Powershell how to count all elements in a multidimensional array











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I've been trying to figure out how to count all elements in an multidimensional array. But .Count only returns the first dimension.



after i gave up to find a proper solution i just created this loop to move all elements to the first dimension and count them. but this is really only a hack.



$mdarr = @((0,1,2,3,4),(5,6,7,8,9),(10,11,12,13,14))
$filecount = New-Object System.Collections.ArrayList
for($i = 0; $i -lt $mdarr.Length; ++$i) {
$filecount += $mdarr[$i]
}
$filecount.Count


How would this be done properly without processing the array first?










share|improve this question


























    up vote
    0
    down vote

    favorite












    I've been trying to figure out how to count all elements in an multidimensional array. But .Count only returns the first dimension.



    after i gave up to find a proper solution i just created this loop to move all elements to the first dimension and count them. but this is really only a hack.



    $mdarr = @((0,1,2,3,4),(5,6,7,8,9),(10,11,12,13,14))
    $filecount = New-Object System.Collections.ArrayList
    for($i = 0; $i -lt $mdarr.Length; ++$i) {
    $filecount += $mdarr[$i]
    }
    $filecount.Count


    How would this be done properly without processing the array first?










    share|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I've been trying to figure out how to count all elements in an multidimensional array. But .Count only returns the first dimension.



      after i gave up to find a proper solution i just created this loop to move all elements to the first dimension and count them. but this is really only a hack.



      $mdarr = @((0,1,2,3,4),(5,6,7,8,9),(10,11,12,13,14))
      $filecount = New-Object System.Collections.ArrayList
      for($i = 0; $i -lt $mdarr.Length; ++$i) {
      $filecount += $mdarr[$i]
      }
      $filecount.Count


      How would this be done properly without processing the array first?










      share|improve this question













      I've been trying to figure out how to count all elements in an multidimensional array. But .Count only returns the first dimension.



      after i gave up to find a proper solution i just created this loop to move all elements to the first dimension and count them. but this is really only a hack.



      $mdarr = @((0,1,2,3,4),(5,6,7,8,9),(10,11,12,13,14))
      $filecount = New-Object System.Collections.ArrayList
      for($i = 0; $i -lt $mdarr.Length; ++$i) {
      $filecount += $mdarr[$i]
      }
      $filecount.Count


      How would this be done properly without processing the array first?







      arrays arraylist multidimensional-array






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 20 at 8:23









      secondplace

      417




      417
























          1 Answer
          1






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          up vote
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          down vote



          accepted










          In the loop you are adding the elements of $mdarr[$i]. You later count the elements of the merge result. Instead of the adding to an ArrayList you could keep a count:



          $xs = @((0,1,2,3,4),(5,6,7,8,9),(10,11,12,13,14))
          $sum = 0;
          foreach ($x in $xs) { $sum += $x.Count }
          $sum // 15

          # alternatively
          $xs | % { $sum += $_.Count }

          # or
          ($xs | % { $_.Count } | Measure-Object -Sum).Sum
          # or
          $xs | % { $_.Count } | Measure-Object -Sum | select -Expand Sum





          share|improve this answer























          • The first options seems to be the fastest after doing some measuring. And the two last ones look a bit cleaner. Thank you for multiple solutions.
            – secondplace
            Nov 20 at 9:32











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          In the loop you are adding the elements of $mdarr[$i]. You later count the elements of the merge result. Instead of the adding to an ArrayList you could keep a count:



          $xs = @((0,1,2,3,4),(5,6,7,8,9),(10,11,12,13,14))
          $sum = 0;
          foreach ($x in $xs) { $sum += $x.Count }
          $sum // 15

          # alternatively
          $xs | % { $sum += $_.Count }

          # or
          ($xs | % { $_.Count } | Measure-Object -Sum).Sum
          # or
          $xs | % { $_.Count } | Measure-Object -Sum | select -Expand Sum





          share|improve this answer























          • The first options seems to be the fastest after doing some measuring. And the two last ones look a bit cleaner. Thank you for multiple solutions.
            – secondplace
            Nov 20 at 9:32















          up vote
          1
          down vote



          accepted










          In the loop you are adding the elements of $mdarr[$i]. You later count the elements of the merge result. Instead of the adding to an ArrayList you could keep a count:



          $xs = @((0,1,2,3,4),(5,6,7,8,9),(10,11,12,13,14))
          $sum = 0;
          foreach ($x in $xs) { $sum += $x.Count }
          $sum // 15

          # alternatively
          $xs | % { $sum += $_.Count }

          # or
          ($xs | % { $_.Count } | Measure-Object -Sum).Sum
          # or
          $xs | % { $_.Count } | Measure-Object -Sum | select -Expand Sum





          share|improve this answer























          • The first options seems to be the fastest after doing some measuring. And the two last ones look a bit cleaner. Thank you for multiple solutions.
            – secondplace
            Nov 20 at 9:32













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          In the loop you are adding the elements of $mdarr[$i]. You later count the elements of the merge result. Instead of the adding to an ArrayList you could keep a count:



          $xs = @((0,1,2,3,4),(5,6,7,8,9),(10,11,12,13,14))
          $sum = 0;
          foreach ($x in $xs) { $sum += $x.Count }
          $sum // 15

          # alternatively
          $xs | % { $sum += $_.Count }

          # or
          ($xs | % { $_.Count } | Measure-Object -Sum).Sum
          # or
          $xs | % { $_.Count } | Measure-Object -Sum | select -Expand Sum





          share|improve this answer














          In the loop you are adding the elements of $mdarr[$i]. You later count the elements of the merge result. Instead of the adding to an ArrayList you could keep a count:



          $xs = @((0,1,2,3,4),(5,6,7,8,9),(10,11,12,13,14))
          $sum = 0;
          foreach ($x in $xs) { $sum += $x.Count }
          $sum // 15

          # alternatively
          $xs | % { $sum += $_.Count }

          # or
          ($xs | % { $_.Count } | Measure-Object -Sum).Sum
          # or
          $xs | % { $_.Count } | Measure-Object -Sum | select -Expand Sum






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 20 at 8:52

























          answered Nov 20 at 8:40









          Micha Wiedenmann

          10.1k1364102




          10.1k1364102












          • The first options seems to be the fastest after doing some measuring. And the two last ones look a bit cleaner. Thank you for multiple solutions.
            – secondplace
            Nov 20 at 9:32


















          • The first options seems to be the fastest after doing some measuring. And the two last ones look a bit cleaner. Thank you for multiple solutions.
            – secondplace
            Nov 20 at 9:32
















          The first options seems to be the fastest after doing some measuring. And the two last ones look a bit cleaner. Thank you for multiple solutions.
          – secondplace
          Nov 20 at 9:32




          The first options seems to be the fastest after doing some measuring. And the two last ones look a bit cleaner. Thank you for multiple solutions.
          – secondplace
          Nov 20 at 9:32


















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