python - 从元组列表中查找空间顺序的最有效方法？

Question

我有一个圆增长算法(带闭合链接的线增长)，其中在每次迭代时在现有点之间添加新点。

每个点的链接信息以元组的形式存储在列表中。该列表会迭代更新。

问题:

• 将这些点的空间顺序作为列表返回的最有效方法是什么？

• 我是否需要在每次迭代时计算整个顺序，或者有没有办法以有序的方式将新点累积插入到该列表中？

我能想到的是以下内容:

tuples = [(1, 4), (2, 5), (3, 6), (1, 6), (0, 7), (3, 7), (0, 8), (2, 8), (5, 9), (4, 9)]

starting_tuple = [e for e in tuples if e[0] == 0 or e[1] == 0][0]
## note: 'starting_tuple' could be either (0, 7) or (0, 8), starting direction doesn't matter

order = list(starting_tuple) if starting_tuple[0] == 0 else [starting_tuple[1], starting_tuple[0]]
## order will always start from point 0


idx = tuples.index(starting_tuple)
## index of the starting tuple


def findNext():
    global idx
    for i, e in enumerate(tuples):
        if order[-1] in e and i != idx:
            ind = e.index(order[-1])
            c = 0 if ind == 1 else 1
            order.append(e[c])
            idx = tuples.index(e)


for i in range(len(tuples)/2):
    findNext()

print order

它可以工作，但既不优雅(非 pythonic)也不高效。 在我看来，递归算法 可能更合适，但遗憾的是我不知道如何实现这样的解决方案。

另外，请注意我使用的是 Python 2，并且只能访问完整的 python 包(没有 numpy)

Answer 1

与递归相比，这对我来说更像是一个字典和生成器问题:

from collections import defaultdict

def findNext(tuples):
    previous = 0
    yield previous  # our first result

    dictionary = defaultdict(list)

    # [(1, 4), (2, 5), (3, 6), ...] -> {0: [7, 8], 1: [4, 6], 2: [5, 8], ...}
    for a, b in tuples:
        dictionary[a].append(b)
        dictionary[b].append(a)

    current = dictionary[0][0]  # dictionary[0][1] should also work
    yield current  # our second result

    while True:
        a, b = dictionary[current]  # possible connections

        following = a if a != previous else b  # only one will move us forward

        if following == 0:  # have we come full circle?
            break

        yield following  # our next result

        previous, current = current, following  # reset for next iteration

tuples = [(1, 4), (2, 5), (3, 6), (1, 6), (7, 0), (3, 7), (8, 0), (2, 8), (5, 9), (4, 9)]

generator = findNext(tuples)

for n in generator:
    print n

输出

% python test.py
0
7
3
6
1
4
9
5
2
8
% 

算法目前假定我们有两个以上的节点。