Pandas groupby/apply has different behaviour with

问题:

I have the following dataframe

   X    Y
0  A   10
1  A    9
2  A    8
3  A    5
4  B  100
5  B   90
6  B   80
7  B   50

and two different functions that are very similar

def func1(x):
    if x.iloc[0]['X'] == 'A':
        x['D'] = 1
    else:
        x['D'] = 0
    return x[['X', 'D']]

def func2(x):
    if x.iloc[0]['X'] == 'A':
        x['D'] = 'u'
    else:
        x['D'] = 'v'
    return x[['X', 'D']]

Now I can groupby/apply these functions

df.groupby('X').apply(func1)
df.groupby('X').apply(func2)

The first line gives me what I want, i.e.

   X  D
0  A  1
1  A  1
2  A  1
3  A  1
4  B  0
5  B  0
6  B  0
7  B  0

But the second line returns something quite strange

   X  D
0  A  u
1  A  u
2  A  u
3  A  u
4  A  u
5  A  u
6  A  u
7  A  u

So my questions are:

• Can anybody explain why the behavior of groupby/apply is different when the type changes?
• How can I get something similar with func2?

回答1:

The problem is simply that a function applied to a GroupBy should never try to change the dataframe it receives. It is implementation dependant whether it is a copy (that can safely be changed but changes will not be seen in original dataframe) or a view. The choice is done by pandas optimizer, and as a user, you should just know that it is forbidden.

The correct way is to force a copy:

def func2(x):
    x = x.copy()
    if x.iloc[0]['X'] == 'A':
        x['D'] = 'u'
    else:
        x['D'] = 'v'
    return x[['X', 'D']]

After that, df.groupby('X').apply(func2).reset_index(level=0, drop=True) gives as expected:

   X  D
0  A  u
1  A  u
2  A  u
3  A  u
4  B  v
5  B  v
6  B  v
7  B  v