用字符串替换最后一个逗号后的最后一个字符 [英] Replacing last characters after last comma with a string

问题描述

我有一个巨大的文本文件，如下所示：

I have a huge text file which look like this:

36,53,90478,0.58699759849,0.33616,4.83449759849,0.0695335954050315,3
36,53,90478,0.58699759849,0.33616,4.83449759849,0.0695335954050315,8
36,53,90478,0.58699759849,0.33616,4.83449759849,0.0695335954050315,14
36,53,15596,0.58454577855,0.26119,2.24878677855,0.116147072052964,12

所需的输出是这样的：

36,53,90478,0.58699759849,0.33616,4.83449759849,0.0695335954050315,MI-03
36,53,90478,0.58699759849,0.33616,4.83449759849,0.0695335954050315,MI-08
36,53,90478,0.58699759849,0.33616,4.83449759849,0.0695335954050315,MI-14
36,53,15596,0.58454577855,0.26119,2.24878677855,0.116147072052964,MI-12

我曾在这里和其他社区尝试过其他相关职位，但无法完全得到我想要的东西。

I have tried other relevant posts here and on other communities but could not exactly get what I want.

更新

这是交叉问题（为此，我既需要Unix / perl答案，又需要批处理/ powershell解决方案。 ）的答案。

This is the cross-question (I wanted both Unix/perl answers and batch/powershell solutions for this.) that has interesting answers.

推荐答案

这是PowerShell的答案，以防您喜欢PS。

Here's a PowerShell answer in case you like PS.

Get-Content C:\Path\To\File.csv | 
    Where{$_ -match '^(.*,)([^,]*)$'} | 
    ForEach { "{0}MI-{1}" -f $Matches[1], $Matches[2].Padleft(2,'0') } |
    Set-Content C:\Path\To\NewFile.csv

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