主页 / 问题 / 1037560
Accepted
GreyKoshak
Asked:2020-10-23 05:34:53 +0000 UTC2020-10-23 05:34:53 +0000 UTC

对元组列表进行排序

  • 772

再会！

有一个元组列表：

temp_list = [(1, 3), (5, 4), (6, 2), (4, 1), (3, 6), (2, 5)]

是否可以以形成链的方式对该列表进行排序：

new_list = [(1, 3), (3, 6), (6, 2), (2, 5), (5, 4), (4, 1)]

有没有一个不错的 python 风格的解决方案？

真诚的，伊戈尔

python

3 个回答

  1. Best Answer

    这个怎么样？

    temp_list = [(1, 3), (5, 4), (6, 2), (4, 1), (3, 6), (2, 5)]
new_list = [min(temp_list)]
for _ in range(len(temp_list) - 1):
    item = new_list[-1]
    new_list.append(
        next(filter(lambda x: item[1] == x[0], temp_list))
    )

print(new_list)
# [(1, 3), (3, 6), (6, 2), (2, 5), (5, 4), (4, 1)]

    通过字典：

    temp_list = [(1, 3), (5, 4), (6, 2), (4, 1), (3, 6), (2, 5)]
d = dict(temp_list)
new_list = [min(temp_list)]
for _ in range(len(temp_list) - 1):
    x = new_list[-1][1]
    new_list.append((x, d.get(x)))

print(new_list)
# [(1, 3), (3, 6), (6, 2), (2, 5), (5, 4), (4, 1)]
    • 4

  2. 一切皆有可能，取决于您的努力程度）但关于美 - 对于每个人来说，有人在效率方面具有美感，有人在代码的简洁中具有美感

    for i in range(len(temp_list)):
    for j in range(len(temp_list)):
        if temp_list[i][1] == temp_list[j][0]:
            try:
                temp_list[i + 1], temp_list[j] = temp_list[j], temp_list[i + 1]
            except IndexError:
                pass
    • 3

  3. 作为一种选择：

    d = { i:j for i,j in [(1, 3), (5, 4), (6, 2), (4, 1), (3, 6), (2, 5)] }

def f(d, k, l):
    nk = d.get(k)
    new_list.append([k, nk])
    k = nk
    l -= 1
    if not l:
        return new_list
    return f(d, k, l)

new_list = []
print(f(d, 1, len(d)))  # [[1, 3], [3, 6], [6, 2], [2, 5], [5, 4], [4, 1]] 

new_list = []
print(f(d, 2, len(d)))  # [[2, 5], [5, 4], [4, 1], [1, 3], [3, 6], [6, 2]]

new_list = []
print(f(d, 3, len(d)))  # [[3, 6], [6, 2], [2, 5], [5, 4], [4, 1], [1, 3]]

new_list = []
print(f(d, 4, len(d)))  # [[4, 1], [1, 3], [3, 6], [6, 2], [2, 5], [5, 4]]

new_list = []
print(f(d, 5, len(d)))  # [[5, 4], [4, 1], [1, 3], [3, 6], [6, 2], [2, 5]]

new_list = []
print(f(d, 6, len(d)))  # [[6, 2], [2, 5], [5, 4], [4, 1], [1, 3], [3, 6]]

new_list = []
print(f(d, 7, len(d)))  # [[7, None], [None, None], [None, None], [None, None], [None, None], [None, None]]
    • 3

相关问题