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])
但这并没有解决我问题的某些部分.
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
我需要以下内容：
>获取具有lvl_2元素的可能组合的新多索引数据框.
>转换我的数据以获得每个元素的比例
这是一个例子：
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
在这里,我使用c的元素的比例：
0.587418 = 0.421075/0.716823
1.154012 = 0.278418/0.241261
0.152608 = 0.117888/0.772491
df = df.groupby(level='lvl_2').transform(lambda x: x[0]/x[1])
但显然,这只会转换每个特定级别的第一个和第二个值.另外,我不知道如何用这些组合建立新的多索引. (a / c,a / b,a / d,c / b,c / d,b / d)
我觉得我走在正确的道路上,但我感到困惑.
解决方法:
如果对于第一级别是相同的其他级别的组合,例如在样本中可以使用reindex的列中的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
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