python-3.x - 根据其他两列的分组值获取均值列

Question

我有一些学校数据:

data = {'name': ['school a', 'school b', 'school c', 'school d', 'school e', 'school f'], 
       'type': ['a', 'a', 'b', 'b', 'a', 'b'],
        'location': ['county a', 'county a', 'county b', 'county b', 'county b', 'county a'], 
        'avg_score': [9, 7, 5, 7, 6, 8]
       }

df = pd.DataFrame(data)


Out:
    name    type    location    avg_score
0   school a    a   county a    9
1   school b    a   county a    7
2   school c    b   county b    5
3   school d    b   county b    7
4   school e    a   county b    6
5   school f    b   county a    8

我想将学校分数与每个地点的学校类型的平均值进行比较。

我可以用 groupby 做到这一点: df.groupby(['type', 'location']).mean().round(2)

Out: 

                avg_score
type location   
a   county a    8
    county b    6
b   county a    8
    county b    6

但是，我想获得一个附加列，其中包含每个位置的这种学校类型的平均值，而不是分组表。

我如何获得这样的 compare_score:

    name       type location avg_score compare_score
0   school a    a   county a    9       8
1   school b    a   county a    7       5
2   school c    b   county b    5       7
3   school d    b   county b    7       7
4   school e    a   county b    6       3
5   school f    b   county a    8       7

我发现了这个问题

Python Pandas average based on condition into new column

并尝试将一些可能的解决方案应用于我的问题:

for atype, alocation in df.groupby('type'):
    df.loc[df.type == type, 'compare'] = (df.where(df['type' == atype]).where(df['location' == alocation]).mean()).avg_score.round(2)```

引发此错误:

---------------------------------------------------------------------------
KeyError                                  Traceback (most recent call last)
~/.local/share/virtualenvs/schule-jwiURUl3/lib/python3.6/site-packages/pandas/core/indexes/base.py in get_loc(self, key, method, tolerance)
   2656             try:
-> 2657                 return self._engine.get_loc(key)
   2658             except KeyError:

pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc()

pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc()

pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()

pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()

KeyError: False

During handling of the above exception, another exception occurred:

KeyError                                  Traceback (most recent call last)
<ipython-input-27-4b3cf2b7aaf6> in <module>
      1 for atype, alocation in df.groupby('type'):
----> 2     df.loc[df.type == type, 'compare'] = (df.where(df['type' == atype]).where(df['location' == alocation]).mean()).avg_score.round(2)

~/.local/share/virtualenvs/schule-jwiURUl3/lib/python3.6/site-packages/pandas/core/frame.py in __getitem__(self, key)
   2925             if self.columns.nlevels > 1:
   2926                 return self._getitem_multilevel(key)
-> 2927             indexer = self.columns.get_loc(key)
   2928             if is_integer(indexer):
   2929                 indexer = [indexer]

~/.local/share/virtualenvs/schule-jwiURUl3/lib/python3.6/site-packages/pandas/core/indexes/base.py in get_loc(self, key, method, tolerance)
   2657                 return self._engine.get_loc(key)
   2658             except KeyError:
-> 2659                 return self._engine.get_loc(self._maybe_cast_indexer(key))
   2660         indexer = self.get_indexer([key], method=method, tolerance=tolerance)
   2661         if indexer.ndim > 1 or indexer.size > 1:

pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc()

pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc()

pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()

pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()

KeyError: False

也许这根本就不是一个好的尝试。你有什么建议吗？
任何提示都受到高度赞赏。

Answer 1

我认为您正在寻找 transform :

df['compare_score']=df.groupby(['type', 'location'])['avg_score'].transform('mean').round(2)
print(df)

---------------

       name type  location  avg_score  compare_score
0  school a    a  county a          9              8
1  school b    a  county a          7              8
2  school c    b  county b          5              6
3  school d    b  county b          7              6
4  school e    a  county b          6              6
5  school f    b  county a          8              8