生成列表中随机元素的一个元素

Question

我试图生成词汇量可以由只'A', 'B', 'C', 'D' or '*'模式的不超过一个重复，人物可以重复任意次数，但美中不足的是，所产生的模式必须有至少一封信。

我试着用随机模块，这是我来到最接近我想要什么：

random.sample(set(vocabulary), 5) 
Out[30]: ['A', 'D', '*', 'B', 'C'] 

理想我想看到的输出，如：

A*** 
ABAB 
ABC* 

等

我该怎么办？

Answer 1

其实你想让你的列表元素的产品，你可以使用itertools.product：

>>> from itertools import product 
>>> voc=['A', 'B', 'C', 'D', '*'] 
>>> for pro in product(voc,repeat=5): 
... print ''.join(pro) 
*AAAA 
*AAAB 
*AAAC 
*AAAD 
*AAA* 
*AABA 
*AABB 
*AABC 
. 
. 

如果你只是想获得平等的子集，你可以使用下面的列表理解的顺风车，例如：

>>> voc=['C', 'D', '*'] 
>>> list(product(voc,repeat=3)) 
[('C', 'C', 'C'), ('C', 'C', 'D'), ('C', 'C', '*'), ('C', 'D', 'C'), ('C', 'D', 'D'), ('C', 'D', '*'), ('C', '*', 'C'), ('C', '*', 'D'), ('C', '*', '*'), ('D', 'C', 'C'), ('D', 'C', 'D'), ('D', 'C', '*'), ('D', 'D', 'C'), ('D', 'D', 'D'), ('D', 'D', '*'), ('D', '*', 'C'), ('D', '*', 'D'), ('D', '*', '*'), ('*', 'C', 'C'), ('*', 'C', 'D'), ('*', 'C', '*'), ('*', 'D', 'C'), ('*', 'D', 'D'), ('*', 'D', '*'), ('*', '*', 'C'), ('*', '*', 'D'), ('*', '*', '*')] 
>>> list(i for i in product(voc,repeat=3)if len(set(i))>1) 
[('C', 'C', 'D'), ('C', 'C', '*'), ('C', 'D', 'C'), ('C', 'D', 'D'), ('C', 'D', '*'), ('C', '*', 'C'), ('C', '*', 'D'), ('C', '*', '*'), ('D', 'C', 'C'), ('D', 'C', 'D'), ('D', 'C', '*'), ('D', 'D', 'C'), ('D', 'D', '*'), ('D', '*', 'C'), ('D', '*', 'D'), ('D', '*', '*'), ('*', 'C', 'C'), ('*', 'C', 'D'), ('*', 'C', '*'), ('*', 'D', 'C'), ('*', 'D', 'D'), ('*', 'D', '*'), ('*', '*', 'C'), ('*', '*', 'D')] 

Answer 2

chars=['A', 'B', 'C', 'D', '*'] 
s="" 
L = len(chars) 
for i in range(0,5): 
    s += chars[random.randrange(0,L)] 

# now ensure that a character is present by setting a random character 
s[random.randrange(0,5)] = chars[random.randrange(0,L-1)]