我有一個客戶列表，每個客戶都可以通過四種不同的方式“激活”：n= 1000df = pd.DataFrame(list(range(0,n)), columns = ['Customer_ID'])df['A'] = np.random.randint(2, size=n)df['B'] = np.random.randint(2, size=n)df['C'] = np.random.randint(2, size=n)每個客戶都可以在“A”或“B”或“C”上激活，并且僅當與激活類型相關的布爾值等于 1 時。在輸入中，我有最終激活的計數。es：Target_A = 500Target_B = 250Target_C = 250代碼中的隨機值是優化器的輸入，代表以這種方式激活客戶端的可能性。我如何才能僅將客戶與其中之一聯系起來以尊重最終目標？如何最小化實際激活計數與輸入數據之間的距離？
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慕尼黑的夜晚無繁華
TA貢獻1864條經驗 獲得超6個贊
你有任何經過測試的例子嗎？我認為這可能有效但不確定：
import pandas as pd
import numpy as np
from pulp import LpProblem, LpVariable, LpMinimize, LpInteger, lpSum, value
prob = LpProblem("problem", LpMinimize)
n= 1000
df = pd.DataFrame(list(range(0,n)), columns = ['Customer_ID'])
df['A'] = np.random.randint(2, size=n)
df['B'] = np.random.randint(2, size=n)
df['C'] = np.random.randint(2, size=n)
Target_A = 500
Target_B = 250
Target_C = 250
A = LpVariable.dicts("A", range(0, n), lowBound=0, upBound=1, cat='Boolean')
B = LpVariable.dicts("B", range(0, n), lowBound=0, upBound=1, cat='Boolean')
C = LpVariable.dicts("C", range(0, n), lowBound=0, upBound=1, cat='Boolean')
O1 = LpVariable("O1", cat='Integer')
O2 = LpVariable("O2", cat='Integer')
O3 = LpVariable("O3", cat='Integer')
#objective
prob += O1 + O2 + O3
#constraints
prob += O1 >= Target_A - lpSum(A)
prob += O1 >= lpSum(A) - Target_A
prob += O2 >= Target_B - lpSum(B)
prob += O2 >= lpSum(B) - Target_B
prob += O3 >= Target_C - lpSum(C)
prob += O3 >= lpSum(C) - Target_C
for idx in range(0, n):
prob += A[idx] + B[idx] + C[idx] <= 1 #cant activate more than 1
prob += A[idx] <= df['A'][idx] #cant activate if 0
prob += B[idx] <= df['B'][idx]
prob += C[idx] <= df['C'][idx]
prob.solve()
print("difference:", prob.objective.value())
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