我有以下嵌套列表:
mynestedlist = [[[], [], [], ['Foo'], [], []], [[], ['Bar'], [], []], ['FOO'], 'BAR']
我想将其展平到最外面的项目,这将使我在主列表中获得4个项目。但是,我只想要带有文本的项目,并想要删除空的括号列表。
所需的输出:
mynestedlist = [[['Bar']], ['FOO'], 'BAR']
我尝试了以下操作:
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']
答案 0 :(得分:5)
您混合使用列表和字符串,两者都是可迭代的。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,如果在空列表中有空列表,则此解决方案会删除除最外面的列表之外的所有:
>>> nested_empty = [[[],[],[],[],[],[]],[[],['Bar'],[], []], ['FOO'], 'BAR']
>>> clean_nested(nested_empty)
[[['Bar']], ['FOO'], 'BAR']
>>> all_nested_empty = [[[],[],[],[],[],[]],[[],[],[], []], []]
>>> clean_nested(all_nested_empty)
[]
答案 1 :(得分:1)
将执行以下操作:
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']
答案 2 :(得分:0)
通过一些递归操作,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