I am looking for a more pythonic/efficient (and short) way to factorize (=enumerate unique instances) of the row-wise set difference in an aggregated dataframe. The below tables illustrate what should not be too complex of a dataframe manipulation:

should be transformed, using list_locations = ['US', 'CA', 'JP', 'BE', 'FR'] into a table of the form

such that every product group has a column rest_of_world_location_list that lists all the items from list_locations that are not part of a product group. Column rest_of_world_index is simply the factorization of rest_of_world_location_list.

MWE input data:

df = pd.DataFrame(
    {
        "product": [10,11,12,13,14,15,16,17,18,19,20],
        "product_group": [1,1,1,2,2,2,3,3,4,4,4],
        "location": ['RoW', 'US', 'CA', 'RoW', 'JP', 'US', 'FR', 'BE', 'RoW', 'US', 'CA']
    }
)
list_locations = ['US', 'CA', 'JP', 'BE', 'FR']

My attempt (works, but likely too complicated):

activity_with_rest_of_world_location = pd.DataFrame(df[df['location'] == 'RoW']['product_group'])
activity_with_rest_of_world_location['index'] = activity_with_rest_of_world_location.index

df_rest_of_world = df[df['product_group'].isin(activity_with_rest_of_world_location['product_group'])]
df_rest_of_world = df_rest_of_world.drop(activity_with_rest_of_world_location['index'])


df_rest_of_world_agg = pd.DataFrame(
    data = df_rest_of_world.groupby('product_group')['location'].apply(tuple))
df_rest_of_world_agg.reset_index(inplace = True)
df_rest_of_world_agg = df_rest_of_world_agg.merge(
    right = activity_with_rest_of_world_location,
    how = 'left',
    on = 'product_group'
)

df_rest_of_world_agg.set_index(keys = 'index', inplace = True)

df_rest_of_world_agg['location_rest_of_world'] = df_rest_of_world_agg.apply(
    lambda row: tuple(set(list_io_countries) - set(row['location'])), 
    axis = 1
)

df_rest_of_world_agg = df_rest_of_world_agg.dropna(subset ='location_rest_of_world')

df_rest_of_world_agg['location'] = pd.factorize(df_rest_of_world_agg['location_rest_of_world'])[0]
df_rest_of_world_agg['location'] = 'RoW_'   df_rest_of_world_agg['location'].astype(str)

CodePudding user response：

IIUC, you can use a single pipeline with 3 steps:

world = set(list_locations)

(df.groupby('product_group', as_index=False)
   # aggregate locations as string and the rest of the from from a set difference
   .agg(**{'location_list': ('location', ', '.join),
           'rest_of_world_location_list': ('location', lambda l: ', '.join(sorted(world.difference(l))))
          })
   # filter the rows without RoW
   .loc[lambda d: d['location_list'].str.contains('RoW')]
   # add category
   .assign(rest_of_world_index=lambda d: 'RoW_' d['location_list'].astype('category').cat.codes.add(1).astype(str)
          )
)

output:

   product_group location_list rest_of_world_location_list rest_of_world_index
0              1   RoW, US, CA                  BE, FR, JP               RoW_2
1              2   RoW, JP, US                  BE, CA, FR               RoW_1
3              4   RoW, US, CA                  BE, FR, JP               RoW_2

CodePudding user response：

Solutions with sets - crete sets per groups, filter out no RoW rows, get differencies with join and last use factorize with frozensets:

list_io_countries = ['US', 'CA', 'JP', 'BE', 'FR']
s = set(list_io_countries) 

df = df.groupby(df['product_group'])['location'].agg(set).reset_index(name='location_list')

df = (df[['RoW' in x for x in df['location_list']]]
       .assign(rest_of_world_location_list = lambda x: x['location_list'].apply(lambda x: ','.join(s - x)),
               rest_of_world_index = lambda x: pd.factorize(x['location_list'].apply(lambda x: frozenset(x - set(['RoW']))))[0]   1,
               location_list = lambda x: x['location_list'].agg(','.join)
               )
       .assign(rest_of_world_index = lambda x: 'RoW_'   x['rest_of_world_index'].astype(str)))

print (df)
   product_group location_list rest_of_world_location_list rest_of_world_index
0              1     RoW,CA,US                    JP,BE,FR               RoW_1
1              2     RoW,JP,US                    CA,BE,FR               RoW_2
3              4     RoW,CA,US                    JP,BE,FR               RoW_1