CODEWITHSUNDEEP
powershell
Nestor
Nestor

Reputation: 13990

Powershell: Combine single arrays into columns

Given:

$column1 = @(1,2,3)
$column2 = @(4,5,6)

How can I combine them into an object $matrix which gets displayed as a matrix with the single arrays as columns:

column1 column2
------- -------
   1       4
   2       5
   3       6

Upvotes: 4

Views: 7083

Answers (5)

mattmess1221
mattmess1221

Reputation: 4454

Here's something I created today. It takes a range of 0 to one of the column lengths, then maps it to a list of hashes. Use the select to turn it into a proper table.

$table = 0..$ColA.Length | % { @{
    ColA = $ColA[$_]
    ColB = $ColB[$_]
}} | Select ColA, ColB

Using the following variables:

$ColA = @(1, 2, 3)
$ColB = @(4, 5, 6)

Results in

ColB ColA
---- ----
   1    4
   2    5
   3    6

Upvotes: 1

John Zabroski
John Zabroski

Reputation: 2357

Here is a combination of mjolinor and Frode F. solutions. I ran into some problems using Frode's object construction trick using select-object. For some reason it would output hash values likely representing object references. I only code in PowerShell a few times a year, so I am just providing this in case anyone else finds it useful (perhaps even my future self).

$column1 = @(1,2,3)
$column2 = @(4,5,6,7)
$column3 = @(2,5,5,2,1,3);

$max = (
    $column1,
    $column2,
    $column3 |
        Measure-Object -Maximum -Property Count).Maximum;

$i=0

0..$max |
foreach {
  new-object psobject -property @{
                                    col1 = $Column1[$i]
                                    col3 = $column3[$i]
                                    col2 = $column2[$i++]
                                   }
  } | ft -auto

Upvotes: 0

mjolinor
mjolinor

Reputation: 68341

Little better, maybe:

$column1 = @(1,2,3)
$column2 = @(4,5,6,7)

$i=0
($column1,$column2 | sort length)[1] |
foreach {
  new-object psobject -property @{
                                    loess = $Column1[$i]
                                    lowess = $column2[$i++]
                                   }
  } | ft -auto

loess lowess
----- ------
    1      4
    2      5
    3      6
           7

Upvotes: 1

Frode F.
Frode F.

Reputation: 54971

It seems that all of my solutions today requires calculated properties. Try:

$column1 = @(1,2,3)
$column2 = @(4,5,6)

0..($column1.Length-1) | Select-Object @{n="Id";e={$_}}, @{n="Column1";e={$column1[$_]}}, @{n="Column2";e={$column2[$_]}}

Id Column1 Column2
-- ------- -------
 0       1       4
 1       2       5
 2       3       6

If the lengths of the arrays are not equal, you could use: 

$column1 = @(1,2,3)
$column2 = @(4,5,6,1)

$max = ($column1, $column2 | Measure-Object -Maximum -Property Count).Maximum    

0..$max | Select-Object @{n="Column1";e={$column1[$_]}}, @{n="Column2";e={$column2[$_]}}

I wasn't sure if you needed the Id, so I included it in the first sample to show how to include it.

Upvotes: 6

Nestor
Nestor

Reputation: 13990

I came up with this.. but it seems too verbose. Anything shorter?

&{
    for ($i=0; $i -lt $y.Length; $i++) {
        New-Object PSObject -Property @{
            y = $y[$i]
            loess = $smooth_loess[$i]
            lowess = $smooth_lowess[$i]
        }
    }
} | Format-Table -AutoSize

Upvotes: 0

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