问题

Suppose we take a pandas dataframe...

    name  age  family
0   john    1       1
1  jason   36       1
2   jane   32       1
3   jack   26       2
4  james   30       2

Then do a groupby() ...

group_df = df.groupby('family')
group_df = group_df.aggregate({'name': name_join, 'age': pd.np.mean})

Then do some aggregate/summarize operation (in my example, my function name_join aggregates the names):

def name_join(list_names, concat='-'):
    return concat.join(list_names)

The grouped summarized output is thus:

        age             name
family                      
1        23  john-jason-jane
2        28       jack-james

Question:

Is there a quick, efficient way to get to the following from the aggregated table?

    name  age  family
0   john   23       1
1  jason   23       1
2   jane   23       1
3   jack   28       2
4  james   28       2

(Note: the age column values are just examples, I don't care for the information I am losing after averaging in this specific example)

The way I thought I could do it does not look too efficient:

create empty dataframe
from every line in group_df, separate the names
return a dataframe with as many rows as there are names in the starting row
append the output to the empty dataframe

回答1:

The rough equivalent is .reset_index(), but it may not be helpful to think of it as the "opposite" of groupby().

You are splitting a string in to pieces, and maintaining each piece's association with 'family'. This old answer of mine does the job.

Just set 'family' as the index column first, refer to the link above, and then reset_index() at the end to get your desired result.

回答2:

There are a few ways to undo DataFrame.groupby, one way is to do DataFrame.groupby.filter(lambda x:True), this gets back to the original DataFrame.

回答3:

Here's a complete example that recovers the original dataframe from the grouped object

def name_join(list_names, concat='-'):
    return concat.join(list_names)

print('create dataframe\n')
df = pandas.DataFrame({'name':['john', 'jason', 'jane', 'jack', 'james'], 'age':[1,36,32,26,30], 'family':[1,1,1,2,2]})
df.index.name='indexer'
print(df)
print('create group_by object')
group_obj_df = df.groupby('family')
print(group_obj_df)

print('\nrecover grouped df')
group_joined_df = group_obj_df.aggregate({'name': name_join, 'age': 'mean'})
group_joined_df


create dataframe

          name  age  family
indexer                    
0         john    1       1
1        jason   36       1
2         jane   32       1
3         jack   26       2
4        james   30       2
create group_by object
<pandas.core.groupby.generic.DataFrameGroupBy object at 0x7fbfdd9dd048>

recover grouped df 
                   name  age
family                      
1       john-jason-jane   23
2            jack-james   28

print('\nRecover the original dataframe')
print(pandas.concat([group_obj_df.get_group(key) for key in group_obj_df.groups]))

Recover the original dataframe
          name  age  family
indexer                    
0         john    1       1
1        jason   36       1
2         jane   32       1
3         jack   26       2
4        james   30       2

来源：https://stackoverflow.com/questions/20122521/is-there-an-ungroup-by-operation-opposite-to-groupby-in-pandas

标签

python