Question

我有每个组的值的总和。

   rf = condition1.groupby(by=['Well Name','Phase'])['Sum of Activity Time 
   (Hr)'].sum()


   Well Name              |   Phase    |       Value   |

   TIGER 55-2-12 LOV 8H   |    INT     |       56.25
                          |    MNH     |       58.25
                          |    SRF     |       34.25
   UNIVERSITY 20 PW 2502H |    INT     |       52.75
                          |    MNH     |       72.50
                          |    SRF     |       28.5
   UNIVERSITY 20 PW UNIT  |    INT     |       64.50
                          |    MNH     |       132.50
                          |    SRF     |       30.00
   UNIVERSITY 20 TG UNIT  |    INT     |       57.00
                          |    MNH     |       129.50
                          |    SRF     |       25.50

我需要有这样的东西：正如您所看到的，三个阶段中每个阶段的最小值就是每个组的总和的最小值。有什么想法吗？

    Well Name              |   Phase    |       Value   | 

    UNIVERSITY 20 PW 2502H |    INT     |       52.75
    TIGER 55-2-12 LOV 8H   |    MNH     |       58.25
    UNIVERSITY 20 TG UNIT  |    SRF     |       25.50

如您所见，只是这些组的总和的最小值。

Answer 1

您可以这样做：

import pandas as pd
import numpy as np
rf=pd.DataFrame(condition1.groupby(by=['Well Name','Phase'])['Sum of Activity Time 
   (Hr)'].sum())
rf=rf[rf.isin(rf.groupby('Well Name').min()['Sum of Activity Time (Hr)'].tolist())].dropna()

还可以尝试：

import pandas as pd
rf=pd.DataFrame(condition1.groupby(by=['Well Name','Phase'])['Sum of Activity Time 
   (Hr)'].sum())
i=0
while i<len(rf.index):
    if rf.loc[rf.index.values[i]][0] != rf.loc[rf.index.values.tolist()[i][0]].apply(min)[0]:
        rf = rf.drop(rf.index.values.tolist()[i])
    else:
        print(i)
        i+=1
rf

输出：

Well Name              |   Phase    |       Value   | 

UNIVERSITY 20 PW 2502H |    INT     |       52.75
TIGER 55-2-12 LOV 8H   |    MNH     |       58.25
UNIVERSITY 20 TG UNIT  |    SRF     |       25.50

有没有一种方法可以获取所有汇总组的最小值？

1 个答案: