从嵌套列表中删除空子列表

Removing empty sublists from a nested list

我有以下嵌套列表:

mynestedlist = [[[], [], [], ['Foo'], [], []], [[], ['Bar'], [], []], ['FOO'], 'BAR']

我想将它展平到最外面的项目,这样主列表中就有 4 个项目。但是,我只想要带有文本的项目并且想要删除空的括号列表。

期望的输出:

mynestedlist = [[['Bar']], ['FOO'], 'BAR']

我尝试了以下方法:

newlist = []
for i in mynestedlist:
    for sub in i:
        if sub != []:
            newlist.append(sub)

但是,我得到以下输出:

[['Foo'], ['bar'], 'FOO', 'B', 'A', 'R']

您混合使用了列表和字符串,它们都是可迭代的。您需要在此处显式测试列表,然后递归或使用堆栈:

def clean_nested(l):
    cleaned = []
    for v in l:
        if isinstance(v, list):
            v = clean_nested(v)
            if not v:
                continue
        cleaned.append(v)
    return cleaned

演示:

>>> mynestedlist = [[[], [], [], ['Foo'], [], []], [[], ['Bar'], [], []], ['FOO'], 'BAR']
>>> clean_nested(mynestedlist)
[[['Foo']], [['Bar']], ['FOO'], 'BAR']

请注意,如果空列表中有空列表,此解决方案会删除除最外层列表以外的所有列表

>>> nested_empty = [[[],[],[],[],[],[]],[[],['Bar'],[], []], ['FOO'], 'BAR']
>>> clean_nested(nested_empty)
[[['Bar']], ['FOO'], 'BAR']
>>> all_nested_empty = [[[],[],[],[],[],[]],[[],[],[], []], []]
>>> clean_nested(all_nested_empty)
[]

执行以下操作:

def del_empty(lst):
    if isinstance(lst, list):
        return [del_empty(sub) for sub in lst if sub != []]
    return lst

>>> del_empty(mynestedlist)
[[['Foo']], [['Bar']], ['FOO'], 'BAR']

通过一些递归,可以在保留字符串元素的同时删除任意深度的空列表。

def remove_nested_null(a):
    if not isinstance(a, list):
        return a

    a = map(remove_nested_null, a)
    a = list(filter(None, a))

    return a