與假說

Question

在假設隨機抽樣，有一個corresponding sampled_from() strategy到random.choice()：

In [1]: from hypothesis import find, strategies as st 

In [2]: find(st.sampled_from(('ST', 'LT', 'TG', 'CT')), lambda x: True) 
Out[2]: 'ST' 

但是，有一種方法來有random.sample()樣的策略，以產生長度爲N個子序列進行序列的？

In [3]: import random 

In [4]: random.sample(('ST', 'LT', 'TG', 'CT'), 2) 
Out[4]: ['CT', 'TG'] 

Answer 1

你可以這樣做：

permutations(elements).map(lambda x: x[:n]) 

Answer 2

感覺這應該可以用lists策略，但我不能讓它工作。通過編寫代碼sampled_from，我能夠做出似乎可行的事情。

from random import sample 
from hypothesis.searchstrategy.strategies import SearchStrategy 
from hypothesis.strategies import defines_strategy 


class SampleMultipleFromStrategy(SearchStrategy): 
    def __init__(self, elements, n): 
     super(SampleMultipleFromStrategy, self).__init__() 
     self.elements = tuple(elements) 
     if not self.elements: 
      raise ValueError 
     self.n = int(n) 

    def do_draw(self, data): 
     return sample(self.elements, self.n) 

@defines_strategy 
def sample_multiple_from(elements, n): 
    return SampleMultipleFromStrategy(elements, n) 

樣品結果：

>>> find(sample_multiple_from([1, 2, 3, 4], 2), lambda x: True) 
[4, 2]