在元组Python中查找索引

Question

tuple = ('e', (('f', ('a', 'b')), ('c', 'd')))

如何获得职位:(二叉树)

[('e', '0'), ('f', '100'), ('a', '1010'), ('b', '1011' ), ('c', '110'), ('d', '111')]

有什么方法可以indexOf吗？

arvore[0] # = e
arvore[1][0][0] # = f
arvore[1][0][1][0] # = a
arvore[1][0][1][1] # = b
arvore[1][1][0] # = c
arvore[1][1][1] # = d

Answer 1

你需要递归遍历元组(如树):

def traverse(t, trail=''):
    if isinstance(t, str):
        yield t, trail
        return
    for i, subtree in enumerate(t):  # left - 0, right - 1
        # yield from traverse(subtree, trail + str(i))   in Python 3.3+
        for x in traverse(subtree, trail + str(i)):
            yield x

用法:

>>> t = ('e', (('f', ('a', 'b')), ('c', 'd')))
>>> list(traverse(t))
[('e', '0'), ('f', '100'), ('a', '1010'), ('b', '1011'), ('c', '110'), ('d', '111')]

顺便说一句,不要tuple用作变量名.它阴影内置类型/功能tuple.