Pandas - Groupby 多索引级别，获取可能的组合，然后转换数据

Question

我一直在努力解决分组、组合和转换的问题。我目前的解决方案是：

df = df.groupby(level='lvl_2').transform(lambda x: x[0]/x[1])

但这并没有解决我的某些问题。

假设代码如下：

import pandas as pd
import numpy as np
import datetime
today = datetime.date.today()
today_1 = datetime.date.today() - datetime.timedelta(1)
today_2 = datetime.date.today() - datetime.timedelta(2)
ticker_date = [('first', 'a',today), ('first', 'a',today_1), ('first', 'a',today_2),
               ('first', 'c',today), ('first', 'c',today_1), ('first', 'c',today_2),
               ('first', 'b',today), ('first', 'b',today_1), ('first', 'b',today_2),
               ('first', 'd',today), ('first', 'd',today_1), ('first', 'd',today_2)]
index_df = pd.MultiIndex.from_tuples(ticker_date,names=['lvl_1','lvl_2','lvl_3'])
df = pd.DataFrame(np.random.rand(12), index_df, ['idx'])

输出是：

                          idx
lvl_1 lvl_2 lvl_3               
first a     2018-02-14  0.421075
            2018-02-13  0.278418
            2018-02-12  0.117888
      c     2018-02-14  0.716823
            2018-02-13  0.241261
            2018-02-12  0.772491
      b     2018-02-14  0.681738
            2018-02-13  0.636927
            2018-02-12  0.668964
      d     2018-02-14  0.770797
            2018-02-13  0.11469
            2018-02-12  0.877965

我需要以下物品：

1. 使用可能的 lvl_2 元素组合获取新的多索引数据帧。
2. 转换我的数据以获得每个元素的比率

这是一个插图：

在这里，我创建了一个“新”列。

                                new
lvl_1   lvl_2       lvl_3   
first   a/c     2018-02-14  0.587418372
                2018-02-13  1.154011631
                2018-02-12  0.152607603
        a/b     2018-02-14  0.617649302
                2018-02-13  0.437127018
                2018-02-12  0.17622473
        a/d     2018-02-14  0.546285209
                2018-02-13  2.427569971
                2018-02-12  0.134274145
        c/b     2018-02-14  1.051464052
                2018-02-13  0.378789092
                2018-02-12  1.154757207
        c/d     2018-02-14  0.929976375
                2018-02-13  2.103592292
                2018-02-12  0.87986537
        b/d     2018-02-14  0.884458554
                2018-02-13  5.553465865
                2018-02-12  0.761948369

进一步解释：

                                    new
    lvl_1   lvl_2       lvl_3   
    first   a/c     2018-02-14  0.587418372
                    2018-02-13  1.154011631
                    2018-02-12  0.152607603

这里，我做a的元素与c的比例：

0.587418 = 0.421075/0.716823
1.154012 = 0.278418/0.241261
0.152608 = 0.117888/0.772491

我尝试了 groupby 和 transform 方法，例如：

df = df.groupby(level='lvl_2').transform(lambda x: x[0]/x[1])

但很明显，这只会转换每个特定级别的第一个和第二个值。另外，我不知道如何使用组合建立新的多索引。 (a/c, a/b, a/d, c/b, c/d, b/d)

我觉得我在正确的道路上，但我觉得卡住了。

Answer 1

如果第一级与示例中的其他级别的组合相同，则可以在具有div 的列中使用reindex 到MultiIndex：

#same as Maarten Fabré answer
np.random.seed(42)

from  itertools import combinations

#get combination of second level values
c = pd.MultiIndex.from_tuples(list(combinations(df.index.levels[1], 2)))

#reshape to unique columns of second level
print (df['idx'].unstack(1))
lvl_2                    a         b         c         d
lvl_1 lvl_3                                             
first 2018-02-12  0.731994  0.601115  0.155995  0.969910
      2018-02-13  0.950714  0.866176  0.156019  0.020584
      2018-02-14  0.374540  0.058084  0.598658  0.708073

#reindex by both levels
df1 = df['idx'].unstack(1).reindex(columns=c, level=0)
print (df1)
                         a                             b                   c
                         b         c         d         c         d         d
lvl_1 lvl_3                                                                 
first 2018-02-12  0.731994  0.731994  0.731994  0.601115  0.601115  0.155995
      2018-02-13  0.950714  0.950714  0.950714  0.866176  0.866176  0.156019
      2018-02-14  0.374540  0.374540  0.374540  0.058084  0.058084  0.598658


df2 = df['idx'].unstack(1).reindex(columns=c, level=1)
print (df2)
                         a                             b                   c
                         b         c         d         c         d         d
lvl_1 lvl_3                                                                 
first 2018-02-12  0.601115  0.155995  0.969910  0.155995  0.969910  0.969910
      2018-02-13  0.866176  0.156019  0.020584  0.156019  0.020584  0.020584
      2018-02-14  0.058084  0.598658  0.708073  0.598658  0.708073  0.708073

---

#divide with flatten MultiIndex    
df3 = df1.div(df2)
df3.columns = df3.columns.map('/'.join)
#reshape back and change order of levels, sorting indices
df3 = df3.stack().reorder_levels([0,2,1]).sort_index()

---

print (df3)
lvl_1       lvl_3     
first  a/b  2018-02-12     1.217727
            2018-02-13     1.097599
            2018-02-14     6.448292
       a/c  2018-02-12     4.692434
            2018-02-13     6.093594
            2018-02-14     0.625632
       a/d  2018-02-12     0.754703
            2018-02-13    46.185944
            2018-02-14     0.528957
       b/c  2018-02-12     3.853437
            2018-02-13     5.551748
            2018-02-14     0.097023
       b/d  2018-02-12     0.619764
            2018-02-13    42.079059
            2018-02-14     0.082031
       c/d  2018-02-12     0.160834
            2018-02-13     7.579425
            2018-02-14     0.845476
dtype: float64

Answer 2

from itertools import combinations
def calc_ratios(data):
    comb = combinations(data.index.get_level_values('lvl_2').unique(), 2)

    ratios = {
        f'{i}/{j}': 
            data.xs(i, level='lvl_2') / 
            data.xs(j, level='lvl_2')
        for i, j in comb
    }
#     print(ratios)
    if ratios:
        return pd.concat(ratios)
result = pd.concat(calc_ratios(data) for group, data in df.groupby('lvl_1'))

    lvl_1 lvl_3   idx
a/b   first   2018-02-14  6.448292467809392
a/b   first   2018-02-13  1.0975992712883451
a/b   first   2018-02-12  1.2177269366284045
a/c   first   2018-02-14  0.6256323575698127
a/c   first   2018-02-13  6.093594353302192
a/c   first   2018-02-12  4.692433684425558
a/d   first   2018-02-14  0.5289572433565499
a/d   first   2018-02-13  46.185944271838835
a/d   first   2018-02-12  0.7547030687230791
b/d   first   2018-02-14  0.08203059119870332
b/d   first   2018-02-13  42.07905879677424
b/d   first   2018-02-12  0.6197637959891664
c/b   first   2018-02-14  10.306839775450461
c/b   first   2018-02-13  0.18012345549282302
c/b   first   2018-02-12  0.25950860865015657
c/d   first   2018-02-14  0.8454761601705119
c/d   first   2018-02-13  7.579425474360648
c/d   first   2018-02-12  0.16083404038888807

（使用np.random.seed(42)生成的数据）