Question

我有一个列表，如果项目/元组包含这样的集合：

a_list = [(a, {x}), (b, {y}), (c, {y,z}), (d, {y,z}), (e, {x,y}), (f, {x,y,z})]

样本模式：

pattern = {x,y}

给定了一些绘制，我想从a_list生成一系列适合模式的项目。模式应该用元组的第二部分即集合来实现。

示例：

draws = 2的结果可能是：

result = [(a, {x}), (b, {y})]
# x, y

或

result = [(e, {x,y}), (a, {x})]
# x, y

或

result = [(e, {x,y}), (b, {y})]
# x, y

两种情况都符合2次抽签中的模式{x，y}。

draws = 1的结果只能是：

result = (e, {x,y})
# x, y

因为只有一个绘图来完成模式，只有项目e完全匹配模式。

draws = 7的结果可能是：

result = [(a, {x}), (b, {y}), (e, {x,y}), (f, {x,y,z}), (c, {y,z})]
# x, y, x, y, x, y, y

可以完成这样的功能，如果，如何？

感谢您的帮助！

松饼

Answer 1

如果我从您的问题和您的评论中理解的是正确的，则下面的函数应该为您提供正确的输出/打印，并且应该与您在问题中所期望的相同。

这是我的解决方案

from itertools import groupby
from random import randint

a_list = [('a', {'x'}), ('b', {'y'}), ('c', {'y','z'}), ('d', {'y','z'}), ('e', {'x','y'}), ('f', {'x','y','z'})]

pattern = {'x', 'y'}


def seq_gen(a = [], pattern = set, draws = 0):
    single, multi = [], []

    for i in a:
        if len(i[1]) == 1:
            single.append(i)
        else:
            multi.append(i)

    final = [j for j in single if list(pattern)[0] in j[1] or list(pattern)[1] in j[1]]
    final += [j for j in multi if pattern == j[1]]
    # Debug
    #print(final)

    if draws == 1:
        for i in final:
            if len(i[1]) == 2:
                # for better use, return a list not a tuple
                return "draw(1) => {0}".format([i])

    if draws > len(final):
        k, f = list(), tuple()
        for _, v in groupby(a, lambda x: x[1]):
            # Debug
            #print(list(v))
            k += list(v)[0]

        return "draw({0}) => {1}".format(draws, [tuple(k[x:x+2]) for x in range(0,len(k), 2)])

    if draws == len(final):
        return "draw({0}) => {1}".format(draws, final)

    else:
        aa = []
        while len(aa) != 2:
            element = final[randint(0, len(final) -1)]
            if element not in aa:
                aa.append(element)
        return "draw({0}) => {1}".format(draws, aa)

for i in range(1,8):
    print(seq_gen(a_list, pattern, i))

输出：

draw(1) => [('e', {'x', 'y'})]
draw(2) => [('e', {'x', 'y'}), ('a', {'x'})]
draw(3) => [('a', {'x'}), ('b', {'y'}), ('e', {'x', 'y'})]
draw(4) => [('a', {'x'}), ('b', {'y'}), ('c', {'z', 'y'}), ('e', {'x', 'y'}), ('f', {'x', 'z', 'y'})]
draw(5) => [('a', {'x'}), ('b', {'y'}), ('c', {'z', 'y'}), ('e', {'x', 'y'}), ('f', {'x', 'z', 'y'})]
draw(6) => [('a', {'x'}), ('b', {'y'}), ('c', {'z', 'y'}), ('e', {'x', 'y'}), ('f', {'x', 'z', 'y'})]
draw(7) => [('a', {'x'}), ('b', {'y'}), ('c', {'z', 'y'}), ('e', {'x', 'y'}), ('f', {'x', 'z', 'y'})]

PS：不要犹豫回复您的反馈意见。如果出现问题，我会尝试使用您的新查询进行修复。

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1 个答案: