Assume I have N lists (vectors) and I want to choose x of them 1<x<[N]
(x is not predetermined)
so I will get the maximum value of func(lists).
For example:
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)
Possible results:
[l1], [l2], [l3], [l1,l2], [l1,l3], [l2,l3], [l1,l2,l3]
If I'll write a naive solution and just run all the combinations it will take forever 2^N.
I'm trying to find a solution using cvxpy or maybe scipy.optimize.minimize but I find it hard to understand the kind of function I need to use for my problem, thought maybe I should try evolutionary algorithm to find an approximate answer, Or maybe I should use portfolio optimization instead.
I chose to use my own version of Evolutionary algorithm Its just more intuitive for me, plus you can play with the population size, generations and the mutation probability:
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
Hope it'll help someone :) cause it took me a while to find this solution.
This can be formulated as a MILP and solved using any MILP solver, but I show the solution here using PuLP.
First, let's see what the answer is for the sample problem by doing all the combinations:
import itertools
allfuncs = sum([[func(combs) for combs in itertools.combinations(mat, r)] for r in range(1, 4)], [])
max(allfuncs)
The answer is -3.3
This solution gives the same answer and should scale to larger problems:
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())
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With