Question

df = pd.DataFrame({'A': [1,2,3,1,2,3], 'B': [10,10,11,10,10,15], 'key1':['a','b','a','b','c','c'],'key2':1})

df1 = pd.DataFrame({'A': [1,2,3,1,2,3], 'B': [100,100,110,100,100,150], 'key1':['a','c','b','a','a','c'],'key2':1})
dfn = pd.merge(df,df1,on='key2')
dfn_grouped = dfn.groupby('key1_y')

the list(dfn_grouped):
[('a',     A_x  B_x key1_x  key2  A_y  B_y key1_y
   0     1   10      a     1    1  100      a
   3     1   10      a     1    1  100      a
   ...    ...          ...      ...
   33    3   15      c     1    1  100      a
   34    3   15      c     1    2  100      a),
  ('b',     A_x  B_x key1_x  key2  A_y  B_y key1_y
    2     1   10      a     1    3  110      b
    8     2   10      b     1    3  110      b
    14    3   11      a     1    3  110      b
    20    1   10      b     1    3  110      b
    26    2   10      c     1    3  110      b
    32    3   15      c     1    3  110      b),
   ('c',     A_x  B_x key1_x  key2  A_y  B_y key1_y
    1     1   10      a     1    2  100      c
    ......        ... ....
    35    3   15      c     1    3  150      c)]

现在我需要通过“key1_x”将dfn_grouped分组，并将其连接到像A_x这样的dict：A_y

    key1_y  key1_x  A_X:A_Y
         b       a   {'10':'110','11':110}   
         b       b   {'10':110}
         b       c   {'10':110,'15':110}
      // if A_x in dict append the A_y like:
      // b       e   {'10':[11,12]}

Answer 1

这是您需要的吗？：

>> grouped = dfn.groupby(['key1_y','key1_x','A_x'])

>> dfg = pd.DataFrame(grouped.apply(lambda x: [a for a in x.A_y])).reset_index()
>> dfg.columns = [u'key1_y', u'key1_x', u'A_x', 'dic_values']
>> dfg['dic'] = [{a:b} for a,b in zip(dfg.A_x.values,dfg.dic_values.values)]
>> dfg.drop(['A_x','dic_values'],1,inplace=True)
>> g_dics = dfg.groupby(['key1_y','key1_x']).apply(lambda x: dict(sum(map(dict.items, [d for d in x.dic]), [])))
>> pd.DataFrame(g_dics).reset_index()

groupby之后的熊猫Groupby

1 个答案: