Conditionally modify array element without resorting to a temporary array

Question

Given an array like this

$array = @('A', 'B', 'C', 'D', 'E')

I can append something to the end of each item in the array like so

$array = $array | Foreach-Object {"$_ *"}

And if I want to append something DIFFERENT to each member I can do so like this

$tempArray = @()
$count = 1
foreach ($item in $array) {
    $tempArray += "$item $count"
    $count ++
}
$array = $tempArray

But that use of a temporary array seems... clumsy. I feel like I am missing something that would allow conditionally changing the values of the original array instead. Not adding new items, just changing the value of existing items.

You are wanting to change or append? What do you hope to end up with? — Doug Maurer
– Doug Maurer, Commented Aug 12, 2020 at 17:23
@doug-maurer, I want to append to the end of the current value. So the final array is A 1, B 2, C 3, D 4, E 5 — Gordon
– Gordon, Commented Aug 12, 2020 at 17:25
I added example for both appending the value and appending a space and a value. — Doug Maurer
– Doug Maurer, Commented Aug 12, 2020 at 18:33

Guenther Schmitz · Accepted Answer · 2020-08-12 18:05:16Z

1

you have two options here

1: set the result of the foreach to the variable you are iterating. this will iterate the array and set the result to the same variable.

$count = 1
$array = foreach ($item in $array) {
    "$item $count"
    $count ++
}

2: set an array element like so (note array elements begin by 0 not 1)

$array[2] = 'foo'

or in a loop like like so. this will set each variable at iteration.

$append = 'a'
for($i=0;$i -lt $array.count;$i++){
    $array[$i] = $array[$i],$append -join ' '
    $append += 'a'
}

answered Aug 12, 2020 at 18:05

Guenther Schmitz

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1 Comment

Gordon Over a year ago

Aha. I think the thing I was missing was your first example, setting the $array = to the result of the loop. I had just been doing the loop and trying to set $item =. For more complex conditions where I am not already creating an index variable, this seems better.

Doug Maurer · Accepted Answer · 2020-08-12 19:29:34Z

1

You don't need to add @(), it's already an array as

$array = 'A', 'B', 'C', 'D', 'E'

If you simply want to append to the string and store back into the array, this is how I would do it. I'm taking advantage of the -Begin, -Process, and -OutVariable parameters of Foreach-Object.

$array | ForEach{$count = 1}{$_ + $count++} -OutVariable array

Output (also stored back in $array)

A1
B2
C3
D4
E5

Now if you did actually want to append a space before the number, then simply add that to the item like this

$array | ForEach{$count = 1}{"$_ " + $count++} -OutVariable array

Output

A 1
B 2
C 3
D 4
E 5

If you need to control the data flow more explicitly I would still lean towards not using for loops. This is equivalent.

$array = 'A', 'B', 'C', 'D', 'E'
$array = 1..$array.Count | ForEach{$array[$_-1] + " $_"}

or you could also write it as

$array = 'A', 'B', 'C', 'D', 'E'
$array = 1..$array.Count | ForEach{"$($array[$_-1]) $_"}

edited Aug 12, 2020 at 19:29

answered Aug 12, 2020 at 18:33

Doug Maurer

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1 Comment

Gordon Over a year ago

Neat, I didn't know the -outVariable trick. I can see that coming in handy. For my immediate needs I thinkGuenther's first solution below is going to allow me to handle the complex conditionals my real code needs. But all three will have their place in other situations.

user7818749 · Accepted Answer · 2020-08-12 17:42:39Z

0

Something like this seems to be what you're after:

$array = @('A','B', 'C', 'D', 'E')
for ($c=1; $c -le $array.count; $c++) {
    Write-host "$($array[$c-1]) $c"
}

answered Aug 12, 2020 at 17:42

user7818749

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