Python pandas groupby 与 lambda

Question

My dataframe:

df = pd.DataFrame({'company': ['A', 'A', 'A', 'A', 'A','B','B','B','B','B'],
                      'offered': [1, 1, 0, 1, 1, 1, 0, 0, 1, 1],
                      'accepted': [0, 1, 0, 1, 1, 0, 0, 0, 1, 0]})

    company offered accepted
0   A       1       0
1   A       1       1
2   A       0       0
3   A       1       1
4   A       1       1
5   B       1       0
6   B       0       0
7   B       0       0
8   B       1       1
9   B       1       0

I want my final result looks like this:

df2 = df.groupby('company')[['offered', 'accepted']].agg('sum')
df2['accept_rate'] = df2['accepted']/df2['offered']
df2

    offered accepted    accept_rate
company         
A   4       3           0.750000
B   3       1           0.333333

However, I would like to do this in one shot (eg, using lambda). Here's what I have tried

df['accept_rate'] = df['accepted'] / df['offered']
df.groupby('company')[['offered', 'accepted', 'accept_rate']].agg({'offered': 'sum',
                                                    'accepted': 'sum',
                                                    'accept_rate': lambda x: df['accepted'].sum()/df['offered'].sum()})

    offered accepted    accept_rate
company         
A   4       3           0.571429
B   3       1           0.571429

As you can see the accept_rate = 0.571429 is for the total/combined companies.

How can I make the accept_rate to look like my desired final result?

Thanks in advance!

Answer 1

你要assign

df.groupby('company').agg({'offered': 'sum', 'accepted': 'sum'}).assign(accept_rate=lambda x: x['accepted']/x['offered'])