有没有办法避免在遗传算法中选择两个相同的parents？

Question

我正在 GA 中选择两个 parents，问题是当我实现“tournament_selection”函数时，它总是 returns 两个相同的 parents.

初始人口为：

num_emitter_coefficients = 782
sol_per_pop_1=1
sol_per_pop_2=3
sol_per_pop=sol_per_pop_1+sol_per_pop_2   

pop_size_1 = [sol_per_pop_1, num_emitter_coefficients]
pop_size_2 = [sol_per_pop_2, num_emitter_coefficients]

initial_population_1 = np.random.uniform(low=0.0, high=0.0, size=pop_size_1)
initial_population_2 = np.random.uniform(low=0.0, high=0.0, size=pop_size_2)

for row in initial_population_2:
    locations_used_in_this_row = 0
    while locations_used_in_this_row != 5:
        column = np.random.randint(num_emitter_coefficients)
        if row[column] == 0.0:
            row[column] = np.random.rand()*10
            locations_used_in_this_row += 1

population=np.vstack([initial_population_1, initial_population_2])
print('Population: ',population)

假设每个人的分数是：

[44,56,63,34]

而在我的项目中，每个人的得分越高，每个人的适应度就越低。 parents选择功能如下图：

def tournament_selection(population, scores, k=4):
    first_pick = np.random.randint(0,len(population))  
    for second_pick in np.random.randint(0,len(population),k-1):
        if scores[second_pick] < scores[first_pick]:
            winner = second_pick
        else:
            winner = first_pick
    return population[winner,:]

当我实现功能时，

for i in range(0, len(population),2):
    parent_1 = tournament_selection(population, scores)
    parent_2 = tournament_selection(population, scores)

它总是返回两个相同的 parents。你能告诉我如何避免这个问题吗？

也许如果我可以从人口列表中排除'winner'，可以解决选择两个相同parents的问题，但我不知道如何用代码实现它，你能请给我想法？非常感谢。

Answer 1

我认为列表理解是你的答案，无需任何更多信息（我知道遗传分析会变得很麻烦）。

我在您的函数开头添加了列表理解，以从您的主列表中过滤掉以前的获胜者。请记住 .index() 只会 select 在列表中第一次出现，所以如果你有很长的列表并且可能有重复项，你将需要另一种方法来

import numpy as np

num_emitter_coefficients = 782
sol_per_pop_1=1
sol_per_pop_2=3
sol_per_pop=sol_per_pop_1+sol_per_pop_2   

pop_size_1 = [sol_per_pop_1, num_emitter_coefficients]
pop_size_2 = [sol_per_pop_2, num_emitter_coefficients]

initial_population_1 = np.random.uniform(low=0.0, high=0.0, size=pop_size_1)
initial_population_2 = np.random.uniform(low=0.0, high=0.0, size=pop_size_2)

for row in initial_population_2:
    locations_used_in_this_row = 0
    while locations_used_in_this_row != 5:
        column = np.random.randint(num_emitter_coefficients)
        if row[column] == 0.0:
            row[column] = np.random.rand()*10
            locations_used_in_this_row += 1

population=np.vstack([initial_population_1, initial_population_2]).tolist()

prevwinner = []

scores = [44,56,63,34]

def tournament_selection(population, scores, k=4):
    # filter previous winners out
    selectable_parents = [nonwinner for nonwinner in population if nonwinner not in prevwinner]
    if len(selectable_parents) > 1:
        first_pick, second_pick = np.random.randint(len(selectable_parents), size=2)
        first_pick_score = scores[population.index(selectable_parents[first_pick])]
        while second_pick == first_pick:
            second_pick = np.random.randint(0,len(selectable_parents))
        second_pick_score = scores[population.index(selectable_parents[second_pick])]
        if second_pick_score < first_pick_score:
            winner = second_pick
            loser = first_pick
        else:
            winner = first_pick
            loser = second_pick

        # append winner to prevwinner list
        prevwinner.append(population[population.index(selectable_parents[winner])])
        prevwinner.append(population[population.index(selectable_parents[loser])])
        return population[population.index(selectable_parents[winner])], population[population.index(selectable_parents[loser])]
    else: 
        # there is one parent left
        return selectable_parents[0]

for i in range(0, len(population),2):
    parent_1, parent_2 = tournament_selection(population, scores)
    print('parent_1: ', np.unique(np.array(parent_1)))
    print('parent_2: ', np.unique(np.array(parent_2)))

Answer 2

与其从总体中随机选择两次，不如使用 np.random.choice 和 replace=False。

但貌似 random.choice 在 numpy 中只支持一维数组，所以你必须在 np.random.randint. See Numpy: Get random set of rows from 2D array

中使用大小参数

import numpy as np

population= [[0., 1., 0., 0., 0., 0.],
             [0., 0., 0., 0., 1., 0.],
             [0., 0., 1., 0., 0., 0.],
             [0., 0., 0., 1., 0., 0.]]

parent_1, parent_2 = np.random.randint(len(population), size=2)

有没有办法避免在遗传算法中选择两个相同的parents？

Is there a way to aviod picking two same parents in genetic aglorithm?

python

genetic-algorithm