Delete zeros from a sparse array [duplicate]












4












$begingroup$



This question already has an answer here:




  • How to prevent a SparseArray entry from being specified when it coincides with the default value?

    1 answer




Why doesn't SparseArray automatically remove zero entries? The code



a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];

b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];

r = DeleteCases[a-b,0]

r //ArrayRules


returns




{{1}->-11,{3}->0,{5}->0,{7}->0,{9}->-99,{11}->0,{2}->22,{13}->1313,{_}->0}




instead of




{{1}->-11,{9}->-99,{2}->22,{13}->1313,{_}->0}.




Is there a nice way to get rid of zero entries? For instance, when I perform Gaussian elimination on $A!in!mathbb{Z}^{mtimes n}_2$, I use A = Mod[A, 2];, and suddenly many entries are zero. I don't want to waste memory on them.



Of course I could use 



A = SparseArray[DeleteCases[ArrayRules[A], #[[2]] == 0&], {m,n}]


but it seems clumsy to use this often.



Also, when I have a very long sparse row that gets updated frequently and fills up somewhat (Gaussian elimination), is it better to use SparseArray with + or Association with Merge?






matrix associations sparse-arrays rule







share|improve this question











$endgroup$



marked as duplicate by Henrik Schumacher, Lukas Lang, Niki Estner, LCarvalho, anderstood Jan 4 at 14:28


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 3




    $begingroup$
    It does not do it automatically because rebuilding a sparse array is expensive. It is left to you to decide when you want to rebuild it and eliminate unnecessary entries. Just apply SparseArray again.
    $endgroup$
    – Szabolcs
    Dec 30 '18 at 9:11
















4












$begingroup$



This question already has an answer here:




  • How to prevent a SparseArray entry from being specified when it coincides with the default value?

    1 answer




Why doesn't SparseArray automatically remove zero entries? The code



a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];

b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];

r = DeleteCases[a-b,0]

r //ArrayRules


returns




{{1}->-11,{3}->0,{5}->0,{7}->0,{9}->-99,{11}->0,{2}->22,{13}->1313,{_}->0}




instead of




{{1}->-11,{9}->-99,{2}->22,{13}->1313,{_}->0}.




Is there a nice way to get rid of zero entries? For instance, when I perform Gaussian elimination on $A!in!mathbb{Z}^{mtimes n}_2$, I use A = Mod[A, 2];, and suddenly many entries are zero. I don't want to waste memory on them.



Of course I could use 



A = SparseArray[DeleteCases[ArrayRules[A], #[[2]] == 0&], {m,n}]


but it seems clumsy to use this often.



Also, when I have a very long sparse row that gets updated frequently and fills up somewhat (Gaussian elimination), is it better to use SparseArray with + or Association with Merge?






matrix associations sparse-arrays rule







share|improve this question











$endgroup$



marked as duplicate by Henrik Schumacher, Lukas Lang, Niki Estner, LCarvalho, anderstood Jan 4 at 14:28


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 3




    $begingroup$
    It does not do it automatically because rebuilding a sparse array is expensive. It is left to you to decide when you want to rebuild it and eliminate unnecessary entries. Just apply SparseArray again.
    $endgroup$
    – Szabolcs
    Dec 30 '18 at 9:11














4












4








4





$begingroup$



This question already has an answer here:




  • How to prevent a SparseArray entry from being specified when it coincides with the default value?

    1 answer




Why doesn't SparseArray automatically remove zero entries? The code



a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];

b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];

r = DeleteCases[a-b,0]

r //ArrayRules


returns




{{1}->-11,{3}->0,{5}->0,{7}->0,{9}->-99,{11}->0,{2}->22,{13}->1313,{_}->0}




instead of




{{1}->-11,{9}->-99,{2}->22,{13}->1313,{_}->0}.




Is there a nice way to get rid of zero entries? For instance, when I perform Gaussian elimination on $A!in!mathbb{Z}^{mtimes n}_2$, I use A = Mod[A, 2];, and suddenly many entries are zero. I don't want to waste memory on them.



Of course I could use 



A = SparseArray[DeleteCases[ArrayRules[A], #[[2]] == 0&], {m,n}]


but it seems clumsy to use this often.



Also, when I have a very long sparse row that gets updated frequently and fills up somewhat (Gaussian elimination), is it better to use SparseArray with + or Association with Merge?






matrix associations sparse-arrays rule







share|improve this question











$endgroup$





This question already has an answer here:




  • How to prevent a SparseArray entry from being specified when it coincides with the default value?

    1 answer




Why doesn't SparseArray automatically remove zero entries? The code



a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];

b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];

r = DeleteCases[a-b,0]

r //ArrayRules


returns




{{1}->-11,{3}->0,{5}->0,{7}->0,{9}->-99,{11}->0,{2}->22,{13}->1313,{_}->0}




instead of




{{1}->-11,{9}->-99,{2}->22,{13}->1313,{_}->0}.




Is there a nice way to get rid of zero entries? For instance, when I perform Gaussian elimination on $A!in!mathbb{Z}^{mtimes n}_2$, I use A = Mod[A, 2];, and suddenly many entries are zero. I don't want to waste memory on them.



Of course I could use 



A = SparseArray[DeleteCases[ArrayRules[A], #[[2]] == 0&], {m,n}]


but it seems clumsy to use this often.



Also, when I have a very long sparse row that gets updated frequently and fills up somewhat (Gaussian elimination), is it better to use SparseArray with + or Association with Merge?





This question already has an answer here:




  • How to prevent a SparseArray entry from being specified when it coincides with the default value?

    1 answer







matrix associations sparse-arrays rule




matrix associations sparse-arrays rule






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 30 '18 at 16:51









m_goldberg

87.7k872198




87.7k872198










asked Dec 30 '18 at 2:44









LeonLeon

405212




405212




marked as duplicate by Henrik Schumacher, Lukas Lang, Niki Estner, LCarvalho, anderstood Jan 4 at 14:28


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Henrik Schumacher, Lukas Lang, Niki Estner, LCarvalho, anderstood Jan 4 at 14:28


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 3




    $begingroup$
    It does not do it automatically because rebuilding a sparse array is expensive. It is left to you to decide when you want to rebuild it and eliminate unnecessary entries. Just apply SparseArray again.
    $endgroup$
    – Szabolcs
    Dec 30 '18 at 9:11














  • 3




    $begingroup$
    It does not do it automatically because rebuilding a sparse array is expensive. It is left to you to decide when you want to rebuild it and eliminate unnecessary entries. Just apply SparseArray again.
    $endgroup$
    – Szabolcs
    Dec 30 '18 at 9:11








3




3




$begingroup$
It does not do it automatically because rebuilding a sparse array is expensive. It is left to you to decide when you want to rebuild it and eliminate unnecessary entries. Just apply SparseArray again.
$endgroup$
– Szabolcs
Dec 30 '18 at 9:11




$begingroup$
It does not do it automatically because rebuilding a sparse array is expensive. It is left to you to decide when you want to rebuild it and eliminate unnecessary entries. Just apply SparseArray again.
$endgroup$
– Szabolcs
Dec 30 '18 at 9:11










1 Answer
1






active

oldest

votes


















8












$begingroup$

You can just apply SparseArray to a SparseArray object to normalize it. For your example:



a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];

b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];

r = DeleteCases[a-b,0];



SparseArray[r] //ArrayRules



{{1} -> -11, {2} -> 22, {9} -> -99, {13} -> 1313, {_} -> 0}







share|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    You can just apply SparseArray to a SparseArray object to normalize it. For your example:



    a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];
    
b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];
    
r = DeleteCases[a-b,0];
    

    
SparseArray[r] //ArrayRules



    {{1} -> -11, {2} -> 22, {9} -> -99, {13} -> 1313, {_} -> 0}







    share|improve this answer









    $endgroup$


















      8












      $begingroup$

      You can just apply SparseArray to a SparseArray object to normalize it. For your example:



      a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];
      
b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];
      
r = DeleteCases[a-b,0];
      

      
SparseArray[r] //ArrayRules



      {{1} -> -11, {2} -> 22, {9} -> -99, {13} -> 1313, {_} -> 0}







      share|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        You can just apply SparseArray to a SparseArray object to normalize it. For your example:



        a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];
        
b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];
        
r = DeleteCases[a-b,0];
        

        
SparseArray[r] //ArrayRules



        {{1} -> -11, {2} -> 22, {9} -> -99, {13} -> 1313, {_} -> 0}







        share|improve this answer









        $endgroup$



        You can just apply SparseArray to a SparseArray object to normalize it. For your example:



        a = SparseArray[{2->22,3->33,5->55,7->77,11->1111,13->1313},13];
        
b = SparseArray[{1->11,3->33,5->55,7->77,9->99,11->1111},13];
        
r = DeleteCases[a-b,0];
        

        
SparseArray[r] //ArrayRules



        {{1} -> -11, {2} -> 22, {9} -> -99, {13} -> 1313, {_} -> 0}








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Dec 30 '18 at 3:06









        Carl WollCarl Woll

        71.2k394185




        71.2k394185















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