有没有办法避免这么多的list(chain(* list_of_list))?

问题描述

如果我有两个字符串的元组列表的列表.我想将其展平为一个非嵌套的元组列表，我可以这样做:

If I have a list of list of list of tuples of two strings. I want to flatten it out to a non-nested list of tuples, I could do this:

>>> from itertools import chain
>>> lst_of_lst_of_lst_of_tuples = [ [[('ab', 'cd'), ('ef', 'gh')], [('ij', 'kl'), ('mn', 'op')]], [[('qr', 'st'), ('uv', 'w')], [('x', 'y'), ('z', 'foobar')]] ]
>>> lllt = lst_of_lst_of_lst_of_tuples
>>> list(chain(*list(chain(*lllt))))
[('ab', 'cd'), ('ef', 'gh'), ('ij', 'kl'), ('mn', 'op'), ('qr', 'st'), ('uv', 'w'), ('x', 'y'), ('z', 'foobar')]

但是，还有没有嵌套list(chain(*lst_of_lst))到非嵌套元组列表的另一种方法吗?

But is there another way of unpacking to the non-nested list of tuples withou the nested list(chain(*lst_of_lst))?

推荐答案

您可以继续打开包装，直到遇到元组为止:

You could keep unpacking until you hit tuples:

def unpack_until(data, type_):
    for entry in data:
        if isinstance(entry, type_):
            yield entry
        else:
            yield from unpack_until(entry, type_)

然后:

>>> list(unpack_until(lllt, tuple))
[('ab', 'cd'),
 ('ef', 'gh'),
 ('ij', 'kl'),
 ('mn', 'op'),
 ('qr', 'st'),
 ('uv', 'w'),
 ('x', 'y'),
 ('z', 'foobar')]

这篇关于有没有办法避免这么多的list(chain(* list_of_list))?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！