我有一个客户列表,每个客户都可以通过四种不同的方式“激活”:
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)每个客户都可以在"A“或"B”或"C“上激活,而且只有在与激活类型相关的布尔值等于1时才能激活。
在input中,我有最终激活的计数。es:
Target_A = 500
Target_B = 250
Target_C = 250代码中的随机值是优化器的输入,表示以这种方式激活客户端的可能性。为了尊重最终目标,我如何才能将客户端与其中之一相关联?如何最小化实际激活计数和输入数据之间的距离?
发布于 2019-04-11 03:59:08
你有没有测试过的例子?我认为这可能会起作用,但不确定:
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())https://stackoverflow.com/questions/55610242
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