python - Python中的中餐厅流程实现

Question

我已经用 Python 编写了 CRP 问题的代码。问题本身可以在这里找到： http ://cog.brown.edu/~mj/classes/cg168/slides/ChineseRestaurants.pdf

并对其进行简短描述：假设我们要将进入餐厅的人分配给可能无限数量的桌子。如果 $z_i$ 表示为进入餐厅的第 $i$' 个人分配的随机变量，则以下应成立：

对于 $n_a>0$ 的概率 $p(z_i=a|z_1,...,z_{i-1})=\frac{n_a}{i-1+\alpha}，第 $i$' 个人将坐在桌 $a$ 并且概率 $p(z_i=a|z_1,...,z_{i-1})=\frac{\alpha}{i-1+\alpha} $i$'th 人将围坐在一张新桌子旁。

我不太确定我的代码是否正确，因为我很惊讶最终的表格数量有多小。

如果有人能说出实施是否正确，如果是，是否有任何可能的改进，我会很高兴。

import numpy as np
def CRP(alpha,N):
    """Chinese Restaurant Process with alpha as concentration parameter and N 
    the number of sample"""
    #Array which will save for each i, the number of people people sitting
    #until table i
    summed=np.ones(1) #first person assigned to the first table
    for i in range(1,N):
        #A loop that assigns the people to tables

        #randind represent the random number from the interval [1,i-1+alpha]
        randind=(float(i)+alpha)*np.random.uniform(low=0.0, high=1.0, size=1)
        #update is the index for the table that the person should be placed which
        #if greater than the total number, will be placed in a new table
        update=np.searchsorted(summed,randind,side='left')
        if randind>i:
            summed=np.append(summed,i+1)
        else:
            zerovec=np.zeros(update)
            onevec=np.ones(summed.size-update)
            summed+=np.append(zerovec,onevec)
    #This part converts summed array to tables array which indicates the number
    #of persons assigned to that table
    tables=np.zeros(summed.size)
    tables[0]=summed[0]
    for i in range(1,summed.size):
        tables[i]=summed[i]-summed[i-1]
    return tables
a=CRP(0.9999,1000)
print a

Answer 1

Suggestion. Forget about the code you have written. Construct declarative tests of the code. By taking that approach, you start with examples for which you know the correct answer. That would have answered Brainiac's question, for example.

Then write your program. You will likely find that if you start approaching problems this way, you may create sub-problems first, for which you can also write tests. Until they all pass, there is no need to rush on to the full problem.