Powershell break a long array into a array of array with length of N in one line?

Question

For example, given a list 1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8 and a number 4, it returns a list of list with length of 4, that is (1, 2, 3, 4), (5, 6, 7, 8), (1, 2, 3, 4), (5, 6, 7, 8).

Basically I want to implement the following Python code in Powershell.

s = 1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8
z = zip(*[iter(s)]*4)  # Here N is 4
# z is (1, 2, 3, 4), (5, 6, 7, 8), (1, 2, 3, 4), (5, 6, 7, 8)

The following script returns 17 instead of 5.

$a = 1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,0
$b = 0..($a.Length / 4) | % {  @($a[($_*4)..($_*4 + 4 - 1)]) } 
$b.Length

Nice solution. How would you handle a number of elements not divisible by N? Which is 4 in this case. $a = 1..18 — Doug Finke
– Doug Finke, Commented Dec 15, 2012 at 19:44
@DougFinke Creating an variable $n = 4, and replace all the 4s with $n should work. — ca9163d9
– ca9163d9, Commented Dec 15, 2012 at 20:02
Agreed. I guess what I'm asking is. What do you do with the left overs when you have a array of #'s that is not divisible by N? — Doug Finke
– Doug Finke, Commented Dec 15, 2012 at 20:33
@DougFinke I used 0..($a.Length / 4) instead of 1..($a.Length /4). So basically it added one more group. There will be an empty array if the length is divided by 4. — ca9163d9
– ca9163d9, Commented Dec 15, 2012 at 20:44

Dave Wyatt · Accepted Answer · 2014-11-10 18:14:24Z

60

This is a bit old, but I figured I'd throw in the method I use for splitting an array into chunks. You can use Group-Object with a constructed property:

$bigList = 1..1000

$counter = [pscustomobject] @{ Value = 0 }
$groupSize = 100

$groups = $bigList | Group-Object -Property { [math]::Floor($counter.Value++ / $groupSize) }

$groups will be a collection of GroupInfo objects; in this case, each group will have exactly 100 elements (accessible as $groups[0].Group, $groups[1].Group, and so on.) I use an object property for the counter to avoid scoping issues inside the -Property script block, since a simple $i++ doesn't write back to the original variable. Alternatively, you can use $script:counter = 0 and $script:counter++ and get the same effect without a custom object.

answered Nov 10, 2014 at 18:14

Dave Wyatt

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4 Comments

Ross Presser Over a year ago

I would like to use this but I need to group by an additional property first. How can I have $counter reset each time the basedomain property changes?

felixfbecker Over a year ago

I think this is the most elegant answer as it uses the pipeline.

Bitcoin Murderous Maniac Over a year ago

Thank you @DaveWyatt, I used this sort of logic for an array of strings with a bit of adjusted to this logic you shared:

$div = [Math]::Round(($List.Count / 4),0); $counter = [pscustomobject] @{ Value = 0 }; $groups = $List | Group-Object -Property { [math]::Floor($counter.Value++ / $div) }; ($groups | ? {$_.Name -eq 0}).Group; ($groups | ? {$_.Name -eq 1}).Group; ($groups | ? {$_.Name -eq 2}).Group; ($groups | ? {$_.Name -eq 3}).Group;

FranciscoNabas Over a year ago

This is old, but here from 2022 I was able to divide a group of 43K computer names in 8 seconds. Thank you sir!

Doug Finke · Accepted Answer · 2012-12-15 01:20:41Z

12

Wrote this in 2009 PowerShell Split-Every Function

Probably can be improved.

Function Split-Every($list, $count=4) {
    $aggregateList = @()

    $blocks = [Math]::Floor($list.Count / $count)
    $leftOver = $list.Count % $count
    for($i=0; $i -lt $blocks; $i++) {
        $end = $count * ($i + 1) - 1

        $aggregateList += @(,$list[$start..$end])
        $start = $end + 1
    }    
    if($leftOver -gt 0) {
        $aggregateList += @(,$list[$start..($end+$leftOver)])
    }

    $aggregateList    
}

$s = 1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8

$r = Split-Every $s 4

$r[0]
""
$r[1]
""
$r[2]
""
$r[3]

answered Dec 15, 2012 at 1:20

Doug Finke

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2 Comments

lovedatsnow Over a year ago

If the length of the list is < $count then $start and $end are used in the $aggregateList code uninitialized (or set to unknown values). Fix is to add $start=$end=0 above the for loop.

jbg Over a year ago

I had an issue with this one, it was ripping apart strings when I had one item in my $list, so I my fix was to add if ($list.Count -eq 1) { return @($list) } above the $aggregateList declaration.

Shay Levy · Accepted Answer · 2012-12-15 10:35:31Z

7

PS> $a = 1..16
PS> $z=for($i=0; $i -lt $a.length; $i+=4){ ,($a[$i]..$a[$i+3])}
PS> $z.count
4    

PS> $z[0]
1
2
3
4

PS> $z[1]
5
6
7
8

PS> $z[2]
9
10
11
12

PS> $z[3]
13
14
15
16

answered Dec 15, 2012 at 10:35

Shay Levy

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2 Comments

Peter Reavy Over a year ago

The accepted solution only works for consecutive integers. For instance when $i is 0, $a[$i] is 1 and $a[$i+3] is 4. Hence 1 .. 4 will work. But the correct solution is $z=for($i=0; $i -lt $a.length; $i+=4){ ,($a[$i .. ($i+3)])}

Julian Dominguez Over a year ago

@PeterReavy great answer!

Haoshu · Accepted Answer · 2021-04-08 19:04:55Z

6

Providing a solution using select. It doesn't need to worry whether $list.Count can be divided by $chunkSize.

function DivideList {
    param(
        [object[]]$list,
        [int]$chunkSize
    )

    for ($i = 0; $i -lt $list.Count; $i += $chunkSize) {
        , ($list | select -Skip $i -First $chunkSize)
    }
}

DivideList -list @(1..17) -chunkSize 4 | foreach { $_ -join ',' }

Output:

1,2,3,4
5,6,7,8
9,10,11,12
13,14,15,16
17

answered Apr 8, 2021 at 19:04

Haoshu

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1 Comment

pavol.kutaj Over a year ago

The advantage of this solution is that it is applicable on an array of strings as well.

TheCodingMatt · Accepted Answer · 2016-07-18 12:56:03Z

3

@Shay Levy Answer: if you change the value of a to 1..15 then your solution not working anymore ( Peter Reavy comment )

So this worked for me:

$a = 1..15
$z=for($i=0; $i -lt $a.length; $i+=4){if ($a.length -gt ($i+3)) { ,($a[$i]..$a[$i+3])} else { ,($a[$i]..$a[-1])}}
$z.count

answered Jul 18, 2016 at 12:56

TheCodingMatt

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Comments

ca9163d9 · Accepted Answer · 2012-12-17 19:16:15Z

1

$a = 1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,0
$b = 0..([Math]::ceiling($a.Length / 4) - 1) | 
    % {  @(, $a[($_*4)..($_*4 + 4 - 1)]) }

Don't know why I had to put a comma after (.

edited Dec 17, 2012 at 19:16

answered Dec 15, 2012 at 3:15

ca9163d9

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Comments

henrycarteruk · Accepted Answer · 2018-04-23 16:20:51Z

1

Clear-Host

$s = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18

$count = $s.Length

$split = $count/2

$split --

$b = $s[0..$split]

$split ++

$a = $s[$split..$count]

write-host "first array"

$b

write-host "next array"

$a

#clean up
Get-Variable -Exclude PWD,*Preference | Remove-Variable -EA 0

edited Apr 23, 2018 at 16:20

henrycarteruk

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answered Jul 15, 2014 at 7:45

Chris

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1 Comment

KhorneHoly Over a year ago

Could you please add more details to this answer? It's just code only, so a bit of explaination would be great

Santiago Squarzon · Accepted Answer · 2023-06-04 19:43:01Z

A simple method using a List<T> to collect input objects elements and output without enumerating using $PSCmdlet.WriteObject. This is compatible with Windows PowerShell 5.1.

function Split-Collection {
    [CmdletBinding()]
    param(
        [Parameter(Mandatory, ValueFromPipeline)]
        [object[]] $InputObject,

        [Parameter(Position = 0)]
        [ValidateRange(1, [int]::MaxValue)]
        [int] $ChunkSize = 5
    )

    begin {
        $list = [System.Collections.Generic.List[object]]::new()
    }
    process {
        foreach($item in $InputObject) {
            $list.Add($item)
            if($list.Count -eq $ChunkSize) {
                $PSCmdlet.WriteObject($list.ToArray())
                $list.Clear()
            }
        }
    }
    end {
        if($list.Count) {
            $PSCmdlet.WriteObject($list.ToArray())
        }
    }
}

Usage:

PS ..\pwsh> 0..10 | Split-Collection 3 | ForEach-Object { "[$_]" }

[0 1 2]
[3 4 5]
[6 7 8]
[9 10]


PS ..\pwsh> Split-Collection -InputObject (0..10) 3 | ForEach-Object { "[$_]" }

[0 1 2]
[3 4 5]
[6 7 8]
[9 10]

An even simpler way to do it if using PowerShell 7+ is with Enumerable.Chunk from LINQ.

PS ..\pwsh> [System.Linq.Enumerable]::Chunk([int[]] (0..10), 3) | ForEach-Object { "[$_]" }

[0 1 2]
[3 4 5]
[6 7 8]
[9 10]

Gennadii Saltyshchak · Accepted Answer · 2023-06-03 13:43:37Z

If collection is big enough (e.g. millions of elements), then performance may matter. None of the solutions proposed before satisfied me first of all from performance standpoints. Here is my solution:

Function Chunk {
    param(
        [object[]] $source,
        [int] $size = 1)
    $chunkCount = [Math]::Ceiling($source.Count / $size)
    0 .. ($chunkCount - 1) `
    | ForEach-Object {
        $startIndex = $_ * $size
        $endIndex = [Math]::Min(($_ + 1) * $size, $source.Count)
        ,$source[$startIndex .. ($endIndex - 1)]
    }
}

Example of usage:

Chunk @(1..17) 4 | foreach { $_ -join ',' }

Output:

1,2,3,4
5,6,7,8
9,10,11,12
13,14,15,16
17

Performance measure:

(Measure-Command { Chunk @(1..1000000) 1000 }).TotalMilliseconds

Output: 140.467 (in milliseconds)

posh · Accepted Answer · 2024-10-18 22:19:25Z

0

$chunk_size = 10
@(1..100) | % { $c = @(); $i = 0 } { if ($i % $chunk_size -eq 0) { $c += ,@() }; $c[-1] += $_; $i++ } 
$c

$chunk_size = 10 # Set desired chunk size
@(1..100) | % `
    { # Initialize loop variables
        $c = @(); # Cumulating array of chunks
        $i = 0;   # Incrementing index
    } {
        if ($i % $chunk_size -eq 0) { # If the index is divisible by chunk size
            $c += ,@() # Add another chunk to the accumulator
        }
        $c[-1] += $_; # Add current element to the last chunk
        $i++          # Increment
    }
$c

answered Oct 18, 2024 at 22:19

posh

1

1 Comment

K. Frank Over a year ago

please avoid aliases for scripts, even though % has kinda a long history of being there on all versions it makes the code less readable. And the main goal for scripts is readability, as opposed to ad-hoc within the shell.

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