Question

我有以下代码

# Random Strategy Selection: ramdomly choose a strategy within each player's strategy profile for each game in the input list

def RandomStrategySelection():    # Return the vectors with the selected
     P = pool_of_games
     j=10    #number of iterations
     s=1    #current round
     random_strategy=[]    #combination of randomly chosen strategies for each player 
     random_pool=[]    #pool of selected random strategies for the input vector games
     rp=random_pool
     while s<=j:
         for game in range (0, len(P)):    
             p1=random.choice(P[game][0][0:3])    #random choice within p1(row)'s strategy profile
             p2=random.choice(P[game][0][4:8])    #random choice within p2(column)'s strategy profile
             random_strategy=[p1,p2]
             random_pool.append(random_strategy)
         s=s+1
     return(rp)

def FitnessEvaluation():         # Return the rank of fitness of all evaluated games
    for game in range (0,len(rp)):
        pf1=rp[game][0]
        pf2=rp[game][1]
        fitness=payoff1+payoff2
    return(fitness)


    #fitness: f(G)=(F(G)*j+s)/j - F(G)=pf1+pf2

RandomStrategySelection 会生成一个对象列表，例如

[[0,2][3,1]]

FitnessEvaluation 应该使用该列表，但我不能让它运行。 FitnessEvaluation 似乎无法识别创建的列表，即使我将其存储在 rp 变量中也是如此。有什么想法吗？谢谢！

Answer 1

您将其存储在本地 rp 变量中。这与 RandomStrategySelection 中的本地 rp 变量不同。

处理此问题的标准方法是保存返回值（在调用程序中）并将其传递给下一个，例如：

pool = RandomStrategySelection()
how_fit = FitnessEvaluation(pool)

...并为第二个函数提供一个声明参数的签名：

FitnessEvaluation(rp):
    for game in range (0,len(rp)):
        ....

如何使用函数中生成的变量（列表）作为下一个函数的输入？

1 个答案: