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

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

Although not as neat as you might want but is cleaner in the way PowerShell works ...

(@(1,2,3,4)) | Select-Object -Skip 1

returns ...

2
3
4

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.

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

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

$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,

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

Unfortunately piping your array to Select-Object -Skip $skipStart | Select-Object -SkipLast $skipEnd is the only foolproof idiomatic way to get the correct items out.

(@wensveen got there first with this strategy, but SO won't let me comment yet, and I think giving an explanation warrants the detail of a full reply.)

Calculating ranges doesn't work

Ranging from $a[3..$a.count-3] and so on doesn't work if your list is shorter than you expect: supposing $a has 4 items, the range you'd end up with is 3..1, i.e. @(3, 2, 1), three items in reverse order, whereas in the Python example a[3:-3] would return zero elements, because the ascending range that starts three from the start and ends three from the end is empty. (In Python, range order is an additional explicit parameter, a[-1:1:-1] permits a reverse-ordered result.)

Output format

As per usual, if you need to force the output to be an array, you can wrap the pipeline in a @(...) array coercion. @RiverHeart mentioned wanting to preserve the input array type e.g. Int32[], but as far as I can tell that doesn't work with ordinary indexing either, I might be missing something.

Getting the right items

Another brief mention: the numbers you use are sliiightly different for the skip operation than the range notation. If you're starting at index 1, you're also skipping 1 item from the start in a zero-indexed array, so that one's the same; but skipping 1 at the end is equivalent to index -2, or equally $a.count-2, for the second-last element.

Adding array indexes

Finally, adding range literals together is a nice feature. It doesn't generalise for the reasons above, it won't stop you from accidentally getting the same item twice, say, if your ranges overlap, but it's nice to be able to go $a[0..2+-3..-1] at an interactive prompt to get the first two and last three, more succinct than making and combining separate arrays and still quite clear. I guess if you want to thumb your nose at Python, that's one thing it doesn't do quite as simply!

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)