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根据索引值的条件选择Pandas DataFrame中的行

[英]Selecting rows in a Pandas DataFrame based on conditions of the index values

Let's say I have the following multi-index DataFrame: 假设我有以下多索引DataFrame:

import pandas as pd
df = pd.DataFrame({'Index0':[0,1,2,3,4,5],'Index1':[100,200,300,400,500,600],'A':[5,2,5,8,1,2]})

示例DataFrame

Now I want to select all the rows where Index1 is less than 400. Everybody knows how that works if Index1 was a regular column: 现在我想选择Index1小于400的所有行。如果Index1是常规列,每个人都知道它是如何工作的:

df[df['Index1'] < 400]

So one method would be to reset_index , perform the selection, then set the index again. 因此,一种方法是reset_index ,执行选择,然后再次设置索引。 This seems quite redundant. 这似乎很多余。

My question is: Is there a way to do this directly? 我的问题是:有没有办法直接这样做? And how to do this when the DataFrame has a row multiindex? 当DataFrame有行多索引时如何做到这一点?

Simpliest here is use query : 这里最简单的是使用query

df1 = df.query('Index1 < 400')
print (df1)
               A
Index0 Index1   
0      100     5
1      200     2
2      300     5

Or get_level_values for select level of MultiIndex with boolean indexing : 或者get_level_values用于具有boolean indexing选择级别的MultiIndex

df1 = df[df.index.get_level_values('Index1') < 400]

Detail : 细节

print (df.index.get_level_values('Index1'))
Int64Index([100, 200, 300, 400, 500, 600], dtype='int64', name='Index1')

If levels have no names select by positions, for query use special keyword ilevel_ with position: 如果级别没有按位置选择名称,则查询使用特殊关键字ilevel_ with position:

df.index.names = [None, None]
print (df)
       A
0 100  5
1 200  2
2 300  5
3 400  8
4 500  1
5 600  2

df1 = df.query('ilevel_1 < 400')

df1 = df[df.index.get_level_values(1) < 400]
print (df1)
       A
0 100  5
1 200  2
2 300  5

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