如何将 DataFrame 索引和列与字典键和多个值进行匹配？

Question

我如何修改下面的字典理解以考虑到 s 列也应该是匹配条件？

import pandas as pd 

dct = {'NNI' : pd.DataFrame({'s': [-1, -1, -1, 1, 1],
                             'count': [13, 11, 10,12, 16]},
                            index =['2007-07-13', '2019-09-18', '2016-08-01', '2021-04-05','2017-01-04' ]),
       'NVEC' : pd.DataFrame({'s': [-1, -1, -1, 1, 1],
                              'count': [12, 10, 9,14,5]},
                             index =['2012-10-09', '2018-10-01', '2022-02-01', '2020-03-20','2016-04-06'])
      }

df = pd.DataFrame({'Date': ['2022-02-14', '2022-02-14', '2022-02-14', '2022-02-14', '2022-02-14'], 
                   's': [-1,-1,-1,1,1], 
                   'count': [10, 10, 10, 9, 9]}, 
                  index = ['NNI', 'NVEC', 'IPA', 'LYTS', 'MYN'])

df:

            Date  s  count
NNI   2022-02-14 -1     10
NVEC  2022-02-14 -1     10
IPA   2022-02-14 -1     10
LYTS  2022-02-14  1      9
MYN   2022-02-14  1      9

dct:

{'NNI':       s  count
 2007-07-13  -1     13
 2019-09-18  -1     11
 2016-08-01  -1     10
 2021-04-05   1     12
 2017-01-04   1     16,

 'NVEC':      s  count
 2012-10-09  -1     12
 2018-10-01  -1     10
 2022-02-01  -1      9
 2020-03-20   1     14
 2016-04-06   1      5}

这是我目前拥有的：

df = df.assign(ratio=pd.Series({k: v['count'].gt(df.loc[k, 'count']).sum() / 
v['count'].ge(df.loc[k, 'count']).sum() for k,v in dct.items()})).fillna(0)

df
            Date  s  count     ratio
NNI   2022-02-14 -1     10  0.800000
NVEC  2022-02-14 -1     10  0.666667
IPA   2022-02-14 -1     10  0.000000
LYTS  2022-02-14  1      9  0.000000
MYN   2022-02-14  1      9  0.000000

想要的结果是：

df
            Date  s  count     ratio
NNI   2022-02-14 -1     10  0.666667
NVEC  2022-02-14 -1     10  0.500000
IPA   2022-02-14 -1     10  0.000000
LYTS  2022-02-14  1      9  0.000000
MYN   2022-02-14  1      9  0.000000

Answer 1

您可以将其添加为布尔掩码，例如：

v.loc[v['s'] == df.loc[k, 's'], 'count']

所以代码变成：

df = df.assign(ratio=pd.Series({k: v.loc[v['s'] == df.loc[k, 's'], 'count'].gt(df.loc[k, 'count']).sum() / 
                                v.loc[v['s'] == df.loc[k, 's'], 'count'].ge(df.loc[k, 'count']).sum() 
                                for k,v in dct.items()})).fillna(0)

输出：

            Date  s  count     ratio
NNI   2022-02-14 -1     10  0.666667
NVEC  2022-02-14 -1     10  0.500000
IPA   2022-02-14 -1     10  0.000000
LYTS  2022-02-14  1      9  0.000000
MYN   2022-02-14  1      9  0.000000

只是一个建议，但在这里使用辅助函数可能会有所帮助，因为那里的除法有点不可读，尤其是在添加索引之后。您可以使用：

def get_ratio(df_row, v):
    msk = v['s'] == df_row['s']
    numerator = v.loc[msk, 'count'].gt(df_row['count']).sum()
    denominator = v.loc[msk, 'count'].ge(df_row['count']).sum()
    return numerator / denominator

df = df.assign(ratio = pd.Series({k: get_ratio(df.loc[k], v) for k,v in dct.items()})).fillna(0)

如何将 DataFrame 索引和列与字典键和多个值进行匹配？

How to match DataFrame index and column against dictionary key and multiple values?

python

dataframe

python-3.x

pandas

dictionary-comprehension