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Python Pandas Expand a Column of List of Lists to Two New Column


Pandas split column of lists into multiple columnsFinding the index of an item given a list containing it in PythonConvert two lists into a dictionary in PythonPython join: why is it string.join(list) instead of list.join(string)?Getting the last element of a list in PythonHow do I get the number of elements in a list in Python?How do I concatenate two lists in Python?Renaming columns in pandasAdding new column to existing DataFrame in Python pandasDelete column from pandas DataFrame by column nameSelect rows from a DataFrame based on values in a column in pandas






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6















I have a DF which looks like this.



name id apps
john 1 [[app1, v1], [app2, v2], [app3,v3]]
smith 2 [[app1, v1], [app4, v4]]


I want to expand the apps column such that it looks like this.



name id app_name app_version
john 1 app1 v1
john 1 app2 v2
john 1 app3 v3
smith 2 app1 v1
smith 2 app4 v4


Any help is appreciated










share|improve this question




























    6















    I have a DF which looks like this.



    name id apps
    john 1 [[app1, v1], [app2, v2], [app3,v3]]
    smith 2 [[app1, v1], [app4, v4]]


    I want to expand the apps column such that it looks like this.



    name id app_name app_version
    john 1 app1 v1
    john 1 app2 v2
    john 1 app3 v3
    smith 2 app1 v1
    smith 2 app4 v4


    Any help is appreciated










    share|improve this question
























      6












      6








      6








      I have a DF which looks like this.



      name id apps
      john 1 [[app1, v1], [app2, v2], [app3,v3]]
      smith 2 [[app1, v1], [app4, v4]]


      I want to expand the apps column such that it looks like this.



      name id app_name app_version
      john 1 app1 v1
      john 1 app2 v2
      john 1 app3 v3
      smith 2 app1 v1
      smith 2 app4 v4


      Any help is appreciated










      share|improve this question














      I have a DF which looks like this.



      name id apps
      john 1 [[app1, v1], [app2, v2], [app3,v3]]
      smith 2 [[app1, v1], [app4, v4]]


      I want to expand the apps column such that it looks like this.



      name id app_name app_version
      john 1 app1 v1
      john 1 app2 v2
      john 1 app3 v3
      smith 2 app1 v1
      smith 2 app4 v4


      Any help is appreciated







      python pandas list






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 3 hours ago









      ImsaImsa

      392424




      392424






















          3 Answers
          3






          active

          oldest

          votes


















          2














          You can .apply(pd.Series) twice to get what you need as an intermediate step, then merge back to the original dataframe.



          import pandas as pd

          df = pd.DataFrame(
          'name': ['john', 'smith'],
          'id': [1, 2],
          'apps': [[['app1', 'v1'], ['app2', 'v2'], ['app3','v3']],
          [['app1', 'v1'], ['app4', 'v4']]]
          )

          dftmp = df.apps.apply(pd.Series).T.melt().dropna()
          dfapp = (dftmp.value
          .apply(pd.Series)
          .set_index(dftmp.variable)
          .rename(columns=0:'app_name', 1:'app_version')
          )

          df[['name', 'id']].merge(dfapp, left_index=True, right_index=True)
          # returns:
          name id app_name app_version
          0 john 1 app1 v1
          0 john 1 app2 v2
          0 john 1 app3 v3
          1 smith 2 app1 v1
          1 smith 2 app4 v4





          share|improve this answer























          • Instead of .apply(pd.Series) (which is awfully slow), use pd.DataFrame(df.apps.tolist())

            – RafaelC
            1 hour ago











          • Either way you are pulling it out of the C-backed API into Python. .apply hides a for loop, while tolist pushes the encapsulated object back to Python. I have not done any tests to see which is faster.

            – James
            1 hour ago











          • I have, that's why I commented.

            – RafaelC
            47 mins ago











          • Can also refer here for details

            – RafaelC
            46 mins ago






          • 1





            Wow, thanks. That is like 30% faster.

            – James
            44 mins ago


















          2














          You can always have a brute force solution. Something like:



          name, id, app_name, app_version = [], [], [], []
          for i in range(len(df)):
          for v in df.loc[i,'apps']:
          app_name.append(v[0])
          app_version.append(v[1])
          name.append(df.loc[i, 'name'])
          id.append(df.loc[i, 'id'])
          df = pd.DataFrame('name': name, 'id': id, 'app_name': app_name, 'app_version': app_version)


          will do the work.



          Note that I assumed df['apps'] is lists of strings if df['apps'] is strings then you need: eval(df.loc[i,'apps']) instead of df.loc[i,'apps']






          share|improve this answer

























          • Even though this works, it is probably unfeasible for large data frames. In pandas, one for loop is already bad enough, so imagine two nested for loops ;} Always try to avoid direct iteration !

            – RafaelC
            1 hour ago


















          1














          My suggestion (there may be easier ways) is using DataFrame.apply alongside pd.concat:



          import pandas as pd

          def expand_row(row):
          apps = row.apps
          apps_names = [app[0] for app in apps]
          apps_versions = [app[1] for app in apps]
          return pd.DataFrame(
          'name': row.name,
          'id': row.id,
          'app_name': apps_names,
          'app_version': apps_versions
          )

          df = pd.DataFrame(
          'name': ['john', 'smith'],
          'id': [1, 2],
          'apps': [[['app1', 'v1'], ['app2', 'v2'], ['app3','v3']],
          [['app1', 'v1'], ['app4', 'v4']]]
          )
          temp_dfs = df.apply(expand_row, axis=1).tolist()
          expanded = pd.concat(temp_dfs)

          print(expanded)

          # name id app_name app_version
          # 0 0 1 app1 v1
          # 1 0 1 app2 v2
          # 2 0 1 app3 v3
          # 0 1 2 app1 v1
          # 1 1 2 app4 v4





          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









            2














            You can .apply(pd.Series) twice to get what you need as an intermediate step, then merge back to the original dataframe.



            import pandas as pd

            df = pd.DataFrame(
            'name': ['john', 'smith'],
            'id': [1, 2],
            'apps': [[['app1', 'v1'], ['app2', 'v2'], ['app3','v3']],
            [['app1', 'v1'], ['app4', 'v4']]]
            )

            dftmp = df.apps.apply(pd.Series).T.melt().dropna()
            dfapp = (dftmp.value
            .apply(pd.Series)
            .set_index(dftmp.variable)
            .rename(columns=0:'app_name', 1:'app_version')
            )

            df[['name', 'id']].merge(dfapp, left_index=True, right_index=True)
            # returns:
            name id app_name app_version
            0 john 1 app1 v1
            0 john 1 app2 v2
            0 john 1 app3 v3
            1 smith 2 app1 v1
            1 smith 2 app4 v4





            share|improve this answer























            • Instead of .apply(pd.Series) (which is awfully slow), use pd.DataFrame(df.apps.tolist())

              – RafaelC
              1 hour ago











            • Either way you are pulling it out of the C-backed API into Python. .apply hides a for loop, while tolist pushes the encapsulated object back to Python. I have not done any tests to see which is faster.

              – James
              1 hour ago











            • I have, that's why I commented.

              – RafaelC
              47 mins ago











            • Can also refer here for details

              – RafaelC
              46 mins ago






            • 1





              Wow, thanks. That is like 30% faster.

              – James
              44 mins ago















            2














            You can .apply(pd.Series) twice to get what you need as an intermediate step, then merge back to the original dataframe.



            import pandas as pd

            df = pd.DataFrame(
            'name': ['john', 'smith'],
            'id': [1, 2],
            'apps': [[['app1', 'v1'], ['app2', 'v2'], ['app3','v3']],
            [['app1', 'v1'], ['app4', 'v4']]]
            )

            dftmp = df.apps.apply(pd.Series).T.melt().dropna()
            dfapp = (dftmp.value
            .apply(pd.Series)
            .set_index(dftmp.variable)
            .rename(columns=0:'app_name', 1:'app_version')
            )

            df[['name', 'id']].merge(dfapp, left_index=True, right_index=True)
            # returns:
            name id app_name app_version
            0 john 1 app1 v1
            0 john 1 app2 v2
            0 john 1 app3 v3
            1 smith 2 app1 v1
            1 smith 2 app4 v4





            share|improve this answer























            • Instead of .apply(pd.Series) (which is awfully slow), use pd.DataFrame(df.apps.tolist())

              – RafaelC
              1 hour ago











            • Either way you are pulling it out of the C-backed API into Python. .apply hides a for loop, while tolist pushes the encapsulated object back to Python. I have not done any tests to see which is faster.

              – James
              1 hour ago











            • I have, that's why I commented.

              – RafaelC
              47 mins ago











            • Can also refer here for details

              – RafaelC
              46 mins ago






            • 1





              Wow, thanks. That is like 30% faster.

              – James
              44 mins ago













            2












            2








            2







            You can .apply(pd.Series) twice to get what you need as an intermediate step, then merge back to the original dataframe.



            import pandas as pd

            df = pd.DataFrame(
            'name': ['john', 'smith'],
            'id': [1, 2],
            'apps': [[['app1', 'v1'], ['app2', 'v2'], ['app3','v3']],
            [['app1', 'v1'], ['app4', 'v4']]]
            )

            dftmp = df.apps.apply(pd.Series).T.melt().dropna()
            dfapp = (dftmp.value
            .apply(pd.Series)
            .set_index(dftmp.variable)
            .rename(columns=0:'app_name', 1:'app_version')
            )

            df[['name', 'id']].merge(dfapp, left_index=True, right_index=True)
            # returns:
            name id app_name app_version
            0 john 1 app1 v1
            0 john 1 app2 v2
            0 john 1 app3 v3
            1 smith 2 app1 v1
            1 smith 2 app4 v4





            share|improve this answer













            You can .apply(pd.Series) twice to get what you need as an intermediate step, then merge back to the original dataframe.



            import pandas as pd

            df = pd.DataFrame(
            'name': ['john', 'smith'],
            'id': [1, 2],
            'apps': [[['app1', 'v1'], ['app2', 'v2'], ['app3','v3']],
            [['app1', 'v1'], ['app4', 'v4']]]
            )

            dftmp = df.apps.apply(pd.Series).T.melt().dropna()
            dfapp = (dftmp.value
            .apply(pd.Series)
            .set_index(dftmp.variable)
            .rename(columns=0:'app_name', 1:'app_version')
            )

            df[['name', 'id']].merge(dfapp, left_index=True, right_index=True)
            # returns:
            name id app_name app_version
            0 john 1 app1 v1
            0 john 1 app2 v2
            0 john 1 app3 v3
            1 smith 2 app1 v1
            1 smith 2 app4 v4






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 2 hours ago









            JamesJames

            14.4k11733




            14.4k11733












            • Instead of .apply(pd.Series) (which is awfully slow), use pd.DataFrame(df.apps.tolist())

              – RafaelC
              1 hour ago











            • Either way you are pulling it out of the C-backed API into Python. .apply hides a for loop, while tolist pushes the encapsulated object back to Python. I have not done any tests to see which is faster.

              – James
              1 hour ago











            • I have, that's why I commented.

              – RafaelC
              47 mins ago











            • Can also refer here for details

              – RafaelC
              46 mins ago






            • 1





              Wow, thanks. That is like 30% faster.

              – James
              44 mins ago

















            • Instead of .apply(pd.Series) (which is awfully slow), use pd.DataFrame(df.apps.tolist())

              – RafaelC
              1 hour ago











            • Either way you are pulling it out of the C-backed API into Python. .apply hides a for loop, while tolist pushes the encapsulated object back to Python. I have not done any tests to see which is faster.

              – James
              1 hour ago











            • I have, that's why I commented.

              – RafaelC
              47 mins ago











            • Can also refer here for details

              – RafaelC
              46 mins ago






            • 1





              Wow, thanks. That is like 30% faster.

              – James
              44 mins ago
















            Instead of .apply(pd.Series) (which is awfully slow), use pd.DataFrame(df.apps.tolist())

            – RafaelC
            1 hour ago





            Instead of .apply(pd.Series) (which is awfully slow), use pd.DataFrame(df.apps.tolist())

            – RafaelC
            1 hour ago













            Either way you are pulling it out of the C-backed API into Python. .apply hides a for loop, while tolist pushes the encapsulated object back to Python. I have not done any tests to see which is faster.

            – James
            1 hour ago





            Either way you are pulling it out of the C-backed API into Python. .apply hides a for loop, while tolist pushes the encapsulated object back to Python. I have not done any tests to see which is faster.

            – James
            1 hour ago













            I have, that's why I commented.

            – RafaelC
            47 mins ago





            I have, that's why I commented.

            – RafaelC
            47 mins ago













            Can also refer here for details

            – RafaelC
            46 mins ago





            Can also refer here for details

            – RafaelC
            46 mins ago




            1




            1





            Wow, thanks. That is like 30% faster.

            – James
            44 mins ago





            Wow, thanks. That is like 30% faster.

            – James
            44 mins ago













            2














            You can always have a brute force solution. Something like:



            name, id, app_name, app_version = [], [], [], []
            for i in range(len(df)):
            for v in df.loc[i,'apps']:
            app_name.append(v[0])
            app_version.append(v[1])
            name.append(df.loc[i, 'name'])
            id.append(df.loc[i, 'id'])
            df = pd.DataFrame('name': name, 'id': id, 'app_name': app_name, 'app_version': app_version)


            will do the work.



            Note that I assumed df['apps'] is lists of strings if df['apps'] is strings then you need: eval(df.loc[i,'apps']) instead of df.loc[i,'apps']






            share|improve this answer

























            • Even though this works, it is probably unfeasible for large data frames. In pandas, one for loop is already bad enough, so imagine two nested for loops ;} Always try to avoid direct iteration !

              – RafaelC
              1 hour ago















            2














            You can always have a brute force solution. Something like:



            name, id, app_name, app_version = [], [], [], []
            for i in range(len(df)):
            for v in df.loc[i,'apps']:
            app_name.append(v[0])
            app_version.append(v[1])
            name.append(df.loc[i, 'name'])
            id.append(df.loc[i, 'id'])
            df = pd.DataFrame('name': name, 'id': id, 'app_name': app_name, 'app_version': app_version)


            will do the work.



            Note that I assumed df['apps'] is lists of strings if df['apps'] is strings then you need: eval(df.loc[i,'apps']) instead of df.loc[i,'apps']






            share|improve this answer

























            • Even though this works, it is probably unfeasible for large data frames. In pandas, one for loop is already bad enough, so imagine two nested for loops ;} Always try to avoid direct iteration !

              – RafaelC
              1 hour ago













            2












            2








            2







            You can always have a brute force solution. Something like:



            name, id, app_name, app_version = [], [], [], []
            for i in range(len(df)):
            for v in df.loc[i,'apps']:
            app_name.append(v[0])
            app_version.append(v[1])
            name.append(df.loc[i, 'name'])
            id.append(df.loc[i, 'id'])
            df = pd.DataFrame('name': name, 'id': id, 'app_name': app_name, 'app_version': app_version)


            will do the work.



            Note that I assumed df['apps'] is lists of strings if df['apps'] is strings then you need: eval(df.loc[i,'apps']) instead of df.loc[i,'apps']






            share|improve this answer















            You can always have a brute force solution. Something like:



            name, id, app_name, app_version = [], [], [], []
            for i in range(len(df)):
            for v in df.loc[i,'apps']:
            app_name.append(v[0])
            app_version.append(v[1])
            name.append(df.loc[i, 'name'])
            id.append(df.loc[i, 'id'])
            df = pd.DataFrame('name': name, 'id': id, 'app_name': app_name, 'app_version': app_version)


            will do the work.



            Note that I assumed df['apps'] is lists of strings if df['apps'] is strings then you need: eval(df.loc[i,'apps']) instead of df.loc[i,'apps']







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 1 hour ago

























            answered 2 hours ago









            MaPyMaPy

            16726




            16726












            • Even though this works, it is probably unfeasible for large data frames. In pandas, one for loop is already bad enough, so imagine two nested for loops ;} Always try to avoid direct iteration !

              – RafaelC
              1 hour ago

















            • Even though this works, it is probably unfeasible for large data frames. In pandas, one for loop is already bad enough, so imagine two nested for loops ;} Always try to avoid direct iteration !

              – RafaelC
              1 hour ago
















            Even though this works, it is probably unfeasible for large data frames. In pandas, one for loop is already bad enough, so imagine two nested for loops ;} Always try to avoid direct iteration !

            – RafaelC
            1 hour ago





            Even though this works, it is probably unfeasible for large data frames. In pandas, one for loop is already bad enough, so imagine two nested for loops ;} Always try to avoid direct iteration !

            – RafaelC
            1 hour ago











            1














            My suggestion (there may be easier ways) is using DataFrame.apply alongside pd.concat:



            import pandas as pd

            def expand_row(row):
            apps = row.apps
            apps_names = [app[0] for app in apps]
            apps_versions = [app[1] for app in apps]
            return pd.DataFrame(
            'name': row.name,
            'id': row.id,
            'app_name': apps_names,
            'app_version': apps_versions
            )

            df = pd.DataFrame(
            'name': ['john', 'smith'],
            'id': [1, 2],
            'apps': [[['app1', 'v1'], ['app2', 'v2'], ['app3','v3']],
            [['app1', 'v1'], ['app4', 'v4']]]
            )
            temp_dfs = df.apply(expand_row, axis=1).tolist()
            expanded = pd.concat(temp_dfs)

            print(expanded)

            # name id app_name app_version
            # 0 0 1 app1 v1
            # 1 0 1 app2 v2
            # 2 0 1 app3 v3
            # 0 1 2 app1 v1
            # 1 1 2 app4 v4





            share|improve this answer





























              1














              My suggestion (there may be easier ways) is using DataFrame.apply alongside pd.concat:



              import pandas as pd

              def expand_row(row):
              apps = row.apps
              apps_names = [app[0] for app in apps]
              apps_versions = [app[1] for app in apps]
              return pd.DataFrame(
              'name': row.name,
              'id': row.id,
              'app_name': apps_names,
              'app_version': apps_versions
              )

              df = pd.DataFrame(
              'name': ['john', 'smith'],
              'id': [1, 2],
              'apps': [[['app1', 'v1'], ['app2', 'v2'], ['app3','v3']],
              [['app1', 'v1'], ['app4', 'v4']]]
              )
              temp_dfs = df.apply(expand_row, axis=1).tolist()
              expanded = pd.concat(temp_dfs)

              print(expanded)

              # name id app_name app_version
              # 0 0 1 app1 v1
              # 1 0 1 app2 v2
              # 2 0 1 app3 v3
              # 0 1 2 app1 v1
              # 1 1 2 app4 v4





              share|improve this answer



























                1












                1








                1







                My suggestion (there may be easier ways) is using DataFrame.apply alongside pd.concat:



                import pandas as pd

                def expand_row(row):
                apps = row.apps
                apps_names = [app[0] for app in apps]
                apps_versions = [app[1] for app in apps]
                return pd.DataFrame(
                'name': row.name,
                'id': row.id,
                'app_name': apps_names,
                'app_version': apps_versions
                )

                df = pd.DataFrame(
                'name': ['john', 'smith'],
                'id': [1, 2],
                'apps': [[['app1', 'v1'], ['app2', 'v2'], ['app3','v3']],
                [['app1', 'v1'], ['app4', 'v4']]]
                )
                temp_dfs = df.apply(expand_row, axis=1).tolist()
                expanded = pd.concat(temp_dfs)

                print(expanded)

                # name id app_name app_version
                # 0 0 1 app1 v1
                # 1 0 1 app2 v2
                # 2 0 1 app3 v3
                # 0 1 2 app1 v1
                # 1 1 2 app4 v4





                share|improve this answer















                My suggestion (there may be easier ways) is using DataFrame.apply alongside pd.concat:



                import pandas as pd

                def expand_row(row):
                apps = row.apps
                apps_names = [app[0] for app in apps]
                apps_versions = [app[1] for app in apps]
                return pd.DataFrame(
                'name': row.name,
                'id': row.id,
                'app_name': apps_names,
                'app_version': apps_versions
                )

                df = pd.DataFrame(
                'name': ['john', 'smith'],
                'id': [1, 2],
                'apps': [[['app1', 'v1'], ['app2', 'v2'], ['app3','v3']],
                [['app1', 'v1'], ['app4', 'v4']]]
                )
                temp_dfs = df.apply(expand_row, axis=1).tolist()
                expanded = pd.concat(temp_dfs)

                print(expanded)

                # name id app_name app_version
                # 0 0 1 app1 v1
                # 1 0 1 app2 v2
                # 2 0 1 app3 v3
                # 0 1 2 app1 v1
                # 1 1 2 app4 v4






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 1 hour ago

























                answered 1 hour ago









                araraonlineararaonline

                643313




                643313



























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