Python遺傳算法的二進制數

我被要求制定一個遺傳算法的目標，以確定一個8位字符串與最1和0的。 eval函數將返回改變的1數加上所以，例如00000000返回1，00011100回報3，和01100101回報6.這是我有：Python遺傳算法的二進制數

def random_population(): 
    from random import choice 

    pop = ''.join(choice(('0','1')) for _ in range(8)) 
    return pop 

def mutate(dna): 
    """ For each gene in the DNA, there is a 1/mutation_chance chance 
    that it will be switched out with a random character. This ensures 
    diversity in the population, and ensures that is difficult to get stuck in 
    local minima. """ 
    dna_out = "" 
    mutation_chance = 100 
    for c in xrange(DNA_SIZE): 
     if int(random.random()*mutation_chance) == 1: 
      dna_out += random_char() 
     else: 
      dna_out += dna[c] return dna_out 

def crossover(dna1, dna2): 
    """ Slices both dna1 and dna2 into two parts at a random index within their 
    length and merges them. Both keep their initial sublist up to the crossover 
    index, but their ends are swapped. """ 
    pos = int(random.random()*DNA_SIZE) 
    return (dna1[:pos]+dna2[pos:], dna2[:pos]+dna1[pos:]) 

def eval(dna): 
    changes = 0 
    for index, bit in enumerate(dna): 
     if(index == 0): 
      prev = bit 
     else: 
      if(bit != prev): 
       changes += 1 
     prev = bit 
    return changes+1 


#============== End Functions =======================# 


#============== Main ================# changes = 0 

prev = 0 
dna = random_population() 
print "dna: " 
print dna 
print eval(dna)

我無法真正搞清楚遺傳算法部分（交叉/變異）。我應該隨機配對數字，然後隨機選擇一對，保持一對不變，然後在隨機點交叉。然後它將隨着整個人口中的一點點隨機變異而結束。目前的交叉和變異編碼僅僅來自我發現並試圖理解的遺傳算法示例。歡迎任何fhelp。

來源

2013-04-13 Nick

一個人口由許多個人組成 - 我只看到一個「dna」。交叉有助於收斂基因的「子程序」，而突變有助於創造出達到目標所需的錯誤。 – User

此外，你需要一個fittness功能，確定一個人如何可能進行交叉和重組。您可以使用輪盤賭輪http://en.wikipedia.org/wiki/Fitness_proportionate_selection來確定哪些人可以交叉並創建進入下一代的孩子。 – User

我建議的一部分：

該代碼不起作用，但可能傳輸信息。我發現，我喜歡做

# a population consists of many individuals 
def random_population(population_size = 10): 
    from random import choice 

    pop = [''.join(choice(('0','1')) for _ in range(8)) for i in range(population_size)] 
    return pop 

# you need a fitness function 
def fitness(individual): 
    return # a value from 0 up 

def algorithm(): 
    # a simple algorithm somehow alike 
    # create population 
    population = random_population() 
    # this loop must stop after some rounds because the best result may never be reached 
    while goal_not_reached(population) and not time_is_up(): 
     # create the roulette wheel 
     roulette_wheel = map(fitness, population) 
     # highest value of roulette wheel 
     max_value = sum(roulette_wheel) 
     # the new generation 
     new_population = [] 
     for i in range(len(population) - len(new_population)): 
      # create children from the population 
       # choose 2 random values from 0 to max_value and find the individuals 
       # for it in the roulette wheel, combine them to new individuals 
      new_population.append(new_individual) 
     # mutate the population 
     population = map(mutate, new_population)    # a new generation is created

來源

2013-04-13 20:07:44 User

有一件事情是這樣的：

選擇最後一批的熱門人選讓說5
有1個伴侶與2，3， 4，5
有2個隊友與3，4，5
有3個隊友與4和5
有4緊密配合5.一般來說你的人口都已經滿befor如果你讓原來的5進入下一代並且每次交配產生2個後代，你就達到了這一點。一個交配及其相反的雙胞胎。
只要我喜歡在40％到60％的長度上隨機切割染色體的實際交叉點，那麼在下一次交配時，我會選擇另一個隨機點。
我將它們交配後，我會在我的染色體中檢查每一點，並給它大概一個！5％的翻轉或變異機率
有時我也會讓一些最糟糕的兩個伴侶降低我的機會本地最大值或最小值

我希望這對你有一點幫助。

-Jeff

編輯：哦，我這是在4月份的問道。對不起，嚴重挖掘。

來源

2013-07-13 23:13:21

Python遺傳算法的二進制數

回答

相關問題