巧：

perl -pe 's/\bA+\b/x.++$i/ge' file

输出：

x1 this is some content.
This is x2 some more content x3. x4
This is yet x5 some more [x6] content.

如何用awk中的前缀字符串用递增数字替换所有匹配项

3 个答案: