Python Pandas按二级索引(或任何其他级别)切片multiindex [英] Python Pandas slice multiindex by second level index (or any other level)

问题描述

有很多帖子将多索引的级别[0]按级别范围进行切片 1 级别[0]索引值的索引

There are many postings on slicing the level[0] of a multiindex by a range of level1. However, I cannot find a solution for my problem; that is, I need a range of the level1 index for level[0] index values

数据帧:第一个是A到Z，等级是1到400；我需要每个level [0]的前2个和后2个(第一个)，但不需要在同一步骤中.

dataframe: First is A to Z, Rank is 1 to 400; I need the first 2 and last 2 for each level[0] (First), but not in the same step.

           Title Score
First Rank 
A     1    foo   100
      2    bar   90
      3    lime  80
      4    lame  70
B     1    foo   400
      2    lime  300
      3    lame  200
      4    dime  100

我正在尝试获取每个级别的最后2行 1 带有以下代码的索引，但仅对第一个level [0]值正确地切片.

I am trying to get the last 2 rows for each level1 index with the below code, but it slices properly only for the first level[0] value.

[IN]  df.ix[x.index.levels[1][-2]:]
[OUT] 
               Title Score
    First Rank 
    A     3    lime  80
          4    lame  70
    B     1    foo   400
          2    lime  300
          3    lame  200
          4    dime  100

我通过交换索引获得了前2行，但是我无法使其适用于后2行.

The first 2 rows I get by swapping the indices, but I cannot make it work for the last 2 rows.

df.index = df.index.swaplevel("Rank", "First")
df= df.sortlevel() #to sort by Rank
df.ix[1:2] #Produces the first 2 ranks with 2 level[1] (First) each.
           Title Score
Rank First 
1     A    foo   100
      B    foo   400
2     A    bar   90
      B    lime  300

我当然可以换回去得到这个:

Of course I can swap this back to get this:

df2 = df.ix[1:2]
df2.index = ttt.index.swaplevel("First","rank") #change the order of the indices back.
df2.sortlevel()
               Title Score
    First Rank 
    A     1    foo   100
          2    bar   90
    B     1    foo   400
          2    lime  300

希望通过相同的步骤获得任何帮助:

Any help is appreciated to get with the same procedure:

• 最后2行用于索引 1 (排名)
• 获得前两行的更好方法

• Last 2 rows for index1 (Rank)
• And a better way to get the first 2 rows

通过@ako编辑以下反馈:

Edit following feedback by @ako:

真正使用pd.IndexSlice可以很容易地切片任何级别的索引.这里是一个更通用的解决方案，下面是我逐步采用的方法来获取第一行和最后两行.此处的更多信息: http://pandas.pydata.org/pandas -docs/stable/advanced.html#using-slicers

Using pd.IndexSlice truly makes it easy to slice any level index. Here a more generic solution and below my step-wise approach to get the first and last two rows. More information here: http://pandas.pydata.org/pandas-docs/stable/advanced.html#using-slicers

"""    
Slicing a dataframe at the level[2] index of the
major axis (row) for specific and at the level[1] index for columns.

"""
    df.loc[idx[:,:,['some label','another label']],idx[:,'yet another label']]

"""
Thanks to @ako below is my solution, including how I
get the top and last 2 rows.
"""
    idx = pd.IndexSlice
    # Top 2
    df.loc[idx[:,[1,2],:] #[1,2] is NOT a row index, it is the rank label. 
    # Last 2
    max = len(df.index.levels[df.index.names.index("rank")]) # unique rank labels
    last2=[x for x in range(max-2,max)]
    df.loc[idx[:,last2],:] #for last 2 - assuming all level[0] have the same lengths.

推荐答案

使用索引器将任意值切成任意维度-只需传递带有该维度所需级别/值的列表即可.

Use an indexer to slice arbitrary values in arbitrary dimensions--just pass a list with whatever the desired levels / values are for that dimension.

idx = pd.IndexSlice
df.loc[idx[:,[3,4]],:]

           Title  Score
First Rank             
A     3     lime     80
      4     lame     70
B     3     lame    200
      4     dime    100

用于再现数据:

from StringIO import StringIO

s="""
First Rank Title Score
A      1    foo   100
A      2    bar   90
A      3    lime  80
A      4    lame  70
B      1    foo   400
B      2    lime  300
B      3    lame  200
B      4    dime  100
"""
df = pd.read_csv(StringIO(s),
                 sep='\s+',
                 index_col=["First", "Rank"])

这篇关于Python Pandas按二级索引(或任何其他级别)切片multiindex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IT屋！