What is the idiomatic way to slice an array relative to both of its ends?

Question

Powershell\'s array notation has rather bizarre, albeit documented, behavior for slicing the end of arrays. This section from the official documentation sums up the bizarreness

Answer 1

I believe this is the right way to do it. all other methods require more code.

$a[1..($a.Count-1)]

Also, if array is converted to string it becomes easy to get data as below:

$a = 0,1,2,3
[string]::Concat($a).Substring(1)

Answer 2

This could be the most idiomatic way to slice an array with both of its ends:

$array[start..stop] where stop is defined by taking the length of the array minus a value to offset from the end of the array:

$a = 1,2,3,4,5,6,7,8,9
$start = 2
$stop = $a.Length-3
$a[$start..$stop]

This will return 3 4 5 6 7

The start value starts counting with zero, so a start value of '2' gives you the third element of the array. The stop value is calculated with ($a.Length-3), this will drop the last two values because $a.Length-3 itself is included in the slice.

I have defined $start and $stop for clarity, obviously you can also write it like this:

$a = 1,2,3,4,5,6,7,8,9
$a[2..($a.Length-3)]

This will also return 3 4 5 6 7

Answer 3

$arr = @(10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

$arr | Select-Object -First 5 | Select-Object -Index (@(0..4) | Where-Object { $_ % 2 -eq 0}) 
$arr | Select-Object -Last 5
$arr | Select-Object -Unique
$arr | Sort-Object | Select-Object -Unique
$arr | Where-Object {  $_ % 5 -eq 0 } | Sort-Object | Select-Object -Unique
$arr | Select-Object -First ($arr.Count - 3)

Actually code speaks for itself. I event don't need to explain.

However, 1) Provide the first five elements, but each second of those five. Equal to arr[:5:2] in Python 2) Get the last five elements. 3) Gives unique elements 4) Firstly sort and then provide unique 5) Gives only elements which equal 0 by applying modulo of 5, sort, unique. 5) Provide the first count of elements in that array minus three elements only.

Answer 4

If you want to get n elements from the end of an array simply fetch the elements from -n to -1:

PS C:\> $a = 0,1,2,3
PS C:\> $n = 2
PS C:\> $a[-$n..-1]
2
3

Edit: PowerShell doesn't support indexing relative to both beginning and end of the array, because of the way $a[$i..$j] works. In a Python expression a[i:j] you specify i and j as the first and last index respectively. However, in a PowerShell .. is the range operator, which generates a sequence of numbers. In an expression $a[$i..$j] the interpreter first evaluates $i..$j to a list of integers, and then the list is used to retrieve the array elements on these indexes:

PS C:\> $a = 0,1,2,3
PS C:\> $i = 1; $j = -1
PS C:\> $index = $i..$j
PS C:\> $index
1
0
-1
PS C:\> $a[$index]
1
0
3

If you need to emulate Python's behavior, you must use a subexpression:

PS C:\> $a = 0,1,2,3
PS C:\> $i = 1; $j = -1
PS C:\> $a[$i..($a.Length+$j-1)]
1
2

Answer 5

Combine Select-Object -Skip and Select-Object -SkipLast like:

$a = 0,1,2,3
$a | Select-Object -Skip 1 | Select-Object -SkipLast 1

Returns:

1
2

Not as elegant as Python, but at least you don't have to use Count or Length, meaning this also works if the array isn't stored in a variable.

Answer 6

If you are looking for, say, the first three and last three elements in an array, with the results in an array, a little array addition will take care of the need.

[array]$A = (([int][char]'A')..([int][char]'Z')) | ForEach-Object {[char]$_}
$B = $A[0..2]+$A[-3..-1]
Clear-Host
Write-Host "Original List"
Write-Host $A -NoNewline -Separator ', '
Write-Host
Write-Host "First three and last three"
Write-Host $B -NoNewline -Separator ', '

Yields:

Original List
A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z
First three and last three
A, B, C, X, Y, Z