sed，用该行的第一部分全局替换一个分隔符 [英] sed, replace globally a delimiter with the first part of the line

问题描述

假设我有以下几行:

<前>1:a:b:c2:d:e:f3:a:b4:a:b:c:d:e:f

我如何使用 sed(或 perl)编辑它以便阅读:

<前>1a1b1c2d2e2f3a3b4a4b4c4d4e4f

我用过这样的 awk:

<前>awk -F':' '{gsub(/:/, $1, $0);打印 $0}'

但需要很长时间才能完成！所以寻找更快的东西.

解决方案

'这有点棘手，但可以用 sed 完成(假设文件 data 包含样本输入):

$ sed '/^\(.\):/{s//\1/: 重试s/^\(.\)\([^:]*\):/\1\2\1/重试}' 数据1a1b1c2d2e2f3a3b4a4b4c4d4e4f$

您可以将脚本用分号拼成一行；MacOS X 上的 sed 有时有点胡思乱想，并且反对某些部分，所以它分为 6 行.第一行匹配以单个字符和冒号开头的行，并在识别时启动一系列操作.例如，第一个替换将1:"替换为1".: retry 也是分支的标签——这是其中的关键部分.下一个替换复制第一个冒号上的行中的第一个字符.t retry 如果替代品改变了任何东西，则返回标签.最后一行分隔了最初匹配行的整个操作序列.

Lets say I have the following lines:

1:a:b:c
2:d:e:f
3:a:b
4:a:b:c:d:e:f

how can I edit this with sed (or perl) in order to read:

1a1b1c
2d2e2f
3a3b
4a4b4c4d4e4f

I have done with awk like this:

awk -F':' '{gsub(/:/, $1, $0); print $0}'

but takes ages to complete! So looking for something faster.

解决方案

'Tis a tad tricky, but it can be done with sed (assuming the file data contains the sample input):

$ sed '/^\(.\):/{
s//\1/
: retry
s/^\(.\)\([^:]*\):/\1\2\1/
t retry
}' data
1a1b1c
2d2e2f
3a3b
4a4b4c4d4e4f
$

You may be able to flatten the script to one line with semi-colons; sed on MacOS X is a bit cranky at times and objected to some parts, so it is split out into 6 lines. The first line matches lines starting with a single character and a colon and starts a sequence of operations for when that is recognized. The first substitute replaces, for example, '1:' by just '1'. The : retry is a label for branching too - a key part of this. The next substitution copies the first character on the line over the first colon. The t retry goes back to the label if the substitute changed anything. The last line delimits the entire sequence of operations for the initially matched line.

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