有没有办法得到所有求和组的最小值?

2024-09-22 20:34:14 发布

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我有每个组的值的总和。你知道吗

   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

我需要这样的东西:3个阶段中的每一个的最小值,你可以看到,每一个值都是每组总和的最小值。有什么想法吗?你知道吗

    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

如你所见,这只是群的最小和。你知道吗


Tags: namevalueunittgintpwtigerwell
1条回答
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1楼 · 发布于 2024-09-22 20:34:14

你可以做:

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()

您也可以尝试:

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

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