Check if an strings of list exists in other list

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DenisPasternak-3587 asked Nov 19, '20 AndreasBaumgarten commented Nov 20, '20

Check if an strings of list exists in other list

Hi,

I want to find object it array using where-object. But I don't want to use long filter for this.

For Example:

 $list = '10','20','30','40'
 ($list -contains '50') -or ($list -contains '20') -or ($list -contains '60')
 True

But it will be false in this cases:

 $list = '10','20','30','40'
 $filterList = '50','20','60'
 ($list -contains $filterList)
 False

 $list = '10','20','30','40'
 $filterList = '50','20','60'
 ($list -in $filterList)
 False

How can I reduce the number of comparison operators like "-or" or "-and" in this condition?

windows-server-powershell

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Answer 1 · 2020-11-20T03:43:24.75Z

RichMatheisen-8856 answered Nov 20, '20 RichMatheisen-8856 edited Nov 20, '20

This isn't shorter in the sense that there's less code, but it might run faster and it doesn't require writing out each condition.

BTW, you can make the code you posted run a bit faster by moving the test for '777' to the front of the list E.g., if ($list -contains '777' -and (($list -contains '50') -or . . .))

 $ReferenceList = '50', '20', '60', '1', '2', '3', '4', 'any'
 $ListToBeChecked = '0','3','6','9','12','15','18','21','24','27','30','33','36','39'
 if ($ListToBeChecked -contains '777'){
     $Pass = $False
 }
 Else{
     $Pass = $false
     ForEach($v in $ReferenceList){
         if ($ListToBeChecked -contains $v){
             $Pass = $true
             break
         }
     }
 }
 $Pass

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Answer 2 · 2020-11-19T21:45:53.827Z

BillStewart answered Nov 19, '20 DenisPasternak-3587 commented Nov 19, '20

What are you expecting $list -contains $filterList to do?

Are you asking if $list contains all members in $filterList, or are you asking if $list contains any of the members in $filterList?

· 1

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DenisPasternak-3587 · Nov 19, 2020 at 09:54 PM

Sorry:
if $list contains any of the members in $filterList

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Answer 3 · 2020-11-19T21:56:00.157Z

DenisPasternak-3587 answered Nov 19, '20 DenisPasternak-3587 edited Nov 19, '20

Sorry, the question:
I want to get True if $list containe any string in $filterList

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Answer 4 · 2020-11-19T21:58:31.207Z

AndreasBaumgarten answered Nov 19, '20 AndreasBaumgarten edited Nov 19, '20

Maybe this helps:

 $list = '10','20','30','40'
 $filterList = '50','20','60'
 Compare-Object -Referenceobject $list -DifferenceObject $filterlist -IncludeEqual

SideIndicators:
== Value is in both objects
=> Value is in the DifferenceObject but not in the ReferenceObject
<= Value is in the ReferenceObject but not in the DifferenceObject

It's possible to filter the result "value is in both lists":

 $list = '10','20','30','40'
 $filterList = '50','20','60'
 $compareResult = Compare-Object -Referenceobject $list -DifferenceObject $filterlist -IncludeEqual | Where-Object{$_.sideIndicator -eq "=="}
 if ($compareResult) {$compareResult = $true}
 else {$compareResult = $false}
 $compareResult

(If the reply was helpful please don't forget to upvote and/or accept as answer, thank you)

Regards
Andreas Baumgarten

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DenisPasternak-3587 · Nov 19, 2020 at 10:08 PM

Sorry, but I don't need to compare. I need to find an object in array.
The expression will be more complex, this is just an example. All I want is to shorten the line.

Thank you.

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Answer 5 · 2020-11-19T22:20:26.787Z

RichMatheisen-8856 answered Nov 19, '20 RichMatheisen-8856 commented Nov 20, '20

I agree with AndreasBaumgarten's answer, but I'll build a little bit on it:

 $RefList = '10','20','30','40'
 $DiffFilterList = '50','20','60'
 if (([array]$x = Compare-Object -Referenceobject $Reflist -DifferenceObject $DiffFilterlist  -excludedifferent -IncludeEqual).count -gt 0){
     "Items in `$DiffFilterList that are in `$RefList"
     $x.InputObject
 }
 else{
     "Nothing in `$DiffFilterList was found in `$RefList"
 }

· 2

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AndreasBaumgarten · Nov 19, 2020 at 10:39 PM

Hi Rich,

may I ask you what is the difference between your approach and my approach (beside the fact your are building $a as an array)?

Please don't get me wrong: I am still learning and like to improve my PowerShell approaches! This is no criticism or something else!

Kind regards
Andreas Baumgarten

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RichMatheisen-8856 AndreasBaumgarten · Nov 20, 2020 at 12:14 AM

I tried to answer this in a comment, but I don't think comments like more than a few lines of code, so I added another answer.

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Answer 6 · 2020-11-19T22:40:04.523Z

BillStewart answered Nov 19, '20

Here's another approach:

 function Compare-ListContainsAny {
   # outputs $true if $list1 contains any item in $list2
   param(
     $list1,
     $list2
   )
   foreach ( $item in $list2 ) {
     if ( $list1 -contains $item ) {
       return $true        
     }
   }
   return $false
 }

With this function defined, Compare-ListContainsAny $list $filterList would output $true.

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Answer 7 · 2020-11-20T00:12:20.157Z

RichMatheisen-8856 answered Nov 20, '20 AndreasBaumgarten commented Nov 20, '20

I don't reject or object to criticism, and I certainly don't take your question as criticism in any case.

Your code certainly answered the OP's requirement of providing a true/false answer. Mine just identified which items produced the result.

However (since you asked), if the number of elements in the reference array is very large, the Compare-Object can be very slow because it searches from the beginning of the reference array. If that's the case I'd probably not use the PowerShell [array] to define that reference list, but use a .Net class that allowed for the use of a binary search. It's a bit more work, and the object types being compared have to be the same, but it's much faster!

 [System.Collections.ArrayList]$RefList = '10','20','30','40'
 $DiffFilterList = '50','20','60'
 $Found = @()
 $DiffFilterList |
     ForEach-Object{
         if ($RefList.BinarySearch($_) -gt -1){
             $Found += $_
         }
     }
 if ($Found.Count -gt 0){
     "Found: $($Found)"
 }
 else {
     "Didn't find any matches"
 }

· 1

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AndreasBaumgarten · Nov 20, 2020 at 07:34 AM

Thank you!

Kind regards
Andreas

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Answer 8 · 2020-11-20T00:38:37.71Z

DenisPasternak-3587 answered Nov 20, '20 DenisPasternak-3587 edited Nov 20, '20

Thank you.

The question:

How I can make more this shortly ?

( ($list -contains '50') -or ($list -contains '20') -or ($list -contains '60') -or ($list -contains '1') -or ($list -contains '2') -or ($list -contains '3') -or ($list -contains '4') -or ($list -contains 'any') ) -and ($list -notcontains '777')

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