Averaging over columns while ignoring zero entries












5












$begingroup$


I have:



list = {{1, 3, 4, 5, 6, 0, 9}, {4, 0, 3, 5, 0, 2, 0}, 
{1, 0, 2, 2, 2, 4, 0}}


I want to calculate the average but without considering the 0s.



The result should be:



Do[
array = list[[All, i]];
total = Total[array];
cnt = Count[array, 0];
result[[i]] = total/(Length@array - cnt);,
{i, 1, Length@result}
];

result // N

{2., 3., 3., 4., 4., 3., 9.}


How can I replace the Do loop?










share|improve this question











$endgroup$








  • 1




    $begingroup$
    Mean /@ DeleteCases[Transpose@list, 0, All]
    $endgroup$
    – MarcoB
    13 hours ago










  • $begingroup$
    Thank you for the solution.
    $endgroup$
    – lio
    13 hours ago












  • $begingroup$
    You’re welcome. I’ve added an answer since this solution works for you
    $endgroup$
    – MarcoB
    13 hours ago








  • 4




    $begingroup$
    I'm not at a computer right now, so please try Total[list]/Total[Unitize[list]].
    $endgroup$
    – J. M. is computer-less
    13 hours ago










  • $begingroup$
    @J.M. Well, that is clever! Cool trick :-)
    $endgroup$
    – MarcoB
    10 hours ago
















5












$begingroup$


I have:



list = {{1, 3, 4, 5, 6, 0, 9}, {4, 0, 3, 5, 0, 2, 0}, 
{1, 0, 2, 2, 2, 4, 0}}


I want to calculate the average but without considering the 0s.



The result should be:



Do[
array = list[[All, i]];
total = Total[array];
cnt = Count[array, 0];
result[[i]] = total/(Length@array - cnt);,
{i, 1, Length@result}
];

result // N

{2., 3., 3., 4., 4., 3., 9.}


How can I replace the Do loop?










share|improve this question











$endgroup$








  • 1




    $begingroup$
    Mean /@ DeleteCases[Transpose@list, 0, All]
    $endgroup$
    – MarcoB
    13 hours ago










  • $begingroup$
    Thank you for the solution.
    $endgroup$
    – lio
    13 hours ago












  • $begingroup$
    You’re welcome. I’ve added an answer since this solution works for you
    $endgroup$
    – MarcoB
    13 hours ago








  • 4




    $begingroup$
    I'm not at a computer right now, so please try Total[list]/Total[Unitize[list]].
    $endgroup$
    – J. M. is computer-less
    13 hours ago










  • $begingroup$
    @J.M. Well, that is clever! Cool trick :-)
    $endgroup$
    – MarcoB
    10 hours ago














5












5








5





$begingroup$


I have:



list = {{1, 3, 4, 5, 6, 0, 9}, {4, 0, 3, 5, 0, 2, 0}, 
{1, 0, 2, 2, 2, 4, 0}}


I want to calculate the average but without considering the 0s.



The result should be:



Do[
array = list[[All, i]];
total = Total[array];
cnt = Count[array, 0];
result[[i]] = total/(Length@array - cnt);,
{i, 1, Length@result}
];

result // N

{2., 3., 3., 4., 4., 3., 9.}


How can I replace the Do loop?










share|improve this question











$endgroup$




I have:



list = {{1, 3, 4, 5, 6, 0, 9}, {4, 0, 3, 5, 0, 2, 0}, 
{1, 0, 2, 2, 2, 4, 0}}


I want to calculate the average but without considering the 0s.



The result should be:



Do[
array = list[[All, i]];
total = Total[array];
cnt = Count[array, 0];
result[[i]] = total/(Length@array - cnt);,
{i, 1, Length@result}
];

result // N

{2., 3., 3., 4., 4., 3., 9.}


How can I replace the Do loop?







list-manipulation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 10 hours ago









m_goldberg

87.4k872198




87.4k872198










asked 13 hours ago









liolio

1,108217




1,108217








  • 1




    $begingroup$
    Mean /@ DeleteCases[Transpose@list, 0, All]
    $endgroup$
    – MarcoB
    13 hours ago










  • $begingroup$
    Thank you for the solution.
    $endgroup$
    – lio
    13 hours ago












  • $begingroup$
    You’re welcome. I’ve added an answer since this solution works for you
    $endgroup$
    – MarcoB
    13 hours ago








  • 4




    $begingroup$
    I'm not at a computer right now, so please try Total[list]/Total[Unitize[list]].
    $endgroup$
    – J. M. is computer-less
    13 hours ago










  • $begingroup$
    @J.M. Well, that is clever! Cool trick :-)
    $endgroup$
    – MarcoB
    10 hours ago














  • 1




    $begingroup$
    Mean /@ DeleteCases[Transpose@list, 0, All]
    $endgroup$
    – MarcoB
    13 hours ago










  • $begingroup$
    Thank you for the solution.
    $endgroup$
    – lio
    13 hours ago












  • $begingroup$
    You’re welcome. I’ve added an answer since this solution works for you
    $endgroup$
    – MarcoB
    13 hours ago








  • 4




    $begingroup$
    I'm not at a computer right now, so please try Total[list]/Total[Unitize[list]].
    $endgroup$
    – J. M. is computer-less
    13 hours ago










  • $begingroup$
    @J.M. Well, that is clever! Cool trick :-)
    $endgroup$
    – MarcoB
    10 hours ago








1




1




$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
$endgroup$
– MarcoB
13 hours ago




$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
$endgroup$
– MarcoB
13 hours ago












$begingroup$
Thank you for the solution.
$endgroup$
– lio
13 hours ago






$begingroup$
Thank you for the solution.
$endgroup$
– lio
13 hours ago














$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
13 hours ago






$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
13 hours ago






4




4




$begingroup$
I'm not at a computer right now, so please try Total[list]/Total[Unitize[list]].
$endgroup$
– J. M. is computer-less
13 hours ago




$begingroup$
I'm not at a computer right now, so please try Total[list]/Total[Unitize[list]].
$endgroup$
– J. M. is computer-less
13 hours ago












$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
10 hours ago




$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
10 hours ago










2 Answers
2






active

oldest

votes


















8












$begingroup$

Mean /@ DeleteCases[Transpose@list, 0, All]



{2,3,3,4,4,3,9}







share|improve this answer









$endgroup$





















    7












    $begingroup$

    I'm posting this as a CW answer, so J.M.'s very nice answer, made in a comment above, gets recorded as a real answer.



    list = {{1, 3, 4, 5, 6, 0, 9}, {4, 0, 3, 5, 0, 2, 0}, {1, 0, 2, 2, 2, 4, 0}};
    Total[list]/Total[Unitize[list]]



    {2, 3, 3, 4, 4, 3, 9}







    share|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      8












      $begingroup$

      Mean /@ DeleteCases[Transpose@list, 0, All]



      {2,3,3,4,4,3,9}







      share|improve this answer









      $endgroup$


















        8












        $begingroup$

        Mean /@ DeleteCases[Transpose@list, 0, All]



        {2,3,3,4,4,3,9}







        share|improve this answer









        $endgroup$
















          8












          8








          8





          $begingroup$

          Mean /@ DeleteCases[Transpose@list, 0, All]



          {2,3,3,4,4,3,9}







          share|improve this answer









          $endgroup$



          Mean /@ DeleteCases[Transpose@list, 0, All]



          {2,3,3,4,4,3,9}








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 13 hours ago









          MarcoBMarcoB

          37.2k556113




          37.2k556113























              7












              $begingroup$

              I'm posting this as a CW answer, so J.M.'s very nice answer, made in a comment above, gets recorded as a real answer.



              list = {{1, 3, 4, 5, 6, 0, 9}, {4, 0, 3, 5, 0, 2, 0}, {1, 0, 2, 2, 2, 4, 0}};
              Total[list]/Total[Unitize[list]]



              {2, 3, 3, 4, 4, 3, 9}







              share|improve this answer











              $endgroup$


















                7












                $begingroup$

                I'm posting this as a CW answer, so J.M.'s very nice answer, made in a comment above, gets recorded as a real answer.



                list = {{1, 3, 4, 5, 6, 0, 9}, {4, 0, 3, 5, 0, 2, 0}, {1, 0, 2, 2, 2, 4, 0}};
                Total[list]/Total[Unitize[list]]



                {2, 3, 3, 4, 4, 3, 9}







                share|improve this answer











                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  I'm posting this as a CW answer, so J.M.'s very nice answer, made in a comment above, gets recorded as a real answer.



                  list = {{1, 3, 4, 5, 6, 0, 9}, {4, 0, 3, 5, 0, 2, 0}, {1, 0, 2, 2, 2, 4, 0}};
                  Total[list]/Total[Unitize[list]]



                  {2, 3, 3, 4, 4, 3, 9}







                  share|improve this answer











                  $endgroup$



                  I'm posting this as a CW answer, so J.M.'s very nice answer, made in a comment above, gets recorded as a real answer.



                  list = {{1, 3, 4, 5, 6, 0, 9}, {4, 0, 3, 5, 0, 2, 0}, {1, 0, 2, 2, 2, 4, 0}};
                  Total[list]/Total[Unitize[list]]



                  {2, 3, 3, 4, 4, 3, 9}








                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  answered 10 hours ago


























                  community wiki





                  m_goldberg































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