PYTHON, if 문이 첫 번째 조건 만 충족합니다. PuLP

Question

나는 시스템을 최적화하기 위해 PuLP를 사용하여 비용을 최소화하려고합니다. 나는 여러 개의 If를 사용하고 있는데 문제는 항상 첫 번째 조건을 충족한다는 것입니다. 여기 내 코드가있다. 나는이 언어에 대해 배우기 시작한 누군가가 나를 도울 수 있기를 바랍니다.

import numpy as np 
import pandas as pd 
from pulp import * 

idx = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23] 
d = { 
'day': pd.Series(['01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14'], index=idx), 
'hour':pd.Series(['00:00:00', '01:00:00', '02:00:00', '03:00:00', '04:00:00', '05:00:00', '06:00:00', '07:00:00', '08:00:00', '09:00:00', '10:00:00', '11:00:00', '12:00:00', '13:00:00', '14:00:00', '15:00:00', '16:00:00', '17:00:00', '18:00:00', '19:00:00', '20:00:00', '21:00:00', '22:00:00', '23:00:00'], index=idx), 
'output':pd.Series([0,0,0,0.087,0.309,0.552,0.682,0.757,0.783,0.771,0.715,0.616,0.466,0.255,0.022,0,0,0,0,0,0,0,0,0], index=idx)} 
cfPV = pd.DataFrame(d) 


idx = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23] 
d1 = { 
'day': pd.Series(['01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14'], index=idx), 
'hour':pd.Series(['00:00:00', '01:00:00', '02:00:00', '03:00:00', '04:00:00', '05:00:00', '06:00:00', '07:00:00', '08:00:00', '09:00:00', '10:00:00', '11:00:00', '12:00:00', '13:00:00', '14:00:00', '15:00:00', '16:00:00', '17:00:00', '18:00:00', '19:00:00', '20:00:00', '21:00:00', '22:00:00', '23:00:00'], index=idx), 
'output':pd.Series([0.528,0.512,0.51,0.448,0.62,0.649,0.601,0.564,0.541,0.515,0.502,0.522,0.57,0.638,0.66,0.629,0.589,0.544,0.506,0.471,0.448,0.438,0.443,0.451], index=idx)} 
cfWT = pd.DataFrame(d1) 


prob = LpProblem ("System", LpMinimize) 

CPV = LpVariable ("PVCapacity",0) #PV Capacity in kW 
CWT = LpVariable ("WTurCapacity",0) #WT Capacity in kW 
CBA = LpVariable ("BatteryCapacity",0) #Battery Capacity kW 

prob+= 63.128*CPV + 88.167*CWT + 200*CBA, "TotalCostSystem" 

xEne = 0 
xREin = 0 
xBin = 0 
xBout = 0 
SOCB = 0 
xPEMin = 0 
xOvEn = 0 
xSum = 0 

CPEM = 230 

for i in idx: 

    xEne = (CPV*cfPV['output'][i]+CWT*cfWT['output'][i]) 

    #Low limit for Variables 
    prob += (CPV*cfPV['output'][i]+CWT*cfWT['output'][i]) >= 0 
    prob += xREin >= 0 
    prob += xBin >= 0 
    prob += xBout >= 0 
    prob += SOCB >= 0 
    prob += xPEMin >= 0 
    prob += xOvEn >= 0 
    prob += xSum >= 0 
    prob += CBA >= SOCB 
    prob += xBin <= (CBA - SOCB) 
    prob += xBout <= SOCB 

    #Cases 

    #Case 1 xEne > CPEM 
    if xEne >= CPEM: 

     xREin = CPEM 
     xBout = 0 
     xOvEn = xEne - CPEM 

     #Case 1.1 xOvEn < CBA - SOCB 
     if (value(xOvEn) <= (CBA - value(SOCB))): 
      xBin = xOvEn 

     #Case 1.2 xOvEn > CBA -SOCB 
     else: 
      xBin = CBA - SOCB 

    #Case 2 xEne < CPEM 
    else: 
     xREin = xEne 
     xBin = 0 
     xOvEn = 0 

     #Case 2.1 SOCB > CPEM - xREin 
     if (value(SOCB) >= (CPEM - value(xREin))): 
      xBout = (CPEM - xREin) 

     #Case 2.2 SOCB < CPEM - xREin 
     else: 

      xBout = SOCB 

    SOCB = SOCB + xBin - xBout 
    xPEMin = xREin + xBout 

    xSum += xPEMin 

prob += xSum >= 5000 


prob.writeLP("PVWTBattSyste.lp") 

prob.solve() 

주어진 해결책은 항상 첫 번째 조건을 충족합니다. 또한 조건이 충족되지 않으면 (예 : CPEM을 50000000000000으로 변경) if가 true로 작동합니다.

미리 감사드립니다.

Answer 1

if elif else을 서로 구별하려면 elif 문을 사용해야합니다. 또한 이것은 다른 언어로 사용되는 스위치 케이스의 영향을받습니다. 그러나 다른 한편으로 들여 쓰기를 사용하고 있습니까? seriuosly처럼 파이썬은 모두 들여 쓰기에 관한 것입니다. 프로그램이 바로 다음 줄에 빈 if을 뛰어 들여 쓰기없이

은 또한 당신은, 다른 말로의 statemt 경우

#Case 1.2 xOvEn > CBA -SOCB 
else: 
    xBin = CBA - SOCB 

#Case 2 xEne < CPEM 
else: 
    xREin = xEne 
    xBin = 0 
    xOvEn = 0 

없이 하나 else 문이있다.