Compare array with null elements

Question

I have two array's having more that 500,000 values and need to compare to find the difference. Some of the values in the array are $null. When I am trying to do the following code and I getting an error Compare-Object : Cannot bind argument to parameter 'ReferenceObject' because it is null

Is there any work around for this issue or is there a better and faster way to find the difference? I am only interested in the count of variance/difference.

Thanks!

Example Code:

$objSource = @('a', $null, 'c')
$objTarget = @('a', 'b','c')
Compare-Object -ReferenceObject $objSource -DifferenceObject $objTarget | Measure

Error Message:

Compare-Object : Cannot bind argument to parameter 'ReferenceObject' because it is null . At line:4 char:32
+ Compare-Object -ReferenceObject <<<<  $objSource -DifferenceObject $objTarget
    + CategoryInfo          : InvalidData: (:) [Compare-Object], ParameterBindingValid     ationException
    + FullyQualifiedErrorId : ParameterArgumentValidationErrorNullNotAllowed,Microsoft     .PowerShell.Commands.CompareObjectCommand

Matt · Accepted Answer · 2017-07-11 11:58:16Z

3

The documentation states that this would happen if the -ReferenceObject or -DifferenceObject were $null but does not really cover if the any elements of those objects are.

NOTE: If the reference set or the difference set is null ($null), Compare-Object generates a terminating error.

Quick work around would be to replace those $null elements with a control character or something along those lines.

$objSource = @('a', $null, 'c') -replace "^$","null"
$objTarget = @('a', 'b','c') -replace "^$","null"
Compare-Object -ReferenceObject $objSource -DifferenceObject $objTarget

The -replace would remove those empty entries from the array with string "null". So we would have a visual indication of that data now. Unlikely that that text would be a natural element but you would have to be wary just the same. If null is not appropriate I would recommend bagel as a keyword!

edited Jul 11, 2017 at 11:58

answered Jan 8, 2015 at 5:09

Matt

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mjolinor · Accepted Answer · 2015-01-08 12:09:51Z

2

This seems to work:

$objSource = @('a', $null, 'c')
$objTarget = @('a', 'b','c')
Compare-Object -ReferenceObject ([string[]]$objSource) -DifferenceObject ([string[]]$objTarget) | Measure

answered Jan 8, 2015 at 12:09

mjolinor

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1 Comment

Sarang Over a year ago

Yes, this works but the result is always 2. I would like to find the variance count.

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