Check if an strings of list exists in other list

Question

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?

Answer 1

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

Answer 2

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

Answer 3

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"
}

Answer 4

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.

Answer 5

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"
}