pandas ：分组：如何一步重置所有组的索引？

问题描述

我尝试将我的数据框分组到组中

  
传入 as_index = False 
 code>添加到groupby，那么您不需要 reset_index 来再次创建groupby-d列的列：
 在[11]中：grouped = df.groupby（'A'，as_index = False）
 
在[12]中：grouped.get_group 'foo'）
出[12]：
 AB 
 0 foo 1 
 2 foo 3 
 4 foo 5 
 6 foo 7 
 7 foo 8 
  
注意：正如上面的例子中指出的那样，上面的索引是不是  [0，1，2，...] ，我声称这在实践中永远不会起作用 - 如果你打算只能通过一些奇怪的箍筋 - 它会变得更加冗长，不太可读，效率也不高...... 
I've tried to split my dataframe to groups
df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar',
                       'foo', 'bar', 'foo', 'foo'],
                   'B' : ['1', '2', '3', '4',
                       '5', '6', '7', '8'],
                   })

grouped = df.groupby('A')
I get 2 groups
     A  B
0  foo  1
2  foo  3
4  foo  5
6  foo  7
7  foo  8

     A  B
1  bar  2
3  bar  4
5  bar  6
Now I want to reset indexes for each group separately
print grouped.get_group('foo').reset_index()
print grouped.get_group('bar').reset_index()
Finally I get the result
     A  B
0  foo  1
1  foo  3
2  foo  5
3  foo  7
4  foo  8

     A  B
0  bar  2
1  bar  4
2  bar  6
Is there better way how to do this? (For example: automatically call some method for each group)
 解决方案 
Pass in as_index=False to the groupby, then you don't need to reset_index to make the groupby-d columns columns again:
In [11]: grouped = df.groupby('A', as_index=False)

In [12]: grouped.get_group('foo')
Out[12]:
     A  B
0  foo  1
2  foo  3
4  foo  5
6  foo  7
7  foo  8
Note: As pointed out (and seen in the above example) the index above is not [0, 1, 2, ...], I claim that this will never matter in practice - if it does you're going to have to just through some strange hoops - it's going to be more verbose, less readable and less efficient...
                        
这篇关于 pandas ：分组：如何一步重置所有组的索引？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！