How to group the pandas dataframe without aggregating and without missing the actual values?

user_19675781 0
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I have a data frame like this:

df:
ID   Award   Award_ID
1    Ninja    N13
1    Ninja    N19
1    Warrior  W16
2    Alpha    A99
2    Delta    D18
3    Alpha    A101
3    Alpha    A102
3    Alpha    A103
4    Delta    D12

Some of the IDs are repeated here. Whenever the ID has multiple occurrences, I want to make it into a single row by collating all the Award & Award_ID information in this format:

ID Multiple_Awards(Award_Names) Award(Award_IDs)

So, the expected output is:

df:
ID   Award                            Award_ID
1    Multiple_Award(Ninja, Warrior)   Ninja(N13, N19), Warrior(W16)
2    Multiple_Award(Alpha, Delta)     Alpha(A99), Delta(D18)
3    Multiple_Award(Alpha)            Alpha(A101,A102,A103)
4    Delta                            D12

In the case of a single award, I want the row to remain the same.

Can anyone help how to get the data into this format?

pandas dataframe python-3.x python
3 Answers

Check Below code:

def custom_agg(val):
  return val.unique() if val.count()<2 else 'Multiple_Award'+str(tuple(val.unique()))

def custom_agg_1(val):
  return str(tuple(val.unique()))

df.assign(Award_ID = df.groupby(['Id','Award'])['Award_ID'].transform(custom_agg_1),
          Award_count = df.groupby(['Id'])['Award_ID'].transform('count')
          ).\
assign(Award_ID=lambda x:  x.apply(lambda x: (x.Award+x.Award_ID) if x.Award_count > 1 else x.Award_ID.replace(')','').replace(',','').replace('(',''), axis=1)).\
assign(Award=lambda x: x.groupby('Id')['Award'].transform(custom_agg)).\
groupby(['Id','Award']).agg({'Award_ID':custom_agg_1}).reset_index().\
assign(Award_ID = lambda x: x.Award_ID.str[2:-2].str.replace('"', '').replace({'\'': ''}, regex=True).replace({',\)': ')'}, regex=True),
       Award = lambda x: x.Award.replace({'\'': ''}, regex=True).replace({',\)': ')'}, regex=True))

Output:

enter image description here

user_18313360

You can do:

df1 = df.groupby(['ID', 'Award'],as_index=False)['Award_ID'].agg(tuple)
df1 = df1.groupby('ID').agg(tuple)
df1['Award_ID'] = df1[['Award', 'Award_ID']].apply(
     lambda x: tuple(a + str(b) for a, b in zip(x['Award'], x['Award_ID'])), axis=1)
df1['Award'] = df.groupby('ID')['Award'].count().transform(
    lambda x: 'Multiple_Award' if x > 1 else '') + df1['Award'].astype(str)

print(df1):

                                Award                                Award_ID
ID                                                                            
1   Multiple_Award('Ninja', 'Warrior')  (Ninja('N13', 'N19'), Warrior('W16',))
2     Multiple_Award('Alpha', 'Delta')          (Alpha('A99',), Delta('D18',))
3             Multiple_Award('Alpha',)        (Alpha('A101', 'A102', 'A103'),)
4                           ('Delta',)                        (Delta('D12',),)
user_14103030

i know it's not the complite solution but i belive you can do other transformations by yourself:

(df.groupby(['ID','Award'],as_index=False)['Award_ID'].agg(tuple).
              assign(Award_ID=lambda x: x.Award+x.Award_ID.astype(str)).
              groupby('ID').
              apply(lambda x: pd.Series({'Award':x.Award.values,
                                         'Award_ID':x.Award_ID.values})).
              applymap(', '.join).
              reset_index())

if gives you:

   ID           Award                              Award_ID
0   1  Ninja, Warrior  Ninja('N13', 'N19'), Warrior('W16',)
1   2    Alpha, Delta          Alpha('A99',), Delta('D18',)
2   3           Alpha         Alpha('A101', 'A102', 'A103')
3   4           Delta                         Delta('D12',)
user_183445120
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