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How to use Pandas to get the count of every combination inclusive

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How to use Pandas to get the count of every combination inclusive


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8















I am trying to figure out what combination of clothing customers are buying together. I can figure out the exact combination, but the problem I can't figure out is the count that includes the combination + others.



For example, I have:



Cust_num  Item    Rev
Cust1 Shirt1 $40
Cust1 Shirt2 $40
Cust1 Shorts1 $40
Cust2 Shirt1 $40
Cust2 Shorts1 $40


This should result in:



Combo                  Count
Shirt1,Shirt2,Shorts1 1
Shirt1,Shorts1 2


The best I can do is unique combinations:



Combo                 Count
Shirt1,Shirt2,Shorts1 1
Shirt1,Shorts1 1


I tried:



df = df.pivot(index='Cust_num',columns='Item').sum()
df[df.notnull()] = "x"
df = df.loc[:,"Shirt1":].replace("x", pd.Series(df.columns, df.columns))
col = df.stack().groupby(level=0).apply(','.join)
df2 = pd.DataFrame(col)
df2.groupby([0]).size().reset_index(name='counts')


But that is just the unique counts.










share|improve this question







New contributor




Taylor Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2





    I feel like this is one sort of problem pandas would not be suitable for.

    – coldspeed
    1 hour ago


















8















I am trying to figure out what combination of clothing customers are buying together. I can figure out the exact combination, but the problem I can't figure out is the count that includes the combination + others.



For example, I have:



Cust_num  Item    Rev
Cust1 Shirt1 $40
Cust1 Shirt2 $40
Cust1 Shorts1 $40
Cust2 Shirt1 $40
Cust2 Shorts1 $40


This should result in:



Combo                  Count
Shirt1,Shirt2,Shorts1 1
Shirt1,Shorts1 2


The best I can do is unique combinations:



Combo                 Count
Shirt1,Shirt2,Shorts1 1
Shirt1,Shorts1 1


I tried:



df = df.pivot(index='Cust_num',columns='Item').sum()
df[df.notnull()] = "x"
df = df.loc[:,"Shirt1":].replace("x", pd.Series(df.columns, df.columns))
col = df.stack().groupby(level=0).apply(','.join)
df2 = pd.DataFrame(col)
df2.groupby([0]).size().reset_index(name='counts')


But that is just the unique counts.










share|improve this question







New contributor




Taylor Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2





    I feel like this is one sort of problem pandas would not be suitable for.

    – coldspeed
    1 hour ago














8












8








8








I am trying to figure out what combination of clothing customers are buying together. I can figure out the exact combination, but the problem I can't figure out is the count that includes the combination + others.



For example, I have:



Cust_num  Item    Rev
Cust1 Shirt1 $40
Cust1 Shirt2 $40
Cust1 Shorts1 $40
Cust2 Shirt1 $40
Cust2 Shorts1 $40


This should result in:



Combo                  Count
Shirt1,Shirt2,Shorts1 1
Shirt1,Shorts1 2


The best I can do is unique combinations:



Combo                 Count
Shirt1,Shirt2,Shorts1 1
Shirt1,Shorts1 1


I tried:



df = df.pivot(index='Cust_num',columns='Item').sum()
df[df.notnull()] = "x"
df = df.loc[:,"Shirt1":].replace("x", pd.Series(df.columns, df.columns))
col = df.stack().groupby(level=0).apply(','.join)
df2 = pd.DataFrame(col)
df2.groupby([0]).size().reset_index(name='counts')


But that is just the unique counts.










share|improve this question







New contributor




Taylor Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












I am trying to figure out what combination of clothing customers are buying together. I can figure out the exact combination, but the problem I can't figure out is the count that includes the combination + others.



For example, I have:



Cust_num  Item    Rev
Cust1 Shirt1 $40
Cust1 Shirt2 $40
Cust1 Shorts1 $40
Cust2 Shirt1 $40
Cust2 Shorts1 $40


This should result in:



Combo                  Count
Shirt1,Shirt2,Shorts1 1
Shirt1,Shorts1 2


The best I can do is unique combinations:



Combo                 Count
Shirt1,Shirt2,Shorts1 1
Shirt1,Shorts1 1


I tried:



df = df.pivot(index='Cust_num',columns='Item').sum()
df[df.notnull()] = "x"
df = df.loc[:,"Shirt1":].replace("x", pd.Series(df.columns, df.columns))
col = df.stack().groupby(level=0).apply(','.join)
df2 = pd.DataFrame(col)
df2.groupby([0]).size().reset_index(name='counts')


But that is just the unique counts.







python pandas






share|improve this question







New contributor




Taylor Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




Taylor Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






New contributor




Taylor Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 hours ago









Taylor SmithTaylor Smith

442




442




New contributor




Taylor Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Taylor Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Taylor Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 2





    I feel like this is one sort of problem pandas would not be suitable for.

    – coldspeed
    1 hour ago














  • 2





    I feel like this is one sort of problem pandas would not be suitable for.

    – coldspeed
    1 hour ago








2




2





I feel like this is one sort of problem pandas would not be suitable for.

– coldspeed
1 hour ago





I feel like this is one sort of problem pandas would not be suitable for.

– coldspeed
1 hour ago












4 Answers
4






active

oldest

votes


















4














Using pandas.DataFrame.groupby:



grouped_item = df.groupby('Cust_num')['Item']
subsets = grouped_item.apply(lambda x: set(x)).tolist()
Count = [sum(s2.issubset(s1) for s1 in subsets) for s2 in subsets]
combo = grouped_item.apply(lambda x:','.join(x))
combo = combo.reset_index()
combo['Count']=Count


Output:



  Cust_num                   Item  Count
0 Cust1 Shirt1,Shirt2,Shorts1 1
1 Cust2 Shirt1,Shorts1 2





share|improve this answer































    2














    Late answer, but you can use:



    df = df.groupby(['Cust_num'], as_index=False).agg(','.join).drop(columns=['Rev']).set_index(['Item']).rename_axis("combo").rename(columns={"Cust_num": "Count"})
    df['Count'] = df['Count'].str.replace(r'Cust','')




    combo                   Count                 
    Shirt1,Shirt2,Shorts1 1
    Shirt1,Shorts1 2





    share|improve this answer

































      1














      I think you need to create a combination of items first.



      How to get all possible combinations of a list’s elements?



      I used the function from Dan H's answer.



      from itertools import chain, combinations
      def all_subsets(ss):
      return chain(*map(lambda x: combinations(ss, x), range(0, len(ss)+1)))


      Then get the unique items.



      uq_items = df.Item.unique()

      list(all_subsets(uq_items))

      [(),
      ('Shirt1',),
      ('Shirt2',),
      ('Shorts1',),
      ('Shirt1', 'Shirt2'),
      ('Shirt1', 'Shorts1'),
      ('Shirt2', 'Shorts1'),
      ('Shirt1', 'Shirt2', 'Shorts1')]


      And use groupby each customer to get their items combination.



      ls = []

      for _, d in df.groupby('Cust_num', group_keys=False):
      # Get all possible subset of items
      pi = np.array(list(all_subsets(d.Item)))

      # Fliter only > 1
      ls.append(pi[[len(l) > 1 for l in pi]])


      Then convert to Series and use value_counts().



      pd.Series(np.concatenate(ls)).value_counts()

      (Shirt1, Shorts1) 2
      (Shirt2, Shorts1) 1
      (Shirt1, Shirt2, Shorts1) 1
      (Shirt1, Shirt2) 1





      share|improve this answer































        -1














        My version which I believe is easier to understand



        new_df = df.groupby("Cust_num").agg({lambda x: ''.join(x.unique())})

        new_df ['count'] = range(1, len(new_df ) + 1)


        Output:



                                    Item      Rev count
        <lambda> <lambda>
        Cust_num
        Cust1 Shirt1 Shirt2 Shorts1 $40 1
        Cust2 Shirt1 Shorts1 $40 2


        Since you do not need the Rev column, you can drop it:



        new_df = new_df = new_df.drop(columns=["Rev"]).reset_index()

        new_df


        Output:



          Cust_num                    Item count
        <lambda>
        0 Cust1 Shirt1 Shirt2 Shorts1 1
        1 Cust2 Shirt1 Shorts1 2





        share|improve this answer


























        • How is the count in your answer the count of inclusive combination of df['Item']? Making new column with range is not an answer.

          – Chris
          41 mins ago














        Your Answer






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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4














        Using pandas.DataFrame.groupby:



        grouped_item = df.groupby('Cust_num')['Item']
        subsets = grouped_item.apply(lambda x: set(x)).tolist()
        Count = [sum(s2.issubset(s1) for s1 in subsets) for s2 in subsets]
        combo = grouped_item.apply(lambda x:','.join(x))
        combo = combo.reset_index()
        combo['Count']=Count


        Output:



          Cust_num                   Item  Count
        0 Cust1 Shirt1,Shirt2,Shorts1 1
        1 Cust2 Shirt1,Shorts1 2





        share|improve this answer




























          4














          Using pandas.DataFrame.groupby:



          grouped_item = df.groupby('Cust_num')['Item']
          subsets = grouped_item.apply(lambda x: set(x)).tolist()
          Count = [sum(s2.issubset(s1) for s1 in subsets) for s2 in subsets]
          combo = grouped_item.apply(lambda x:','.join(x))
          combo = combo.reset_index()
          combo['Count']=Count


          Output:



            Cust_num                   Item  Count
          0 Cust1 Shirt1,Shirt2,Shorts1 1
          1 Cust2 Shirt1,Shorts1 2





          share|improve this answer


























            4












            4








            4







            Using pandas.DataFrame.groupby:



            grouped_item = df.groupby('Cust_num')['Item']
            subsets = grouped_item.apply(lambda x: set(x)).tolist()
            Count = [sum(s2.issubset(s1) for s1 in subsets) for s2 in subsets]
            combo = grouped_item.apply(lambda x:','.join(x))
            combo = combo.reset_index()
            combo['Count']=Count


            Output:



              Cust_num                   Item  Count
            0 Cust1 Shirt1,Shirt2,Shorts1 1
            1 Cust2 Shirt1,Shorts1 2





            share|improve this answer













            Using pandas.DataFrame.groupby:



            grouped_item = df.groupby('Cust_num')['Item']
            subsets = grouped_item.apply(lambda x: set(x)).tolist()
            Count = [sum(s2.issubset(s1) for s1 in subsets) for s2 in subsets]
            combo = grouped_item.apply(lambda x:','.join(x))
            combo = combo.reset_index()
            combo['Count']=Count


            Output:



              Cust_num                   Item  Count
            0 Cust1 Shirt1,Shirt2,Shorts1 1
            1 Cust2 Shirt1,Shorts1 2






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 1 hour ago









            ChrisChris

            3,710422




            3,710422

























                2














                Late answer, but you can use:



                df = df.groupby(['Cust_num'], as_index=False).agg(','.join).drop(columns=['Rev']).set_index(['Item']).rename_axis("combo").rename(columns={"Cust_num": "Count"})
                df['Count'] = df['Count'].str.replace(r'Cust','')




                combo                   Count                 
                Shirt1,Shirt2,Shorts1 1
                Shirt1,Shorts1 2





                share|improve this answer






























                  2














                  Late answer, but you can use:



                  df = df.groupby(['Cust_num'], as_index=False).agg(','.join).drop(columns=['Rev']).set_index(['Item']).rename_axis("combo").rename(columns={"Cust_num": "Count"})
                  df['Count'] = df['Count'].str.replace(r'Cust','')




                  combo                   Count                 
                  Shirt1,Shirt2,Shorts1 1
                  Shirt1,Shorts1 2





                  share|improve this answer




























                    2












                    2








                    2







                    Late answer, but you can use:



                    df = df.groupby(['Cust_num'], as_index=False).agg(','.join).drop(columns=['Rev']).set_index(['Item']).rename_axis("combo").rename(columns={"Cust_num": "Count"})
                    df['Count'] = df['Count'].str.replace(r'Cust','')




                    combo                   Count                 
                    Shirt1,Shirt2,Shorts1 1
                    Shirt1,Shorts1 2





                    share|improve this answer















                    Late answer, but you can use:



                    df = df.groupby(['Cust_num'], as_index=False).agg(','.join).drop(columns=['Rev']).set_index(['Item']).rename_axis("combo").rename(columns={"Cust_num": "Count"})
                    df['Count'] = df['Count'].str.replace(r'Cust','')




                    combo                   Count                 
                    Shirt1,Shirt2,Shorts1 1
                    Shirt1,Shorts1 2






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 39 mins ago

























                    answered 1 hour ago









                    Pedro LobitoPedro Lobito

                    50.5k16138172




                    50.5k16138172























                        1














                        I think you need to create a combination of items first.



                        How to get all possible combinations of a list’s elements?



                        I used the function from Dan H's answer.



                        from itertools import chain, combinations
                        def all_subsets(ss):
                        return chain(*map(lambda x: combinations(ss, x), range(0, len(ss)+1)))


                        Then get the unique items.



                        uq_items = df.Item.unique()

                        list(all_subsets(uq_items))

                        [(),
                        ('Shirt1',),
                        ('Shirt2',),
                        ('Shorts1',),
                        ('Shirt1', 'Shirt2'),
                        ('Shirt1', 'Shorts1'),
                        ('Shirt2', 'Shorts1'),
                        ('Shirt1', 'Shirt2', 'Shorts1')]


                        And use groupby each customer to get their items combination.



                        ls = []

                        for _, d in df.groupby('Cust_num', group_keys=False):
                        # Get all possible subset of items
                        pi = np.array(list(all_subsets(d.Item)))

                        # Fliter only > 1
                        ls.append(pi[[len(l) > 1 for l in pi]])


                        Then convert to Series and use value_counts().



                        pd.Series(np.concatenate(ls)).value_counts()

                        (Shirt1, Shorts1) 2
                        (Shirt2, Shorts1) 1
                        (Shirt1, Shirt2, Shorts1) 1
                        (Shirt1, Shirt2) 1





                        share|improve this answer




























                          1














                          I think you need to create a combination of items first.



                          How to get all possible combinations of a list’s elements?



                          I used the function from Dan H's answer.



                          from itertools import chain, combinations
                          def all_subsets(ss):
                          return chain(*map(lambda x: combinations(ss, x), range(0, len(ss)+1)))


                          Then get the unique items.



                          uq_items = df.Item.unique()

                          list(all_subsets(uq_items))

                          [(),
                          ('Shirt1',),
                          ('Shirt2',),
                          ('Shorts1',),
                          ('Shirt1', 'Shirt2'),
                          ('Shirt1', 'Shorts1'),
                          ('Shirt2', 'Shorts1'),
                          ('Shirt1', 'Shirt2', 'Shorts1')]


                          And use groupby each customer to get their items combination.



                          ls = []

                          for _, d in df.groupby('Cust_num', group_keys=False):
                          # Get all possible subset of items
                          pi = np.array(list(all_subsets(d.Item)))

                          # Fliter only > 1
                          ls.append(pi[[len(l) > 1 for l in pi]])


                          Then convert to Series and use value_counts().



                          pd.Series(np.concatenate(ls)).value_counts()

                          (Shirt1, Shorts1) 2
                          (Shirt2, Shorts1) 1
                          (Shirt1, Shirt2, Shorts1) 1
                          (Shirt1, Shirt2) 1





                          share|improve this answer


























                            1












                            1








                            1







                            I think you need to create a combination of items first.



                            How to get all possible combinations of a list’s elements?



                            I used the function from Dan H's answer.



                            from itertools import chain, combinations
                            def all_subsets(ss):
                            return chain(*map(lambda x: combinations(ss, x), range(0, len(ss)+1)))


                            Then get the unique items.



                            uq_items = df.Item.unique()

                            list(all_subsets(uq_items))

                            [(),
                            ('Shirt1',),
                            ('Shirt2',),
                            ('Shorts1',),
                            ('Shirt1', 'Shirt2'),
                            ('Shirt1', 'Shorts1'),
                            ('Shirt2', 'Shorts1'),
                            ('Shirt1', 'Shirt2', 'Shorts1')]


                            And use groupby each customer to get their items combination.



                            ls = []

                            for _, d in df.groupby('Cust_num', group_keys=False):
                            # Get all possible subset of items
                            pi = np.array(list(all_subsets(d.Item)))

                            # Fliter only > 1
                            ls.append(pi[[len(l) > 1 for l in pi]])


                            Then convert to Series and use value_counts().



                            pd.Series(np.concatenate(ls)).value_counts()

                            (Shirt1, Shorts1) 2
                            (Shirt2, Shorts1) 1
                            (Shirt1, Shirt2, Shorts1) 1
                            (Shirt1, Shirt2) 1





                            share|improve this answer













                            I think you need to create a combination of items first.



                            How to get all possible combinations of a list’s elements?



                            I used the function from Dan H's answer.



                            from itertools import chain, combinations
                            def all_subsets(ss):
                            return chain(*map(lambda x: combinations(ss, x), range(0, len(ss)+1)))


                            Then get the unique items.



                            uq_items = df.Item.unique()

                            list(all_subsets(uq_items))

                            [(),
                            ('Shirt1',),
                            ('Shirt2',),
                            ('Shorts1',),
                            ('Shirt1', 'Shirt2'),
                            ('Shirt1', 'Shorts1'),
                            ('Shirt2', 'Shorts1'),
                            ('Shirt1', 'Shirt2', 'Shorts1')]


                            And use groupby each customer to get their items combination.



                            ls = []

                            for _, d in df.groupby('Cust_num', group_keys=False):
                            # Get all possible subset of items
                            pi = np.array(list(all_subsets(d.Item)))

                            # Fliter only > 1
                            ls.append(pi[[len(l) > 1 for l in pi]])


                            Then convert to Series and use value_counts().



                            pd.Series(np.concatenate(ls)).value_counts()

                            (Shirt1, Shorts1) 2
                            (Shirt2, Shorts1) 1
                            (Shirt1, Shirt2, Shorts1) 1
                            (Shirt1, Shirt2) 1






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 1 hour ago









                            ResidentSleeperResidentSleeper

                            36210




                            36210























                                -1














                                My version which I believe is easier to understand



                                new_df = df.groupby("Cust_num").agg({lambda x: ''.join(x.unique())})

                                new_df ['count'] = range(1, len(new_df ) + 1)


                                Output:



                                                            Item      Rev count
                                <lambda> <lambda>
                                Cust_num
                                Cust1 Shirt1 Shirt2 Shorts1 $40 1
                                Cust2 Shirt1 Shorts1 $40 2


                                Since you do not need the Rev column, you can drop it:



                                new_df = new_df = new_df.drop(columns=["Rev"]).reset_index()

                                new_df


                                Output:



                                  Cust_num                    Item count
                                <lambda>
                                0 Cust1 Shirt1 Shirt2 Shorts1 1
                                1 Cust2 Shirt1 Shorts1 2





                                share|improve this answer


























                                • How is the count in your answer the count of inclusive combination of df['Item']? Making new column with range is not an answer.

                                  – Chris
                                  41 mins ago


















                                -1














                                My version which I believe is easier to understand



                                new_df = df.groupby("Cust_num").agg({lambda x: ''.join(x.unique())})

                                new_df ['count'] = range(1, len(new_df ) + 1)


                                Output:



                                                            Item      Rev count
                                <lambda> <lambda>
                                Cust_num
                                Cust1 Shirt1 Shirt2 Shorts1 $40 1
                                Cust2 Shirt1 Shorts1 $40 2


                                Since you do not need the Rev column, you can drop it:



                                new_df = new_df = new_df.drop(columns=["Rev"]).reset_index()

                                new_df


                                Output:



                                  Cust_num                    Item count
                                <lambda>
                                0 Cust1 Shirt1 Shirt2 Shorts1 1
                                1 Cust2 Shirt1 Shorts1 2





                                share|improve this answer


























                                • How is the count in your answer the count of inclusive combination of df['Item']? Making new column with range is not an answer.

                                  – Chris
                                  41 mins ago
















                                -1












                                -1








                                -1







                                My version which I believe is easier to understand



                                new_df = df.groupby("Cust_num").agg({lambda x: ''.join(x.unique())})

                                new_df ['count'] = range(1, len(new_df ) + 1)


                                Output:



                                                            Item      Rev count
                                <lambda> <lambda>
                                Cust_num
                                Cust1 Shirt1 Shirt2 Shorts1 $40 1
                                Cust2 Shirt1 Shorts1 $40 2


                                Since you do not need the Rev column, you can drop it:



                                new_df = new_df = new_df.drop(columns=["Rev"]).reset_index()

                                new_df


                                Output:



                                  Cust_num                    Item count
                                <lambda>
                                0 Cust1 Shirt1 Shirt2 Shorts1 1
                                1 Cust2 Shirt1 Shorts1 2





                                share|improve this answer















                                My version which I believe is easier to understand



                                new_df = df.groupby("Cust_num").agg({lambda x: ''.join(x.unique())})

                                new_df ['count'] = range(1, len(new_df ) + 1)


                                Output:



                                                            Item      Rev count
                                <lambda> <lambda>
                                Cust_num
                                Cust1 Shirt1 Shirt2 Shorts1 $40 1
                                Cust2 Shirt1 Shorts1 $40 2


                                Since you do not need the Rev column, you can drop it:



                                new_df = new_df = new_df.drop(columns=["Rev"]).reset_index()

                                new_df


                                Output:



                                  Cust_num                    Item count
                                <lambda>
                                0 Cust1 Shirt1 Shirt2 Shorts1 1
                                1 Cust2 Shirt1 Shorts1 2






                                share|improve this answer














                                share|improve this answer



                                share|improve this answer








                                edited 38 mins ago

























                                answered 48 mins ago









                                Lee MtotiLee Mtoti

                                13110




                                13110













                                • How is the count in your answer the count of inclusive combination of df['Item']? Making new column with range is not an answer.

                                  – Chris
                                  41 mins ago





















                                • How is the count in your answer the count of inclusive combination of df['Item']? Making new column with range is not an answer.

                                  – Chris
                                  41 mins ago



















                                How is the count in your answer the count of inclusive combination of df['Item']? Making new column with range is not an answer.

                                – Chris
                                41 mins ago







                                How is the count in your answer the count of inclusive combination of df['Item']? Making new column with range is not an answer.

                                – Chris
                                41 mins ago












                                Taylor Smith is a new contributor. Be nice, and check out our Code of Conduct.










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                                Taylor Smith is a new contributor. Be nice, and check out our Code of Conduct.













                                Taylor Smith is a new contributor. Be nice, and check out our Code of Conduct.












                                Taylor Smith is a new contributor. Be nice, and check out our Code of Conduct.
















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