Question

背景信息：
每个元素的IndicatorType为“＆＃39; 10＆＃39;或者＆＃39; ND＆＃39;
指标的值为“＆＃39;！＆＃39; ，＆＃39;＆＃39; （空字符串/ null），＆＃39; o＆＃39;，＆＃39; u＆＃39;

问题：
我需要过滤每个＆＃34;元素＆＃34;中的数据。基于此伪代码规则的组，然后将结果导出到文件：
For Each＆＃34; Element＆＃34;分组如果指标=＆＃39;！＆＃39;或者＆＃39;＆＃39;对于BOTH IndicatorTypes（即＆＃39; 10＆＃39;，＆＃39; ND＆＃39;），选择指示符类型= ND的组 ELSE如果指标=＆＃39;！＆＃39;或者＆＃39;＆＃39;对于IndicatorType = 10和指标=＆＃39; o＆＃39;或者＆＃39; u＆＃39; for IndicatorType = ND选择indicatorType = 10

的组

我的源文件：

"Code","IndicatorType","Indicator","Element","Data"
"111","10","","S","0.039"
"111","10","!","Cr ","0.045"
"111","10","","Zn","0.011"
"111","10","!","P","0.013"
"111","10","","Ni ","56.480"
"111","10","!","Co ","1.081"
"111","10","!","Fe","45.655"
"111","10","!","Si","0.364"
"111","10","!","Mn","0.005"
"111","10","!","Al","0.007"
"111","10","!","Cu","0.014"
"111","10","!","Y","0.00"
"111","ND","","S","0.037"
"111","ND","","Cr ","0.039"
"111","ND","","Zn","0.010"
"111","ND","","P","0.013"
"111","ND","o","Ni ","37.107"
"111","ND","o","Co ","0.887"
"111","ND","o","Fe","37.430"
"111","ND","","Si","0.348"
"111","ND","","Mn","0.005"
"111","ND","","Al","0.008"
"111","ND","","Cu","0.013"
"111","ND","","Y","0.00"

按元素分组我的代码：

$myfile = Get-ChildItem -Path $myfileSource *.csv
$myfileData = Import-Csv $myfile.FullName | Group-Object Element | 
              Where-Object -FilterScript {($_.Group.IndicatorType -eq 'ND' -and
                                          $_.Group.Indicator -eq '!' -or 
                                          $_.Group.Indicator -eq '') -or
                                          ($_.Group.IndicatorType -eq '10' -and
                                          $_.Group.Indicator -eq 'o' -or 
                                          $_.Group.Indicator -eq 'u' -or 
                                          $_.Group.Indicator -eq '!' -or
                                          $_.Group.Indicator -eq '')  
                                         }| Export-Csv -Path $myFile.FullName -Force -NoTypeInformation

PROBEM：
问题在于我的Where-Object，它返回所有值。
如何将伪代码规则应用于Where-Object以仅选择所需的组？

Answer 1

$_.Group将返回同一元素的两个对象，因此您的测试将使用-eq右侧的值过滤列表（例如，组中两个对象的IndicatorType），并且条件很可能是真的。例如：

$_.Group.IndicatorType -eq 'ND'
#Translates to array of "IndicatorTypes in the group" -eq 'ND'
10,'ND' -eq 'ND'
#The line above filters the array on the left to show matching values, this returns
'ND'
#A value is always true, so this test will be
$true

您可以通过首先在要执行测试的组中查找对象来解决此问题。

此外，使用Group-Object Element | Where-Object …意味着如果where子句为true，则将保留从Group-Object返回的整个对象（包含Count，Name和elements-group）。您应该使用Foreach-Object，因为您只想为每个已批准的元素组保留一个对象。尝试：

$myfile = Get-ChildItem -Path $myfileSource *.csv

$myfileData = Import-Csv $myfile.FullName | 
#For Each "Element" Group
Group-Object Element | Foreach-Object {
    $Ten = $_.Group | Where-Object { $_.IndicatorType -eq '10' }
    $ND = $_.Group | Where-Object { $_.IndicatorType -eq 'ND' }

    if(('!','' -contains $Ten.Indicator) -and ('!','' -contains $ND.Indicator)) {
        #if Indicator='!' or '' for BOTH IndicatorTypes(ie '10','ND'), Select group where indicatorType=ND
        $ND
    } elseif (('!','' -contains $Ten.Indicator) -and ('o','u' -contains $ND.Indicator)) {
        #ELSE if Indicator='!' or '' for IndicatorType = 10 and Indicator='o' or 'u' for IndicatorType=ND Select group where indicatorType=10
        $Ten
    }
} | Export-Csv -Path $myFile.FullName -Force -NoTypeInformation

Answer 2

Group-Object返回GroupInfo个对象的集合，每个对象都包含组中元素的集合 - 基本上是列表列表。运行cmdlet会生成（基于示例输入）12个组的列表：

PS[1] (191) > $data = import-csv  data2.txt | group-object element
PS[1] (192) > $data.Count
12

查看返回的第一项：

PS[1] > $data[0].Group.Count
2

您可以看到它包含两个看起来像这样的条目：

PS[1] (193) > $data[0].Group

Code          : 111
IndicatorType : 10
Indicator     :
Element       : S
Data          : 0.039

Code          : 111
IndicatorType : ND
Indicator     :
Element       : S
Data          : 0.037

这意味着您的选择条件必须应用于与该组关联的数据集合中的每个/所有条目，以决定是否要返回该组。在您的示例代码中，您似乎假设某个群体是平的而不是。但遗憾的是，我并不完全理解您要尝试的内容，因此我无法提供完整的答案。

在组内过滤 - 组 - 对象

2 个答案: