PYTHON，if 语句只满足第一个条件。 纸浆

Question

I am trying to use PuLP to optimize a system, minimizing the cost of it. I am using multiple If's and the problem is that it always meets the first condition. Here is my code. I hope someone can help me, as I am just starting to learn about this language.

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()

The solution given always meets first condition. Also, when the condition is not met (changing CPEM to 50000000000000, for example) the if works as it is true.

Thank you in advance!

Answer 1

You should use elif statement to differentiate one if elif else one from another. Also this have the effect of a switch case used in other languages. But on the other hand, are you using indentation at all? Like seriuosly Python is all about indentation.

Also you have one else statement without an if statemt

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

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

In other words, without indentation your program is just jumping an empty if to the next line.