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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?

list-manipulation

share|improve this question

edited 2 days ago

m_goldberg

87.4k872198

asked 2 days ago

liolio

1,108217

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the solution.
  $endgroup$
  – lio
  2 days ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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?

list-manipulation

share|improve this question

edited 2 days ago

m_goldberg

87.4k872198

asked 2 days ago

liolio

1,108217

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the solution.
  $endgroup$
  – lio
  2 days ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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?

list-manipulation

share|improve this question

edited 2 days ago

m_goldberg

87.4k872198

asked 2 days ago

liolio

1,108217

$endgroup$

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

        {1, 0, 2, 2, 2, 4, 0}}

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.}

share|improve this question

edited 2 days ago

m_goldberg

87.4k872198

asked 2 days ago

liolio

1,108217

share|improve this question

edited 2 days ago

m_goldberg

87.4k872198

asked 2 days ago

liolio

1,108217

edited 2 days ago

m_goldberg

87.4k872198

edited 2 days ago

m_goldberg

87.4k872198

asked 2 days ago

liolio

1,108217

asked 2 days ago

liolio

1,108217

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

answered 2 days ago

MarcoBMarcoB

37.4k556113

$endgroup$

add a comment | 

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

answered 2 days ago

community wiki

m_goldberg

$endgroup$

add a comment | 

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8

$begingroup$

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

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

share|improve this answer

answered 2 days ago

MarcoBMarcoB

37.4k556113

$endgroup$

add a comment | 

8

$begingroup$

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

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

share|improve this answer

answered 2 days ago

MarcoBMarcoB

37.4k556113

$endgroup$

add a comment | 

$begingroup$

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

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

share|improve this answer

answered 2 days ago

MarcoBMarcoB

37.4k556113

$endgroup$

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

share|improve this answer

answered 2 days ago

MarcoBMarcoB

37.4k556113

answered 2 days ago

MarcoBMarcoB

37.4k556113

answered 2 days ago

MarcoBMarcoB

37.4k556113

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]]

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]]

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]]

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]]