Question

我想计算熊猫数据库中项目的最高得分-最低得分

当前df如下所示：

projectID supplierID score
1         1          50
1         2          60 
1         3          75

我希望它看起来像这样：
max-min => 75-50 = 25

projectID supplierID score max-min
1         1          50    25
1         2          60    25
1         3          75    25

我想对每个projectID进行此操作

Answer 1

使用np.ptp（“峰到峰”）：

df['max-min']=df.groupby('projectID').score.transform(np.ptp)
#df.groupby('projectID').score.transform(np.ptp)
Out[229]: 
0    25
1    25
2    25
Name: score, dtype: int64

Answer 2

您可以使用transform通过传回将max和min减去的lambda函数来广播结果。

df['max-min'] = df.groupby('projectID').score.transform(lambda s: s.max() - s.min())

    projectID   supplierID  score   max-min
0   1           1           50      25
1   1           2           60      25
2   1           3           75      25

Answer 3

您可以使用groupby获取最大值和最小值，然后join将“ projectId”上的结果作为新列：

import pandas as pd

df = pd.DataFrame([[1, 1, 30],
                   [1, 2, 50],
                   [2, 1, 60],
                   [2, 2, 40],
                   [1, 3, 20]],
                  columns=["projectID", "supplierID", "score"])

df.join(  df.groupby(["projectID"])["score"].max() 
        - df.groupby(["projectID"])["score"].min(), 
        on="projectID", rsuffix="_max-min")

Answer 4

您可以使用Declare @YourTable Table ([advisor_rep_id] varchar(50),[individual_url_rep_codes] varchar(50),[split_url_rep_codes] varchar(50)) Insert Into @YourTable Values (57444,'9289','9569') ,(4407,'397','7128, 9226') ,(52779,'8613, 8614, 8616',NULL) ,(56732,NULL,'9193, 4423') ,(56713,'3456','9193, 4423') Select A.[advisor_rep_id] ,URLS = B.RetVal from @YourTable A Cross Apply ( Select RetSeq = Row_Number() over (Order By (Select null)) ,RetVal = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)'))) From (Select x = Cast('<x>' + replace((Select replace(concat([individual_url_rep_codes],',',[split_url_rep_codes]),',','§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml).query('.')) as A Cross Apply x.nodes('x') AS B(i) ) B Where B.RetVal is not null + advisor_rep_id URLS 57444 9289 57444 9569 4407 397 4407 7128 4407 9226 52779 8613 52779 8614 52779 8616 56732 9193 56732 4423 56713 3456 56713 9193 56713 4423：

GroupBy

transform方法将常规g = df.groupby('projectID')['score'] df['max-min'] = g.transform('max') - g.transform('min')的结果与石斑鱼系列进行对齐。

计算最高分数-每个供应商在熊猫中的最低分数

4 个答案: