Python 熊猫在最近的日期出生

我有一个特定主题的报告数据框架。每个报告都有分数，受试者在某些日期有分数，但在其他日期没有分数。我想创建一个新的数据框，它只包含每个主题的最新分数。下面是一个MRE。原始数据帧如下所示：

         Subject   Rpt_date  Score_1  Score_2
0   L. Skywalker 2020-12-01      9.0      NaN
1   L. Skywalker 2020-12-06      NaN      8.0
2   L. Skywalker 2021-01-11      7.0      NaN
3        H. Solo 2020-11-19      NaN      7.0
4        H. Solo 2020-12-15      NaN      5.0
5        H. Solo 2021-01-26      4.0      NaN
6      L. Organa 2020-11-20      6.0      NaN
7      L. Organa 2020-12-01      NaN      6.0
8      L. Organa 2020-12-19      NaN      7.0
9      D. Djarin 2020-12-10      NaN     10.0
10     D. Djarin 2020-12-12     10.0      NaN
11     D. Djarin 2021-01-03      NaN     10.0

        Subject  Score_1  Score_2
0  L. Skywalker      7.0      8.0
1       H. Solo      4.0      5.0
2     L. Organa      6.0      7.0
3     D. Djarin     10.0     10.0

所需的输出如下所示：

         Subject   Rpt_date  Score_1  Score_2
0   L. Skywalker 2020-12-01      9.0      NaN
1   L. Skywalker 2020-12-06      NaN      8.0
2   L. Skywalker 2021-01-11      7.0      NaN
3        H. Solo 2020-11-19      NaN      7.0
4        H. Solo 2020-12-15      NaN      5.0
5        H. Solo 2021-01-26      4.0      NaN
6      L. Organa 2020-11-20      6.0      NaN
7      L. Organa 2020-12-01      NaN      6.0
8      L. Organa 2020-12-19      NaN      7.0
9      D. Djarin 2020-12-10      NaN     10.0
10     D. Djarin 2020-12-12     10.0      NaN
11     D. Djarin 2021-01-03      NaN     10.0

        Subject  Score_1  Score_2
0  L. Skywalker      7.0      8.0
1       H. Solo      4.0      5.0
2     L. Organa      6.0      7.0
3     D. Djarin     10.0     10.0

我的MRE在技术上是有效的，但看起来像是一个乱七八糟的东西，在一个大数据帧中会非常慢

import pandas as pd
import numpy as np

def merge_latest(df, col):
    df_temp = df.dropna(subset=[col]).copy()
    df_temp['Last_rpt'] = df_temp.groupby('Subject')['Rpt_date'].transform('max')
    df_temp.drop(df_temp.loc[df_temp['Rpt_date'] != df_temp['Last_rpt']]
                .index, inplace=True)
    d = dict(zip(df_temp['Subject'], df_temp[c]))
    return d

df_data = {'Subject':['L. Skywalker', 'L. Skywalker', 'L. Skywalker',
                        'H. Solo', 'H. Solo', 'H. Solo',
                        'L. Organa', 'L. Organa', 'L. Organa',
                        'D. Djarin', 'D. Djarin', 'D. Djarin'],
            'Rpt_date':['12/1/2020', '12/6/2020', '1/11/2021',
                        '11/19/2020', '12/15/2020', '1/26/2021',
                        '11/20/2020', '12/1/2020', '12/19/2020',
                        '12/10/2020', '12/12/2020', '1/3/2021'],
            'Score_1':[9, np.nan, 7,
                    np.nan, np.nan, 4,
                    6, np.nan, np.nan,
                    np.nan, 10, np.nan],
            'Score_2':[np.nan, 8, np.nan,
                    7, 5, np.nan,
                    np.nan, 6, 7,
                    10, np.nan, 10]}

df = pd.DataFrame(data=df_data)
df['Rpt_date'] = pd.to_datetime(df['Rpt_date'])

print(df)

fin_df = pd.DataFrame()
fin_df['Subject'] = df['Subject'].unique()

for c in ['Score_1', 'Score_2']:
    merge_dict = merge_latest(df, c)
    fin_df[c] = fin_df['Subject'].map(merge_dict)
    
print(fin_df)

---

首先使用报告日期排序，然后使用。这个解决方案是矢量化的

df['Rpt_date'] = pd.to_datetime(df['Rpt_date'])
fin_df = df.sort_values('Rpt_date').groupby('Subject', as_index=False).last()

输出

        Subject   Rpt_date  Score_1  Score_2
0     D. Djarin 2021-01-03     10.0     10.0
1       H. Solo 2021-01-26      4.0      5.0
2     L. Organa 2020-12-19      6.0      7.0
3  L. Skywalker 2021-01-11      7.0      8.0

---

这太完美了。仅供使用此答案的任何人参考，我碰巧已经按日期对数据进行了排序，但是@Vishnudev正确地建议添加

排序\u值

，以确保

.last（）

采用所需的值。