6

How to keep index when using pandas merge中提出了一个类似的问题，但它不适用于 MultiIndexes，即

a = DataFrame(np.array([1,2,3,4,1,2,3,3]).reshape((4,2)), columns=['col1','to_merge_on'], index=['a','b','a','b'])
id = pd.MultiIndex.from_arrays([[1,1,2,2],['a','b','a','b']], names =['id1','id2'])
a.index = id

In [207]: a
Out[207]: 
         col1  to_merge_on
id1 id2                   
1   a       1            2
    b       3            4
2   a       1            2
    b       3            4

b=DataFrame(data={"col2": [1,2,3], 'to_merge_on' : [1,3,5]})

In [209]: b
Out[209]: 
   col2  to_merge_on
0     1            1
1     2            3
2     3            5

a.reset_index().merge(b, how="left").set_index('index')

In [208]: a.reset_index().merge(b, how="left").set_index('index')
------------------------------------------------------------
Traceback (most recent call last):
  File "<ipython console>", line 1, in <module>
  File "C:\Python27\lib\site-packages\pandas\core\frame.py", line 2054, in set_index
    level = frame[col]
  File "C:\Python27\lib\site-packages\pandas\core\frame.py", line 1458, in __getitem__
    return self._get_item_cache(key)
  File "C:\Python27\lib\site-packages\pandas\core\generic.py", line 294, in _get_item_cache
    values = self._data.get(item)
  File "C:\Python27\lib\site-packages\pandas\core\internals.py", line 625, in get
    _, block = self._find_block(item)
  File "C:\Python27\lib\site-packages\pandas\core\internals.py", line 715, in _find_block
    self._check_have(item)
  File "C:\Python27\lib\site-packages\pandas\core\internals.py", line 722, in _check_have
    raise KeyError('no item named %s' % str(item))
KeyError: 'no item named index'

如何在左侧数据框中保留 MultiIndex 的同时进行合并？

4

1 回答 1

3

临时解决方案：

In [255]: a = a.reset_index()

In [256]: a
Out[256]: 
   id1 id2  col1  to_merge_on
0    1   a     1            2
1    1   b     3            4
2    2   a     1            2
3    2   b     3            4

In [271]: c = pd.merge(a, b, how="left")

In [272]: c
Out[272]: 
   id1 id2  col1  to_merge_on  col2
0    1   a     1            2   NaN
1    2   a     1            2   NaN
2    2   b     3            3     2
3    1   b     3            4   NaN

In [273]: c = c.set_index(['id1','id2'])

In [274]: c
Out[274]: 
         col1  to_merge_on  col2
id1 id2                         
1   a       1            2   NaN
2   a       1            2   NaN
    b       3            3     2
1   b       3            4   NaN
于 2012-11-27T03:19:03.737 回答