如何在python中创建一个包含所有可能对的次数相等的列表序列？

Question

给定字母x,y ，有一种方法可以编写一个python函数，该函数返回一个包含所有可能对（ [x,x], [x,y], [y,x], [y,y] ）的列表序列。相等次数？

例如，我可以编写一个函数来接收输入（例如[x,y] ）和每个可能的对出现的次数（例如2 ），然后返回例如列表序列： [xxxyyxyyx]吗？

优选地，我希望函数生成一个“随机”序列，以便它也可以返回：

[y x y y x x x y y]

在两种情况下， [x,x] [x,y] [y,x]和[y,y]恰好出现两次。

理想情况下，我想找到一种解决方案，该解决方案也适用于例如4个字母[x,y,z,w] 。

Answer 1

根据您的编辑，以下内容将指导您所参考的浏览。

import random
import itertools

def generate_sequence(alphabet, num_repeats):

    # generate the permutations of the list of size 2
    # repeated `num_repeats` times
    perms = [(letter1, letter2) for letter1, letter2 in itertools.product(alphabet, alphabet) for i in xrange(num_repeats)]

    # save the original length of `perm` for later
    perm_len = len(perms)

    # randomly choose a starting point and add it to our result
    curr_s, curr_e = random.choice(perms)
    result = [curr_s, curr_e]

    # remove the starting point from the list... on to the next one
    perms.remove((curr_s, curr_e))

    # while we still have pairs in `perms`...
    while len(perms):

        # get all possible next pairs if the end of the current pair
        # equals the beginning of the next pair
        next_vals = [(s,e) for s,e in perms if s == curr_e]

        # if there aren't anymore, we may have exhausted the pairs that
        # start with `curr_e`, in which case choose a new random pair
        if len(next_vals) != 0:
            next_s, next_e = random.choice(next_vals)
        else:
            next_s, next_e = random.choice(perms)

        # remove the next pair from `perm` and append it to the `result`
        perms.remove((next_s, next_e))
        result.append(next_e)

        # set the current pair to the next pair and continue iterating...
        curr_s, curr_e = next_s, next_e

    return result

alphabet = ('x', 'y', 'z')
num_repeats = 2
print generate_sequence(alphabet, num_repeats)

这个输出

['z', 'z', 'z', 'x', 'x', 'y', 'z', 'y', 'y', 'z', 'y', 'x', 'y', 'x', 'z', 'x', 
 'z', 'y', 'x']

Answer 2

有了这个，您可以选择最常见的对：

>>> from collections import Counter
>>> dict(Counter(zip(a, a[1:])).most_common())
{('z', 'z'): 2, ('z', 'y'): 2, ('x', 'y'): 2, ('z', 'x'): 2, ('y', 'y'): 2, ('x', 'x'): 2, ('y', 'x'): 2, ('x', 'z'): 2, ('y', 'z'): 2}

如果您只关心配对：

>>> [t[0] for t in Counter(zip(a, a[1:])).most_common()]
[('z', 'z'), ('z', 'y'), ('x', 'y'), ('z', 'x'), ('x', 'x'), ('y', 'x'), ('x', 'z'), ('y', 'y'), ('y', 'z')]

如果您只关心哪些对出现2次：

>>> [pair for pair, count in Counter(zip(a, a[1:])).items() if count == 2]
[('z', 'z'), ('z', 'y'), ('x', 'y'), ('z', 'x'), ('x', 'x'), ('y', 'x'), ('x', 'z'), ('y', 'y'), ('y', 'z')]