Updating hash table values in a 'foreach' loop in PowerShell

Question

I'm trying to loop through a hash table and set the value of each key to 5 and PowerShell gives an error:

$myHash = @{}
$myHash["a"] = 1
$myHash["b"] = 2
$myHash["c"] = 3

foreach($key in $myHash.keys){
    $myHash[$key] = 5
}

An error occurred while enumerating through a collection:

Collection was modified; enumeration operation may not execute..
At line:1 char:8
+ foreach <<<< ($key in $myHash.keys){
    + CategoryInfo          : InvalidOperation: (System.Collecti...tableEnumer
   ator:HashtableEnumerator) [], RuntimeException
    + FullyQualifiedErrorId : BadEnumeration

What gives and how do I resolve this problem?

Related question: stackoverflow.com/questions/4726137/…

tenpn
– tenpn

2011-05-04 07:36:33 +00:00
Commented May 4, 2011 at 7:36 — tenpn
– tenpn, Commented May 4, 2011 at 7:36

Andrew Savinykh · Accepted Answer · 2011-05-04 07:36:40Z

56

You can't modify Hashtable while enumerating it. This is what you can do:

$myHash = @{}
$myHash["a"] = 1
$myHash["b"] = 2
$myHash["c"] = 3

$myHash = $myHash.keys | foreach{$r=@{}}{$r[$_] = 5}{$r}

Edit 1

Is this any simpler for you:

$myHash = @{}
$myHash["a"] = 1
$myHash["b"] = 2
$myHash["c"] = 3

foreach($key in $($myHash.keys)){
    $myHash[$key] = 5
}

edited May 4, 2011 at 7:36

answered May 4, 2011 at 7:24

Andrew Savinykh

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8 Comments

ted Over a year ago

Thanks zespri ---- that's brutally painful for what should be a straight forward operation....

Andrew Savinykh Over a year ago

Powershell is very flexible. There are many ways to accomplish the same thing. Admittedly, it takes time to learn, but once you are there this kind of stuff comes easy. I edited my answer to provide a way that may appear more intuitive for you. This resembles more the format you put it in originally.

JasonMArcher Over a year ago

This is the nature of enumerators. If you change the collection while you are enumerating, bad things could happen. Like skipping items, getting the same item twice, or worse.

Andrew Savinykh Over a year ago

@JasonMArcher: now I agree with you, but just for arguments sake, don't you think, that what you've described would be likely to happen if you add / remove items in collection while iterating. This logically should not be a problem for substituting one value to another? On the second thought, with Hashtable, changing value, means that the stored object might need to move to another bucket, and keeping this in mind, this behaviour can make sense.

JasonMArcher Over a year ago

Yes, but it is easier to just not allow any modification, rather than try to figure out if the modification might be a problem or not.

|

Peter Mortensen · Accepted Answer · 2019-07-19 16:28:33Z

21

There is a much simpler way of achieving this. You cannot change the value of a hashtable while enumerating it because of the fact that it's a reference type variable. It's exactly the same story in .NET.

Use the following syntax to get around it. We are converting the keys collection into a basic array using the @() notation. We make a copy of the keys collection, and reference that array instead which means we can now edit the hashtable.

$myHash = @{}
$myHash["a"] = 1
$myHash["b"] = 2
$myHash["c"] = 3

foreach($key in @($myHash.keys)){
    $myHash[$key] = 5
}

edited Jul 19, 2019 at 16:28

Peter Mortensen

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answered Mar 14, 2012 at 21:30

Andy Peaple

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2 Comments

Cheeso Over a year ago

Re: ` It's exactly the same story in .Net.` ... Powershell IS .NET. Just sayin.

Simon Elms Over a year ago

Thank you for describing why you used the @() operator. It appears in another answer but they didn't explain what was going on.

iRon · Accepted Answer · 2020-05-24 17:47:09Z

10

You do not need to clone the whole hashtable for this example. Just enumerating the key collection by forcing it to an array @(...) is enough:

foreach($key in @($myHash.keys)) {...

edited May 24, 2020 at 17:47

answered Jul 8, 2017 at 11:36

iRon

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1 Comment

jaymjarri Over a year ago

or we can use clone of keys (very similar)foreach($key in @($myHash.keys.clone()){..

Peter Mortensen · Accepted Answer · 2019-07-19 16:26:55Z

5

Use clone:

foreach($key in ($myHash.clone()).keys){
    $myHash[$key] = 5
}

Or in the one-liner:

$myHash = ($myHash.clone()).keys | % {} {$myHash[$_] = 5} {$myHash}

edited Jul 19, 2019 at 16:26

Peter Mortensen

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answered May 4, 2011 at 7:44

Emiliano Poggi

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3 Comments

Emiliano Poggi Over a year ago

Can someone explain why I got -2 recently? It would be helpful.

Sarath Rachuri Over a year ago

this works. thanks. can you explain why the hashtable needs to be cloned before modifying the value

Emiliano Poggi Over a year ago

(Powershell v2 don't know for later versions) you cannot modify an hashtable while enumerating on it.

Will Priestley · Accepted Answer · 2013-04-30 11:33:52Z

3

I'm new to PowerShell, but I'm quite a fan of using in-built functions, because I find it more readable. This is how I would tackle the problem, using GetEnumerator and Clone. This approach also allows one to reference to the existing hash values ($_.value) for modifying purposes.

$myHash = @{}
$myHash["a"] = 1 
$myHash["b"] = 2
$myHash["c"] = 3

$myHash.Clone().GetEnumerator() | foreach-object {$myHash.Set_Item($_.key, 5)}

answered Apr 30, 2013 at 11:33

Will Priestley

392 bronze badges

Comments

Peter Mortensen · Accepted Answer · 2019-07-19 16:31:44Z

3

You have to get creative!

$myHash = @{}
$myHash["a"] = 1
$myHash["b"] = 2
$myHash["c"] = 3

$keys = @()
[array] $keys = $myHash.keys

foreach($key in $keys)
{
    $myHash.Set_Item($key, 5)
}

$myHash

Name                         Value
----                         -----
c                              5
a                              5
b                              5

edited Jul 19, 2019 at 16:31

Peter Mortensen

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answered Nov 12, 2015 at 21:31

ScubaZip

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1 Comment

mnencia Over a year ago

You could improve the answer explaining why the OP script doesn't work.

Peter Mortensen · Accepted Answer · 2019-07-19 16:32:37Z

0

As mentioned in a previous answer, clone is the way to go. I had a need to replace any null values in a hash with "Unknown" nd this one-liner does the job.

($record.Clone()).keys | %{if ($record.$_ -eq $null) {$record.$_ = "Unknown"}}

edited Jul 19, 2019 at 16:32

Peter Mortensen

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answered Aug 12, 2016 at 20:16

Scooter

1

Comments

Peter Mortensen · Accepted Answer · 2019-07-19 16:35:08Z

0

$myHash = @{
    Americas = 0;
    Asia = 0;
    Europe = 0;
}

$countries = @("Americas", "Asia", "Europe", "Americas", "Asia")

foreach($key in $($myHash.Keys))
{
    foreach($Country in $countries)
    {
        if($key -eq $Country)
        {
            $myHash[$key] += 1
        }
    }
}

$myHash

edited Jul 19, 2019 at 16:35

Peter Mortensen

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answered Jun 14, 2019 at 7:49

Dhanasekaran M

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1 Comment

Dhanasekaran M Over a year ago

Update Hash value when array and hash key has same value by one. The above code is works fine for me.

Peter Mortensen · Accepted Answer · 2019-07-19 16:35:59Z

0

$myHash = @{
    Americas = 0;
    Asia = 0;
    Europe = 0;
}

$countries = @("Americas", "Asia", "Europe", "Americas", "Asia")

foreach($key in $($myHash.Keys))
{
    foreach($Country in $countries)
    {
        if($key -eq $Country)
        {
            $myHash[$key] += 1
        }
    }
}

Updating a hash value if array elements matched with a hash key.

edited Jul 19, 2019 at 16:35

Peter Mortensen

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answered Jun 14, 2019 at 8:08

Dhanasekaran M

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Comments

Peter Mortensen · Accepted Answer · 2019-07-19 16:26:32Z

-1

It seems when you update the hash table inside the foreach loop, the enumerator invalidates itself. I got around this by populating a new hash table:

$myHash = @{}
$myHash["a"] = 1
$myHash["b"] = 2
$myHash["c"] = 3

$newHash = @{}
foreach($key in $myHash.keys){
    $newHash[$key] = 5
}
$myHash = $newHash

edited Jul 19, 2019 at 16:26

Peter Mortensen

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answered May 4, 2011 at 7:35

tenpn

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