Python groupby 结果计数频率

Question

我有一个数据框

df = pd.DataFrame({'id':['one','one','two','two','three','three','three'],
                   'type':['current','saving','current','current','current','saving','credit']})

我想统计只有'current'的id的数量 应该是这样的:

only_currnt_id_list = ['two']

Answer 1

我认为你需要:

L = df.groupby('id') \
      .filter(lambda x: (x['type'] == 'current').all() and 
                        (x['type'] == 'current').sum() == 1)['id'].tolist()
print (L)

['two']

编辑:

df = pd.DataFrame({'id':['one','one','two','three','three','three'],'type':['current','current','current','current','saving','credit']})
print (df)
      id     type
0    one  current
1    one  current
2    two  current
3  three  current
4  three   saving
5  three   credit

<小时/>

L = df.groupby('id') \
      .filter(lambda x: (x['type'] == 'current').all() and 
                        (x['type'] == 'current').sum() == 1)['id'].tolist()
print (L)
['two']

L = df.groupby('id') \
      .filter(lambda x: (x['type'] == 'current').all())['id'].unique().tolist()
print (L)
['one', 'two']