What is the faster way to count occurrences of equal sublists in a nested list?












6















I have a list of lists in Python and I want to (as fastly as possible : very important...) append to each sublist the number of time it appear into the nested list.



I have done that with some pandas data-frame, but this seems to be very slow and I need to run this lines on very very large scale. I am completely willing to sacrifice nice-reading code to efficient one.



So for instance my nested list is here:



l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]


I need to have:



res = [[1, 3, 2, 2], [1, 3, 5, 1]]


EDIT



Order in res does not matter at all.






python







share|improve this question





























    6















    I have a list of lists in Python and I want to (as fastly as possible : very important...) append to each sublist the number of time it appear into the nested list.



    I have done that with some pandas data-frame, but this seems to be very slow and I need to run this lines on very very large scale. I am completely willing to sacrifice nice-reading code to efficient one.



    So for instance my nested list is here:



    l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]


    I need to have:



    res = [[1, 3, 2, 2], [1, 3, 5, 1]]


    EDIT



    Order in res does not matter at all.






    python







    share|improve this question



























      6












      6








      6








      I have a list of lists in Python and I want to (as fastly as possible : very important...) append to each sublist the number of time it appear into the nested list.



      I have done that with some pandas data-frame, but this seems to be very slow and I need to run this lines on very very large scale. I am completely willing to sacrifice nice-reading code to efficient one.



      So for instance my nested list is here:



      l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]


      I need to have:



      res = [[1, 3, 2, 2], [1, 3, 5, 1]]


      EDIT



      Order in res does not matter at all.






      python







      share|improve this question
















      I have a list of lists in Python and I want to (as fastly as possible : very important...) append to each sublist the number of time it appear into the nested list.



      I have done that with some pandas data-frame, but this seems to be very slow and I need to run this lines on very very large scale. I am completely willing to sacrifice nice-reading code to efficient one.



      So for instance my nested list is here:



      l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]


      I need to have:



      res = [[1, 3, 2, 2], [1, 3, 5, 1]]


      EDIT



      Order in res does not matter at all.






      python




      python






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 hours ago









      Muhammad Ahmad

      2,0891420




      2,0891420










      asked 2 hours ago









      Léo JoubertLéo Joubert

      134210




      134210
























          3 Answers
          3






          active

          oldest

          votes


















          5














          If order does not matter you could use collections.Counter with extended iterable unpacking, as a variant of @Chris_Rands solution:



          from collections import Counter
          

          
l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]
          

          
result = [[*t, count] for t, count in Counter(map(tuple, l)).items()]
          
print(result)


          Output



          [[1, 3, 5, 1], [1, 3, 2, 2]]





          share|improve this answer





















          • 1





            This is a reasonable (although mostly cosmetic) variant of my solution, assuming Python 3 of course

            – Chris_Rands
            2 hours ago



















          4














          This is quite an odd output to want but it is of course possible. I suggest using collections.Counter(), no doubt others will make different suggestions and a timeit style comparison would reveal the fastest of course for particular data sets:



          >>> from collections import Counter
          
>>> l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]
          
>>> [list(k) + [v] for k, v in Counter(map(tuple,l)).items()]
          
[[1, 3, 2, 2], [1, 3, 5, 1]]


          Note to preserve the insertion order prior to CPython 3.6 / Python 3.7, use the OrderedCounter recipe.






          share|improve this answer































            1














            If numpy is an option, you could use np.unique setting axis to 0 and return_counts to True, and concatenate the unique rows and counts using np.vstack:



            l = np.array([[1, 3, 2], [1, 3, 2] ,[1, 3, 5]])
            
x, c = np.unique(l, axis=0, return_counts=True)
            
np.vstack([x.T,c]).T
            

            
array([[1, 3, 2, 2],
            
       [1, 3, 5, 1]])





            share|improve this answer























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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5














              If order does not matter you could use collections.Counter with extended iterable unpacking, as a variant of @Chris_Rands solution:



              from collections import Counter
              

              
l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]
              

              
result = [[*t, count] for t, count in Counter(map(tuple, l)).items()]
              
print(result)


              Output



              [[1, 3, 5, 1], [1, 3, 2, 2]]





              share|improve this answer





















              • 1





                This is a reasonable (although mostly cosmetic) variant of my solution, assuming Python 3 of course

                – Chris_Rands
                2 hours ago
















              5














              If order does not matter you could use collections.Counter with extended iterable unpacking, as a variant of @Chris_Rands solution:



              from collections import Counter
              

              
l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]
              

              
result = [[*t, count] for t, count in Counter(map(tuple, l)).items()]
              
print(result)


              Output



              [[1, 3, 5, 1], [1, 3, 2, 2]]





              share|improve this answer





















              • 1





                This is a reasonable (although mostly cosmetic) variant of my solution, assuming Python 3 of course

                – Chris_Rands
                2 hours ago














              5












              5








              5







              If order does not matter you could use collections.Counter with extended iterable unpacking, as a variant of @Chris_Rands solution:



              from collections import Counter
              

              
l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]
              

              
result = [[*t, count] for t, count in Counter(map(tuple, l)).items()]
              
print(result)


              Output



              [[1, 3, 5, 1], [1, 3, 2, 2]]





              share|improve this answer















              If order does not matter you could use collections.Counter with extended iterable unpacking, as a variant of @Chris_Rands solution:



              from collections import Counter
              

              
l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]
              

              
result = [[*t, count] for t, count in Counter(map(tuple, l)).items()]
              
print(result)


              Output



              [[1, 3, 5, 1], [1, 3, 2, 2]]






              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 1 hour ago

























              answered 2 hours ago









              Daniel MesejoDaniel Mesejo

              16.6k21430




              16.6k21430








              • 1





                This is a reasonable (although mostly cosmetic) variant of my solution, assuming Python 3 of course

                – Chris_Rands
                2 hours ago














              • 1





                This is a reasonable (although mostly cosmetic) variant of my solution, assuming Python 3 of course

                – Chris_Rands
                2 hours ago








              1




              1





              This is a reasonable (although mostly cosmetic) variant of my solution, assuming Python 3 of course

              – Chris_Rands
              2 hours ago





              This is a reasonable (although mostly cosmetic) variant of my solution, assuming Python 3 of course

              – Chris_Rands
              2 hours ago













              4














              This is quite an odd output to want but it is of course possible. I suggest using collections.Counter(), no doubt others will make different suggestions and a timeit style comparison would reveal the fastest of course for particular data sets:



              >>> from collections import Counter
              
>>> l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]
              
>>> [list(k) + [v] for k, v in Counter(map(tuple,l)).items()]
              
[[1, 3, 2, 2], [1, 3, 5, 1]]


              Note to preserve the insertion order prior to CPython 3.6 / Python 3.7, use the OrderedCounter recipe.






              share|improve this answer




























                4














                This is quite an odd output to want but it is of course possible. I suggest using collections.Counter(), no doubt others will make different suggestions and a timeit style comparison would reveal the fastest of course for particular data sets:



                >>> from collections import Counter
                
>>> l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]
                
>>> [list(k) + [v] for k, v in Counter(map(tuple,l)).items()]
                
[[1, 3, 2, 2], [1, 3, 5, 1]]


                Note to preserve the insertion order prior to CPython 3.6 / Python 3.7, use the OrderedCounter recipe.






                share|improve this answer


























                  4












                  4








                  4







                  This is quite an odd output to want but it is of course possible. I suggest using collections.Counter(), no doubt others will make different suggestions and a timeit style comparison would reveal the fastest of course for particular data sets:



                  >>> from collections import Counter
                  
>>> l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]
                  
>>> [list(k) + [v] for k, v in Counter(map(tuple,l)).items()]
                  
[[1, 3, 2, 2], [1, 3, 5, 1]]


                  Note to preserve the insertion order prior to CPython 3.6 / Python 3.7, use the OrderedCounter recipe.






                  share|improve this answer













                  This is quite an odd output to want but it is of course possible. I suggest using collections.Counter(), no doubt others will make different suggestions and a timeit style comparison would reveal the fastest of course for particular data sets:



                  >>> from collections import Counter
                  
>>> l = [[1, 3, 2], [1, 3, 2] ,[1, 3, 5]]
                  
>>> [list(k) + [v] for k, v in Counter(map(tuple,l)).items()]
                  
[[1, 3, 2, 2], [1, 3, 5, 1]]


                  Note to preserve the insertion order prior to CPython 3.6 / Python 3.7, use the OrderedCounter recipe.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 2 hours ago









                  Chris_RandsChris_Rands

                  15.9k53869




                  15.9k53869























                      1














                      If numpy is an option, you could use np.unique setting axis to 0 and return_counts to True, and concatenate the unique rows and counts using np.vstack:



                      l = np.array([[1, 3, 2], [1, 3, 2] ,[1, 3, 5]])
                      
x, c = np.unique(l, axis=0, return_counts=True)
                      
np.vstack([x.T,c]).T
                      

                      
array([[1, 3, 2, 2],
                      
       [1, 3, 5, 1]])





                      share|improve this answer




























                        1














                        If numpy is an option, you could use np.unique setting axis to 0 and return_counts to True, and concatenate the unique rows and counts using np.vstack:



                        l = np.array([[1, 3, 2], [1, 3, 2] ,[1, 3, 5]])
                        
x, c = np.unique(l, axis=0, return_counts=True)
                        
np.vstack([x.T,c]).T
                        

                        
array([[1, 3, 2, 2],
                        
       [1, 3, 5, 1]])





                        share|improve this answer


























                          1












                          1








                          1







                          If numpy is an option, you could use np.unique setting axis to 0 and return_counts to True, and concatenate the unique rows and counts using np.vstack:



                          l = np.array([[1, 3, 2], [1, 3, 2] ,[1, 3, 5]])
                          
x, c = np.unique(l, axis=0, return_counts=True)
                          
np.vstack([x.T,c]).T
                          

                          
array([[1, 3, 2, 2],
                          
       [1, 3, 5, 1]])





                          share|improve this answer













                          If numpy is an option, you could use np.unique setting axis to 0 and return_counts to True, and concatenate the unique rows and counts using np.vstack:



                          l = np.array([[1, 3, 2], [1, 3, 2] ,[1, 3, 5]])
                          
x, c = np.unique(l, axis=0, return_counts=True)
                          
np.vstack([x.T,c]).T
                          

                          
array([[1, 3, 2, 2],
                          
       [1, 3, 5, 1]])






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 2 hours ago









                          yatuyatu

                          6,6911826




                          6,6911826






























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