python - 处理多级列表 (2)

Question

假设我有一个这样的列表:

li = ['0', ['1', '2'], ['3', ['4', '5'], '6', ['7'], '8'], '9']

我怎样才能实现一个以 (elem, indexes) 的形式生成元组的函数:

• elem是元素
• indexes 是到达该元素所需的索引。

例子:

>>> for x in my_func(li):
    print x
('0', [0])
('1', [1, 0])
('2', [1, 1])
('3', [2, 0])
('4', [2, 1, 0])
('5', [2, 1, 1])
('6', [2, 2])
('7', [2, 3, 0])
('8', [2, 4])
('9', [3])

Answer 1

这是一个嵌套版本

li = ['0', ['1', '2'], ['3', ['4', '5'], '6', ['7'], '8'], '9']
def to_int(lst, path=[]):
    for i, j in enumerate(lst):
        if isinstance(j, list):
            yield list(to_int(j, path+[i]))
        else:
            yield int(j), path+[i]

print(list(to_int(li)))

还有一个扁平化的版本(Python3.3+)

li = ['0', ['1', '2'], ['3', ['4', '5'], '6', ['7'], '8'], '9']
def to_int(lst, path=[]):
    for i, j in enumerate(lst):
        if isinstance(j, list):
            yield from to_int(j, path+[i])
        else:
            yield int(j), path+[i]

for item in to_int(li):
    print (item)

如果不能使用yield from(

li = ['0', ['1', '2'], ['3', ['4', '5'], '6', ['7'], '8'], '9']
def to_int(lst, path=[]):
    for i, j in enumerate(lst):
        if isinstance(j, list):
            for item in to_int(j, path+[i]): yield item
        else:
            yield int(j), path+[i]

for item in to_int(li):
    print (item)

请注意，将空列表作为默认参数通常不是一个好主意，但在这里我们注意不要改变它，所以没关系。