假设我有 N 个列表(向量)并且我想选择其中的 x 个 1<x<[N]
(x 未预先确定)
所以我将获得 func(lists) 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]
如果我编写一个简单的解决方案并运行所有组合,它将永远花费 2^N。
我正在尝试使用 cvxpy 找到解决方案或者也许 scipy.optimize.minimize 但我发现很难理解我需要使用哪种函数来解决我的问题,我想也许我应该尝试进化算法来找到一个近似的答案, 或者我应该使用 portfolio optimization相反。
最佳答案
我选择使用我自己的 Evolutionary algorithm 版本它对我来说更直观,而且你可以玩弄人口规模、世代和突变概率:
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
希望它能对某人有所帮助 :) 因为我花了一段时间才找到这个解决方案。
关于python - 寻找具有最大函数值的列表的最佳组合,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35158666/