Question

我有以下脚本来获取文件中的逗号计数。

ls | select -First 10 |
% { 
    $_.FullName;
    cat $_ | Select-String `, -AllMatches | 
    Select-Object LineNumber, @{ n = "Count"; e = { $_.Matches.Count }} |
    Group-Object Count
}

该脚本为每个文件返回以下行。如何获取每个文件最常用的逗号计数？至于以下示例，我将需要传递最多逗号计数 77 ，行数 108 和总行数（ 108 + 8 +5 = 121 < / em>）到另一个程序。

File1.txt Count Name Group ----- ---- ----- 108 77 {@{LineNumber=1; Count=77}, @{LineNumber=2; Count=77}, @{LineNumber=3; Count=77... 8 78 {@{LineNumber=7; Count=78}, @{LineNumber=15; Count=78}, @{LineNumber=22; Count=... 5 79 {@{LineNumber=16; Count=79}, @{LineNumber=32; Count=79}, @{LineNumber=37; Count... File2.txt .....

Answer 1

如果我理解你在问什么，你应该按照计数（降序）对输出进行排序并选择第一个：

ls | select -First 10 |
% { 
    $_.FullName;
    cat $_ | Select-String `, -AllMatches | 
    Select-Object LineNumber, @{ n = "Count"; e = { $_.Matches.Count }} |
    Group-Object Count | Sort Count -Descending | Select-Object -First 1
}

使用逗号测量行数：

ls | select -First 10 |
% { 
    $_.FullName;
    cat $_ | Select-String `, -AllMatches | 
    Select-Object LineNumber, @{ n = "Count"; e = { $_.Matches.Count }} |
    Measure-Object Count
}

Answer 2

您可以执行以下操作，将相关的数据包装到每个文件的自定义对象中：

ls | select -First 10 |
% { 
    $numLines = 0
    $grouped = cat $_ `
               |%{ $numLines++; $_ }  `
               | Select-String ',' -AllMatches `
               | Select-Object LineNumber, @{ n = "Count"; e = { $_.Matches.Count }} `
               | Group-Object Count `
               | sort Count -Desc

    [PsCustomObject] @{
         File = $_.FullName
         TopCommaCount = [int] $grouped[0].Name
         TopCommaCountLines = [int] $grouped[0].Count
         TotalLines = $numLines
     }
}

获取最常见的逗号计数，行数和总行数

2 个答案: