python - 如何在 python 2 中彻底迭代？

Question

这个问题已经有答案了:

Iterating through list of list (11 个回答)

已关闭 7 年前。

f = (3, 4, 5, {3: 4}, [16, 7, 8])
g = (1, 2, [3, 4, [5, 6], {7: 8}], 9, 10, {11: f}, {12: [1, 2, {3, 4}, [5, 6]]})

我正在尝试递归地迭代 g .

如何在Python中递归地迭代每个元素，这适用于任何具有任何嵌套级别的列表？

我尝试过hasattr , __iter__但不适用于未知级别的嵌套。

f=(3,4,5,{3:4},[6,7,8])
g = (1, 2, [3, 4, [5, 6], {7: 8}], 9, 10, {11: (3, 4, 5, {3: 4}, [16, 7, 8])}, {12: [1, 2, set([3, 4]), [5, 6]]})
print g
for each in g:
    print each
    try:
        if hasattr(each,"__iter__"):
            for ind in each:
                print ind
                if hasattr(ind,"__iter__"):
                    for ind1 in ind:
                        print ind1

Answer 1

要递归地迭代某些内容，您当然需要一个递归函数。这是一个可以分解为 list、tuple、set、frozenset 和 字典的值。它返回一个平面生成器，您可以轻松地在单个 for 循环中进行迭代:

def recursive_iterator(iterable):
    for item in iterable:

        # directly iterable types:
        if type(item) in (list, tuple, set, frozenset):
            for child_item in recursive_iterator(item):
                yield child_item

        # other iterable types where we do not want to iterate over the item directly:
        elif type(item) in (dict,):
            for child_item in recursive_iterator(item.values()):
                yield child_item

        # not iterable types which we want to return as they are:
        else: 
            yield item

以下是使用该函数的方法:

f = (3, 4, 5, {3: 4}, [16, 7, 8])
g = (1, 2, [3, 4, [5, 6], {7: 8}], 9, 10, {11: f}, {12: [1, 2, {3, 4}, [5, 6]]})

for x in recursive_iterator(g):
    print(x, end=" ")

输出如下:

1 2 3 4 5 6 8 9 10 3 4 5 4 16 7 8 1 2 3 4 5 6 

See this code running on ideone.com