Question

假设我有N个列表（向量），我想选择其中的x 1<x<[N]（x未预先确定）所以我将获得func（列表）的最大值。

例如：

l1 = [3,4,7,-2]
l2 = [0.5,3,6,2.7]
l3 = [0,5,8,3.6]
mat = [l1, l2, l3]

result = maximize(func, mat)

def func(mat):
    # doing some math between lists. For Example:
    sum_list = list(mat[0])
    for li in mat[1:]:
        sum_list = map(operator.add, sum_list, li)
    accum_min_lst = []
    for i, val in enumerate(sum_list):
        x = sum_list[:i + 1]
        accum_min_lst.append(val - max(x))
    return min(accum_min_lst)

可能的结果：

[l1], [l2], [l3], [l1,l2], [l1,l3], [l2,l3], [l1,l2,l3]

如果我会写一个天真的解决方案并且只运行所有组合，它将需要永远2 ^ N.

我正在尝试使用cvxpy或scipy.optimize.minimize找到解决方案但我觉得很难理解我需要用于我的问题的那种功能，想想也许我应该尝试使用进化算法找到一个近似的答案，或许我应该使用portfolio optimization代替。

Answer 1

我选择使用我自己的版本Evolutionary algorithm它对我来说更直观，而且你可以玩种群大小，世代和变异概率：

from random import choice, random

def stack_overflow_example(self):
    def fitness(trial):
        trial_max = self.func(trial, mat)
        if trial_max > self.best_res:
            self.best_res = trial_max
            return trial_max
        else:
            return -sys.maxint

    def mutate(parent):
        mutation = []
        for p in parent:
            if random() < prob:
                mutation.append(choice([0, 1]))
            else:
                mutation.append(p)
        return mutation

    l1 = [3, 4, 7, -2]
    l2 = [0.5, 3, 6, 2.7]
    l3 = [0, 5, 8, 3.6]
    mat = [l1, l2, l3]

    max_num_of_loops = 1000
    prob = 0.075  # mutation probability
    gen_size = 10  # number of children in each generation
    self.bin_parent = [1] * len(mat)  # first parent all ones
    self.best_res = self.func(self.bin_parent, mat)  # starting point for comparison
    for _ in xrange(max_num_of_loops):
        backup_parent = self.bin_parent
        copies = (mutate(self.bin_parent) for _ in xrange(gen_size))
        self.bin_parent = max(copies, key=fitness)
        res = self.func(self.bin_parent, mat)
        if res >= self.best_res:
            self.best_res = res
            print (">> " + str(res))
        else:
            self.bin_parent = backup_parent
    print("Final result: " + str(self.best_res))
    print("Chosen lists:")
    chosen_lists = self.choose_strategies(self.bin_parent, mat)
    for i, li in enumerate(chosen_lists):
        print(">> list[{}] : values: {}".format(i, li))

def func(self, bin_list, mat):
    chosen_mat = self.bin_list_to_mat(bin_list, mat)
    if len(chosen_mat) == 0:
        return -sys.maxint
    # doing some math between lists:
    sum_list = list(chosen_mat[0])
    for li in chosen_mat[1:]:
        sum_list = map(operator.add, sum_list, li)
    accum_min_lst = []
    for i, val in enumerate(sum_list):
        x = sum_list[:i + 1]
        accum_min_lst.append(val - max(x))
    return min(accum_min_lst)

@staticmethod
def bin_list_to_mat(bin_list, mat):
    chosen_lists = []
    for i, stg in enumerate(mat):
        if bin_list[i] == 1:
            chosen_lists.append(stg)
    return chosen_lists

希望它能帮助别人:)因为我花了一些时间才找到这个解决方案。

Answer 2

这可以表示为MILP并使用任何MILP解算器解决，但我在这里使用PuLP显示解决方案。

首先，让我们通过完成所有组合来看看样本问题的答案：

import itertools

allfuncs = sum([[func(combs) for combs in itertools.combinations(mat, r)] for r in range(1, 4)], [])

max(allfuncs)

答案是-3.3

此解决方案给出了相同的答案，并应扩展到更大的问题：

import pulp

prob = pulp.LpProblem("MaxFunc", pulp.LpMaximize)

allcols = range(0, len(l1))
allrows = range(0, len(mat))

# These will be our selected rows
rowselected = pulp.LpVariable.dicts('rowselected', allrows, cat=pulp.LpBinary)

# Calulate column sums (equivalent to sum_list in the example)
colsums = pulp.LpVariable.dicts('colsums', allcols, cat=pulp.LpContinuous)
for c in allcols:
    prob += colsums[c] == sum(mat[r][c]*rowselected[r] for r in allrows)

# This is our objective - maximimise this
maxvalue = pulp.LpVariable('maxvalue')
prob += maxvalue

# The tricky part - maximise subject to being less than each of these differences
# I'm relatively confident that all these constraints are equivalent
# to calculating the maximum and subtracting that
for c1 in allcols:
    for c2 in allcols[:c1]:
        prob += maxvalue <= colsums[c1] - colsums[c2]

# choose at least one row
prob += pulp.lpSum(rowselected) >= 1

prob.solve()

print(prob.objective.value())

for c in allrows:
    print(rowselected[c].value())

查找具有最大功能值的列表的最佳组合

2 个答案: