在数据框中找到最小值，并在新列上添加标签

Question

我可以对python代码进行哪些改进以提高效率？对于我的情况，我有这个数据帧

In [1]: df = pd.DataFrame({'PersonID': [1, 1, 1, 2, 2, 2, 3, 3, 3],
                           'Name': ["Jan", "Jan", "Jan", "Don", "Don", "Don", "Joe", "Joe", "Joe"],
                           'Label': ["REL", "REL", "REL", "REL", "REL", "REL", "REL", "REL", "REL"],
                           'RuleID': [55, 55, 55, 3, 3, 3, 10, 10, 10],
                           'RuleNumber': [3, 4, 5, 1, 2, 3, 234, 567, 999]})

结果如下：

In [2]: df
Out[2]: 
   PersonID Name Label  RuleID  RuleNumber
0         1  Jan   REL      55          3
1         1  Jan   REL      55          4
2         1  Jan   REL      55          5
3         2  Don   REL       3          1
4         2  Don   REL       3          2
5         2  Don   REL       3          3
6         3  Joe   REL      10        234
7         3  Joe   REL      10        567
8         3  Joe   REL      10        999

这里我需要完成的是将Label列下的字段更新为MAIN，以获得与应用于个人ID和姓名的每个规则ID相关联的最低规则值。因此，结果需要如下所示：

In [3]: df
Out[3]:
   PersonID Name Label  RuleID  RuleNumber
0         1  Jan  MAIN      55           3
1         1  Jan   REL      55           4
2         1  Jan   REL      55           5
3         2  Don  MAIN       3           1
4         2  Don   REL       3           2
5         2  Don   REL       3           3
6         3  Joe  MAIN      10         234
7         3  Joe   REL      10         567
8         3  Joe   REL      10         999

这是我为实现这一点而编写的代码：

In [4]:

df['Label'] = np.where(
        df['RuleNumber'] ==
        df.groupby(['PersonID', 'Name', 'RuleID'])['RuleNumber'].transform('min'),
        "MAIN", df.Label)

是否有更好的方法更新标签列下的值？我觉得我是在用蛮力逼我过去，这可能不是最有效的方法

我使用了以下SO线程来获得结果：

Replace column values within a groupby and condition

Replace values within a groupby based on multiple conditions

https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.idxmin.html

https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.transform.html

Using Pandas to Find Minimum Values of Grouped Rows

如有任何建议，将不胜感激

多谢各位

Answer 1

import pandas as pd

df = pd.DataFrame({'PersonID': [1, 1, 1, 2, 2, 2, 3, 3, 3],
'Name': ["Jan", "Jan", "Jan", "Don", "Don", "Don", "Joe", "Joe", "Joe"],
'Label': ["REL", "REL", "REL", "REL", "REL", "REL", "REL", "REL", "REL"],
'RuleID': [55, 55, 55, 3, 3, 3, 10, 10, 10],
'RuleNumber': [3, 4, 5, 1, 2, 3, 234, 567, 999]})

df.loc[df.groupby('Name')['RuleNumber'].idxmin()[:], 'Label'] = 'MAIN'

Answer 2

似乎您可以按分组的idxmin进行筛选，而不考虑排序顺序，并基于此更新RuleNumber。您可以按如下方式使用loc、np.where、mask或where：

df.loc[df.groupby(['PersonID', 'Name', 'RuleID'])['RuleNumber'].idxmin(), 'Label'] = 'MAIN'

或者在您尝试时使用np.where：

df['Label'] = (np.where((df.index == df.groupby(['PersonID', 'Name', 'RuleID'])
                         ['RuleNumber'].transform('idxmin')), 'MAIN', 'REL'))
df
Out[1]: 
   PersonID Name Label  RuleID  RuleNumber
0         1  Jan  MAIN      55           3
1         1  Jan   REL      55           4
2         1  Jan   REL      55           5
3         2  Don  MAIN       3           1
4         2  Don   REL       3           2
5         2  Don   REL       3           3
6         3  Joe  MAIN      10         234
7         3  Joe   REL      10         567
8         3  Joe   REL      10         999

使用mask或其逆where也可以：

df['Label'] = (df['Label'].mask((df.index == df.groupby(['PersonID', 'Name', 'RuleID'])
                         ['RuleNumber'].transform('idxmin')), 'MAIN'))

或

df['Label'] = (df['Label'].where((df.index != df.groupby(['PersonID', 'Name', 'RuleID'])
                         ['RuleNumber'].transform('idxmin')), 'MAIN'))

Answer 3

在PersonID上使用duplicated：

df.loc[~df['PersonID'].duplicated(),'Label'] = 'MAIN'
print(df)

输出：

   PersonID Name Label  RuleID  RuleNumber
0         1  Jan  MAIN      55           3
1         1  Jan   REL      55           4
2         1  Jan   REL      55           5
3         2  Don  MAIN       3           1
4         2  Don   REL       3           2
5         2  Don   REL       3           3
6         3  Joe  MAIN      10         234
7         3  Joe   REL      10         567
8         3  Joe   REL      10         999