我正在尝试对 pandas 聚合进行排序

Question

I'm trying to aggregate and sort data from my dataset, but I don't know how to do. Can someone help me?

data = {'message_id':  ['1', '1', '1', '1', '2', '2', '2'],
        'to': ['one', 'two', 'three', 'four', 'five', 'six', 'five'],
        'idt': ['1','2','3','4','5','6','5']
        }

df = pd.DataFrame(data, columns = ['message_id','to','idt'])

agg_func_text = {'to': [ set], 'idt': [ set]}

df.sort_values(by=['message_id', 'to'])

df3=df.groupby(['message_id']).agg(agg_func_text)

as result:

message_id  to set              idt set
1       {four, three, one, two}     {2, 3, 1, 4}
2       {five, six}         {5, 6}

but I would like to recevied this as result:

message_id  to set              idt set
1       {one, two, three, four}     {1, 2, 3, 4}
2       {five, six}         {5, 6}

Answer 1

In python set is not defined order, so cannot sorting or change ordering there, possible soution is use dict.fromkeys().keys() trick for remove duplicates and output is tuple (which should be sorted and there is also defined order):

f = lambda x: dict.fromkeys(x).keys()
agg_func_text = {'to': f, 'idt': f}

#if need sorting assign back
df = df.sort_values(by=['message_id', 'idt'])

df3=df.groupby('message_id').agg(agg_func_text)

print (df3)
                                 to           idt
message_id                                       
1           (one, two, three, four)  (1, 2, 3, 4)
2                       (five, six)        (5, 6)

Answer 2

sort using a number in a dictionary, save the results, then translate from the number back to a character representation of the number.

 data = {'message_id':  ['1', '1', '1', '1', '2', '2', '2'],
    'to': ['one', 'two', 'three', 'four', 'five', 'six', 'five'],
    'idt': ['1','2','3','4','5','6','5']
    }

 df = pd.DataFrame(data, columns = ['message_id','to','idt'])
 print(df)
 agg_func_text = {'to': set, 'idt': set}
 df.sort_values(by=['message_id', 'to'])

 grouped=df.groupby(['message_id']).agg(agg_func_text)

 grouped['idt']=grouped['idt'].apply(lambda x: sorted(x))
 dct={'one':1, 'two':2,'three':3,'four':4,'five':5,'six':6,'seven':7,'eight':8,'nine':9}
 dct2={1: 'one',2:'two',3:'three',4:'four',5:'five',6:'six',7:'seven',8:'eight',9:'nine'}
 grouped['to']=grouped['to'].apply(lambda x: sorted([dct[item] for item in x]))
 grouped['to']=grouped['to'].apply(lambda x: [dct2[item] for item in x])
 print(grouped)

 output:
                          to           idt
 message_id                                       
 1           [one, two, three, four]  [1, 2, 3, 4]
 2                       [five, six]        [5, 6]