我有一个df，

cluster_id memo

1 m

1 n

2 m

2 m

2 n

3 m

3 m

3 m

3 n

4 m

4 n

4 n

4 n

我想groupby cluster_id并应用以下功能，

def valid_row_dup(df):

num_real_invs = df[df['memo'] == 'm'].shape[0]

num_reversals_invs = df[df['memo'] == 'n'].shape[0]

if num_real_invs == df.shape[0]:

return True

elif num_reversals_invs == df.shape[0]:

return False

elif abs(num_real_invs - num_reversals_invs) > 0:

# even diff

if abs(num_real_invs - num_reversals_invs) % 2 == 0:

return True

else:

if abs(num_real_invs - num_reversals_invs) == 1:

return False

# odd diff

else:

return True

elif num_real_invs - num_reversals_invs == 0:

return False

将每个groupby对象作为df传入func; 将布尔结果分配回df，

cluster_id memo valid

1 m False

1 n False

2 m False

2 m False

2 n False

3 m True

3 m True

3 m True

3 n True

4 m True

4 n True

4 n True

4 n True

解决办法：应用功能然后合并：

df.merge(df.groupby('cluster_id').apply(valid_row_dup).to_frame(), on='cluster_id')

cluster_id memo 0

0 1 m False

1 1 n False

2 2 m False

3 2 m False

4 2 n False

5 3 m True

6 3 m True

7 3 m True

8 3 n True

9 4 m True

10 4 n True

11 4 n True

12 4 n True