用字符串替換最后一個逗號后的最后一個字符

Question

我有一個巨大的文本文件，看起來像這樣：

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

所需的輸出是這樣的：

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

我在這里和其他社區嘗試了其他相關帖子，但無法完全得到我想要的。

更新

這是交叉問題（我想要 Unix/perl 答案和批處理/powershell 解決方案。）有有趣的答案。

Answer 1

如果您喜歡 PS，這里有一個 PowerShell 答案。

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

Answer 2

下一個代碼執行您想要的操作，除了在小於10時將最后一個標記填充為零，希望它有所幫助。

編輯：我想出了一種在最后一個數字小於 10 時插入前導零的方法。有點難看，但確實如此。 :)

@echo off

setlocal EnableDelayedExpansion

for /F "delims=, tokens=1-8" %%A in (f.txt) do (
    set /a "t=%%H-10"
    if "!t:~0,1!" equ "-" (set "n=0%%H") else (set "n=%%H")
    echo(%%A,%%B,%%C,%%D,%%E,%%F,%%G,MI-!n!>>f.new.txt
)

move /Y f.new.txt f.txt >nul 2>&1

對於文件（在這種情況下為 f.txt）：

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

產生以下結果（也在 f.txt 中）：更新

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

Answer 3

這是一個cmd批處理文件，它依賴於一個不錯的 hack 來拆分逗號分隔列表的最后一項，而與字符串中出現的逗號數量無關。 基本技術如下所示； 請注意，這需要啟用延遲擴展：

set "x=This,is,the,original,list."
set "y=" & set "z=%x:,=" & set "y=!y!,!z!" & set "z=%" & set "y=!y:~1!"
echo ORIGINAL:  %x%
echo LAST ITEM: %z%
echo REMAINDER: %y%

所以這是腳本的代碼，將上述方法保存在名為:GET_LAST_ITEM的子例程中：

@echo off
setlocal EnableExtensions DisableDelayedExpansion

rem // Define constants here:
set "_FILE=%~1" & rem // (specify the CSV file by the first argument)

for /F "usebackq delims=" %%L in ("%_FILE%") do (
    call :GET_LAST_ITEM LAST REST "%%L"
    setlocal EnableDelayedExpansion
    set "LAST=0!LAST!"
    echo(!REST!,MI-!LAST:~-2!
    endlocal
)

endlocal
exit /B


:GET_LAST_ITEM  rtn_last  rtn_without_last  val_string
::This function splits off the last comma-separated item of a string.
::Note that exclamation marks must not occur within the given string.
::PARAMETERS:
::  rtn_last            variable to receive the last item
::  rtn_without_last    variable to receive the remaining string
::  val_string          original string
setlocal EnableDelayedExpansion
set "STR=,%~3"
set "PRE=" & set "END=%STR:,=" & set "PRE=!PRE!,!END!" & set "END=%"
endlocal & set "%~1=%END%" & set "%~2=%PRE:~2%"
exit /B

Answer 4

這就是答案@RomanPerekhrest我的交叉問題提供了（我也求UNIX / perl的解決方案）， 在這里：

帶有 sprintf 函數的 awk 方法（添加前導零）：

 awk -F, -v OFS=',' '$8="MI-"sprintf("%02d",$8);' file

輸出：

 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