Pandas：在多个列中查找具有匹配值的行的python方法（分层条件）

对不起，标题有点不清楚。我无法用语言简明扼要地描述这个问题。希望我下面的描述能帮助澄清。欢迎对标题进行任何澄清性编辑

我正在尝试从一个数据帧创建一个networkx流程图。数据框记录订单如何流经多家公司。dataframe中的大多数行都是连接的，连接显示在多个列中。样本数据如下：

df = pd.DataFrame({'Company': ['A', 'A', 'B', 'B', 'B', 'C', 'C'],
              'event_type':['new', 'route', 'receive', 'execute', 'route', 'receive', 'execute'],
             'event_id': ['110', '120', '200', '210', '220', '300', '310'],
             'prior_event_id': [np.nan, '110', np.nan, '120', '210', np.nan, '300'],
             'route_id': [np.nan, 'foo', 'foo', np.nan, 'bar', 'bar', np.nan]}
             )

数据框如下所示：

  Company event_type event_id prior_event_id route_id
0       A        new      110            NaN      NaN
1       A      route      120            110      foo
2       B    receive      200            NaN      foo
3       B    execute      210            120      NaN
4       B      route      220            210      bar
5       C    receive      300            NaN      bar
6       C    execute      310            300      NaN

订单经过3家公司：A、B、C。在每个公司内，后一个事件可以通过

event\u id

-

previor\u event\u id

对链接到其源事件。但这种方法不适用于属于不同公司的记录。例如，第1行和第2行将仅通过一列

route\u id

进行匹配。因此，我尝试重新创建的链接机制是一种层次结构，因为如果

event\u id

-

previor\u event\u id

列对没有产生任何结果，我将只使用列

route\u id

进行匹配

下图可能有助于说明链接机制：

我的解决方案相当笨拙：

# Make every event unique so as to not confound the linking
df['event_sub'] = df.groupby(df.event_type).cumcount()+1 
df['event'] = df.event_type + ' ' + df.event_sub.astype(str) 

# Find the match based on first matching criterion
replace_dict_event = dict(df[['event_id', 'event']].values)
df['source'] = df['prior_event_id'].apply(lambda x: replace_dict_event.get(x) if replace_dict_event.get(x) else np.nan )
df['target'] = df['event_id'].apply(lambda x: replace_dict_event.get(x) if replace_dict_event.get(x) else np.nan )

# From last step, find the match based on second matching criterion for the unmatched rows 
replace_dict_rtd = dict(df[df.event_type == 'route'][['route_id', 'event']].values)
df.loc[df.event_type == 'receive', 'source'] = df[df.event_type == 'receive']['route_id'].apply(lambda x: replace_dict_rtd.get(x))
df

我基本上使用了两次

apply

来逐步获得匹配。我想知道是否有一种更干净、更像蟒蛇的方法来做这件事

我的结果如下：

以及我由此创建的networkx图：

---

您有两种不同类型的链接：a）通过匹配

先前的事件id

和

事件id

定义的链接，以及b）通过

路由id

定义的链接。使用两组不同的命令来提取两种不同类型的关系是pythonic（或者仅仅是良好的编码实践）

也就是说，由于您处理的是表格数据，因此最好使用合并（特别是内部联接）来提取链接，而不是使用apply进行字典查找。表格数据的数据库针对这类查询进行了优化，而大型数据集的查询速度会慢得多

#!/usr/bin/env python
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd

if __name__ == '__main__':

    df = pd.DataFrame({'Company': ['A', 'A', 'B', 'B', 'B', 'C', 'C'],
                       'event_type':['new', 'route', 'receive', 'execute', 'route', 'receive', 'execute'],
                       'event_id': ['110', '120', '200', '210', '220', '300', '310'],
                       'prior_event_id': [np.nan, '110', np.nan, '120', '210', np.nan, '300'],
                       'route_id': [np.nan, 'foo', 'foo', np.nan, 'bar', 'bar', np.nan]}
    )

    # --------------------------------------------------------------------------------
    # a) links established by matching event_id with prior_event_id
    df2 = pd.merge(df, df, left_on='event_id', right_on='prior_event_id', how='inner')

    #       Company_x event_type_x event_id_x prior_event_id_x route_id_x Company_y event_type_y event_id_y prior_event_id_y route_id_y
    # 0         A          new        110              NaN        NaN         A        route        120              110        foo
    # 1         A        route        120              110        foo         B      execute        210              120        NaN
    # 2         B      execute        210              120        NaN         B        route        220              210        bar
    # 3         C      receive        300              NaN        bar         C      execute        310              300        NaN

    # --------------------------------------------------------------------------------
    # b) links established by matching route_id

    # remove events without route ids
    valid = df['route_id'].notna()
    df3 = df['valid']

    #   Company event_type event_id prior_event_id route_id
    # 1       A      route      120            110      foo
    # 2       B    receive      200            NaN      foo
    # 4       B      route      220            210      bar
    # 5       C    receive      300            NaN      bar

    # join on route_id
    df4 = pd.merge(df3, df3, on='route_id', how='inner')

    #   Company_x event_type_x event_id_x prior_event_id_x route_id Company_y event_type_y event_id_y prior_event_id_y
    # 0         A        route        120              110      foo         A        route        120              110
    # 1         A        route        120              110      foo         B      receive        200              NaN
    # 2         B      receive        200              NaN      foo         A        route        120              110
    # 3         B      receive        200              NaN      foo         B      receive        200              NaN
    # 4         B        route        220              210      bar         B        route        220              210
    # 5         B        route        220              210      bar         C      receive        300              NaN
    # 6         C      receive        300              NaN      bar         B        route        220              210
    # 7         C      receive        300              NaN      bar         C      receive        300              NaN

    # remove cases where a company was matched to itself
    valid = df4['Company_x'] != df4['Company_y']
    df5 = df4[valid]

    #       Company_x event_type_x event_id_x prior_event_id_x route_id Company_y event_type_y event_id_y prior_event_id_y
    # 1         A        route        120              110      foo         B      receive        200              NaN
    # 2         B      receive        200              NaN      foo         A        route        120              110
    # 5         B        route        220              210      bar         C      receive        300              NaN
    # 6         C      receive        300              NaN      bar         B        route        220              210