问题:

Assume we have:

$a = @(1, @(2, @(3)))

I would like to flatten $a to get @(1, 2, 3).

I have found one solution:

@($a | % {$_}).count

But maybe there is a more elegant way?

回答1:

Piping is the correct way to flatten nested structures, so I'm not sure what would be more "elegant". Yes, the syntax is a bit line-noisy looking, but frankly quite serviceable.

2020 Edit

The recommended syntax these days is to expand % to ForEach-Object. A bit more verbose but definitely more readable:

@($a | ForEach-Object {$_}).count

回答2:

Same code, just wrapped in function:

function Flatten($a)
{
    ,@($a | % {$_})
}

Testing:

function AssertLength($expectedLength, $arr)
{
    if($ExpectedLength -eq $arr.length) 
    {
        Write-Host "OK"
    }
    else 
    {
        Write-Host "FAILURE"
    }
}

# Tests
AssertLength 0 (Flatten @())
AssertLength 1 (Flatten 1)
AssertLength 1 (Flatten @(1))
AssertLength 2 (Flatten @(1, 2))
AssertLength 2 (Flatten @(1, @(2)))
AssertLength 3 (Flatten @(1, @(2, @(3))))