我們來看一下莫凡老大的code在一個不知道答案的時候有沒有一種逼近或者最大效率的一種演算法來求得答案呢
在選擇策略的時候到底可以應用在哪，來做一下小筆記，那麼我的理解是
上次說要應用到決策方式，程式碼大概是說這樣 那大概就是說，
一組DNA可能
[000000000000]
產生下一代，將會做隨機分割，還有變異，隨機交配[000000000000]隨機分割
[1000000/00000]
[00000/0000001]
[1000000/00000]
[00000/0000001]
隨機交配
這將會讓基因越來越棒，
也就是[1000000000/00][0000000000/01]
進行交配[1000000000][01]
最後交配完可能得到
[100000000001]
隨機變異
那麼好的跟好的交配一定更好嗎?[1,0,0,0,0,0,0,0,[0/1],0,0,1]
不一定，或許變異後有機率產生更好的DNA也說不定
所以會有基因變異的東東，
交配完後，隨機在DNA裡面挑取一個基因片段進行變異
那麼應用方面應該就是他所周哥所說的fitens
那麼我們要怎樣定義什麼是好的基因呢?
假設產生下一代的話我們對陣列中的元素做計分的動作，
分數越高 ，就代表基因越優
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| """ | |
| Visualize Microbial Genetic Algorithm to find the maximum point in a graph. | |
| Visit my tutorial website for more: https://morvanzhou.github.io/tutorials/ | |
| """ | |
| import numpy as np | |
| import matplotlib.pyplot as plt | |
| DNA_SIZE = 10 # DNA length | |
| POP_SIZE = 20 # population size | |
| CROSS_RATE = 0.6 # mating probability (DNA crossover) | |
| MUTATION_RATE = 0.01 # mutation probability | |
| N_GENERATIONS = 200 | |
| X_BOUND = [0, 5] # x upper and lower bounds | |
| def F(x): return np.sin(10*x)*x + np.cos(2*x)*x # to find the maximum of this function | |
| class MGA(object): | |
| def __init__(self, DNA_size, DNA_bound, cross_rate, mutation_rate, pop_size): | |
| self.DNA_size = DNA_size | |
| DNA_bound[1] += 1 | |
| self.DNA_bound = DNA_bound | |
| self.cross_rate = cross_rate | |
| self.mutate_rate = mutation_rate | |
| self.pop_size = pop_size | |
| # initial DNAs for winner and loser | |
| self.pop = np.random.randint(*DNA_bound, size=(1, self.DNA_size)).repeat(pop_size, axis=0) | |
| def translateDNA(self, pop): | |
| # convert binary DNA to decimal and normalize it to a range(0, 5) | |
| return pop.dot(2 ** np.arange(self.DNA_size)[::-1]) / float(2 ** self.DNA_size - 1) * X_BOUND[1] | |
| def get_fitness(self, product): | |
| return product # it is OK to use product value as fitness in here | |
| def crossover(self, loser_winner): # crossover for loser | |
| cross_idx = np.empty((self.DNA_size,)).astype(np.bool) | |
| for i in range(self.DNA_size): | |
| cross_idx[i] = True if np.random.rand() < self.cross_rate else False # crossover index | |
| loser_winner[0, cross_idx] = loser_winner[1, cross_idx] # assign winners genes to loser | |
| return loser_winner | |
| def mutate(self, loser_winner): # mutation for loser | |
| mutation_idx = np.empty((self.DNA_size,)).astype(np.bool) | |
| for i in range(self.DNA_size): | |
| mutation_idx[i] = True if np.random.rand() < self.mutate_rate else False # mutation index | |
| # flip values in mutation points | |
| loser_winner[0, mutation_idx] = ~loser_winner[0, mutation_idx].astype(np.bool) | |
| return loser_winner | |
| def evolve(self, n): # nature selection wrt pop's fitness | |
| for _ in range(n): # random pick and compare n times | |
| sub_pop_idx = np.random.choice(np.arange(0, self.pop_size), size=2, replace=False) | |
| sub_pop = self.pop[sub_pop_idx] # pick 2 from pop | |
| product = F(self.translateDNA(sub_pop)) | |
| fitness = self.get_fitness(product) | |
| loser_winner_idx = np.argsort(fitness) | |
| loser_winner = sub_pop[loser_winner_idx] # the first is loser and second is winner | |
| loser_winner = self.crossover(loser_winner) | |
| loser_winner = self.mutate(loser_winner) | |
| self.pop[sub_pop_idx] = loser_winner | |
| DNA_prod = self.translateDNA(self.pop) | |
| pred = F(DNA_prod) | |
| return DNA_prod, pred | |
| plt.ion() # something about plotting | |
| x = np.linspace(*X_BOUND, 200) | |
| plt.plot(x, F(x)) | |
| ga = MGA(DNA_size=DNA_SIZE, DNA_bound=[0, 1], cross_rate=CROSS_RATE, mutation_rate=MUTATION_RATE, pop_size=POP_SIZE) | |
| for _ in range(N_GENERATIONS): # 100 generations | |
| DNA_prod, pred = ga.evolve(5) # natural selection, crossover and mutation | |
| # something about plotting | |
| if 'sca' in globals(): sca.remove() | |
| sca = plt.scatter(DNA_prod, pred, s=200, lw=0, c='red', alpha=0.5); plt.pause(0.05) | |
| plt.ioff();plt.show() |
弔詭時間
遺傳演算法最佳化高頻交易策略
額從哪裡下手呢
對策略
接下來要怎麼進行的策略分析.....
對策略
接下來要怎麼進行的策略分析.....
