如何最小化与给定输入分布的距离？

Question

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= 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)每个客户都可以在“A”或“B”或“C”上激活，并且仅当与激活类型相关的布尔值等于 1 时。在输入中，我有最终激活的计数。es：Target_A = 500Target_B = 250Target_C = 250代码中的随机值是优化器的输入，代表以这种方式激活客户端的可能性。我如何才能仅将客户与其中之一联系起来以尊重最终目标？如何最小化实际激活计数与输入数据之间的距离？

Answer 1

你有任何经过测试的例子吗？我认为这可能有效但不确定：

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())