Question

在假设中，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，有没有办法让random.sample() - 像策略一样从序列中产生长度为N的子序列？

In [3]: import random

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

Answer 1

你可以这样做：

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

Answer 2

使用lists策略感觉应该是可行的，但我无法使其发挥作用。通过编写sampled_from代码，我能够制作似乎有用的东西。

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)

示例结果：

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

随机抽样假设

2 个答案: