Powershell how to count all elements in a multidimensional array












0















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



























    0















    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

























      0












      0








      0








      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














      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




      arrays arraylist multidimensional-array






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 20 '18 at 8:23









      secondplacesecondplace

      457




      457
























          1 Answer
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          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 '18 at 9:32











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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1














          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 '18 at 9:32
















          1














          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 '18 at 9:32














          1












          1








          1







          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 '18 at 8:52

























          answered Nov 20 '18 at 8:40









          Micha WiedenmannMicha Wiedenmann

          10.3k1364103




          10.3k1364103













          • 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 '18 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 '18 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 '18 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 '18 at 9:32


















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