Question

我正在尝试使用Python中的类来模拟Markov链。这是我的代码：

class Chain:
    def __init__(self, probabilities, start):
        self.probs = probabilities
        self.start = start
        self.prob_sum = 0
        self.names = list(self.probs.keys())

    def __iter__(self):
        self.pos = self.start
        return self

    def __next__(self):
        import random

        self.random_num = random.randrange(100)
        prob_l = self.probs[self.pos]
        for ind, prob in enumerate(prob_l):
            self.prob_sum += prob
            if self.random_num < self.prob_sum:
                exclude_names = self.names[:ind] + self.names[ind + 1 :]
                self.prob_sum = 0
                self.pos = exclude_names[ind]
                return self.pos
        return self.pos


chain = Chain({"A": [50, 25], "B": [50, 25], "C": [50, 50]}, "A")
chain_iter = iter(chain)
for k in range(100):
    print(next(chain_iter))

这种工作按预期进行，但有时会重复字母C。由于字典中有两个50，因此应该有50/50的几率去A或B。它永远不应重复。（由于存在问题禁令，我正在将此问题改进为新问题，而不是在评论中进行填充。）

注意：如果有帮助，我正在使用repl.it。

注2：也许与索引编制有关？我不确定。

Answer 1

由于当前节点的索引计算不正确，您会看到重复的C。

这是带有注释的更新代码：

class Chain:
  def __init__(self, probabilities, start):  # only called once
    self.probs = probabilities
    self.start = start
    self.names = list(self.probs.keys())
    
  def __iter__(self):  # only called once
    return self
    
  def __next__(self):  # each iteration
    import random
    self.pos = self.start  # next step
    self.random_num = random.randrange(100) # choice percentile must be in here
    i = self.names.index(self.pos) # get index of this node in big list
    prob_l = self.probs[self.pos]  # get probs
    self.prob_sum = 0  # start prob scan
    for ind, prob in enumerate(prob_l): # probs of going to another node
      self.prob_sum += prob  # until 100%
      if self.random_num < self.prob_sum:  # passed percentile, go to another node
        exclude_names = self.names[:i] + self.names[i + 1:]  # big list without this node
        self.start = exclude_names[ind]  # for next iteration
        break   # found percentile in probs
    return self.pos # add current pos to chain
    
chain = Chain({"A": [50, 25], "B": [50, 25], "C": [50, 50]}, "A")
chain_iter = iter(chain)
for k in range(100):
   print(next(chain_iter), end=" ")

输出（包装）

A B A A B B C B A B A C A B C B A C B A B C B B C 
B A B C B A C A B A A B C B B A B B A B A C B C B 
B C A B A C A C B A C A B A B A C A B B C B A B A 
C A C A C B A A C B A A B B B A A A C A A B A B B

索引有什么问题？

1 个答案: