Question

我想用另一个可变长度的列表过滤嵌套列表。如果子列表中的任何项与筛选器列表的任何元素匹配，则应排除子列表。以下代码适用于我，但有一个＆＃34;清洁＆＃34;这项任务的解决方案？

the_list = [['blue'], ['blue', 'red', 'black'], ['green', 'yellow'], ['yellow', 'green'], ['orange'], ['white', 'gray']]
filters = ['blue', 'white']

filtered_list = []
for sublist in the_list:
    for item in sublist:
        if item in filters:
            break
        filtered_list.append(sublist)
        break

预期产出：

filtered_list = [['green', 'yellow'], ['yellow', 'green'], ['orange']]

Answer 1

使用any可能会更加语义化。

for sublist in the_list:
    if any(item in filters_exclude for item in sublist):
        continue
    filtered_list.append(sublist)

也许是矫枉过正，但你甚至可以将其纳入自己的功能然后使用内置filter

def good_list(some_list):
    return not any(item in filters_exclude for item in some_list)

filtered_list = filter(good_list, the_list)

这应该完成你描述的目标。但是，您编写的代码存在潜在问题，如评论中的mentioend。

Answer 2

您可以使用filter和map将其作为＆＃34; one-liner＆＃34;。它没有提高可读性，但它有效：

filters_exclude = [2, 4]
initial_list = [[1, 2, 3, 4], [1, 2, 3], [2, 3, 4, 5]]

final = list(map(lambda x: filter(lambda y: y not in filters_exclude, x), initial_list)

示例：

>>> filters_exclude = [2, 4]
>>> map(lambda x: filter(lambda y: y not in filters_exclude, x), [[1, 2, 3, 4], [1, 2, 3]])
[[1, 3], [1, 3]]

Answer 3

您可以使用列表理解：

the_list = [['blue'], ['blue', 'red', 'black'], ['green', 'yellow'],['orange'], ['white', 'gray']]
filters = ['blue', 'white']
final_l = [i for i in the_list if not any(b in filters for b in i)]

输出：

[['green', 'yellow'], ['orange']]

或者，使用过滤器：

final_l = filter(lambda x:not any(b in filters for b in x), the_list)

Answer 4

您可以使用条件列表理解。

>>> [sublist for sublist in the_list 
     if all(filter not in set(sublist) for filter in filters)]
[['green', 'yellow'], ['orange']]

Answer 5

您还可以使用filter() Set intersection列出与列表filters没有交集的列表：

>>> the_list = [['blue'], ['blue', 'red', 'black'], ['green', 'yellow'], ['yellow', 'green'], ['orange'], ['white', 'gray']]
>>> filters = ['blue', 'white']
>>> list(filter(lambda x: not set(x).intersection(filters), the_list))
[['green', 'yellow'], ['yellow', 'green'], ['orange']]

或者理解：

>>> [x for x in the_list if not set(x).intersection(filters)]
[['green', 'yellow'], ['yellow', 'green'], ['orange']]

Answer 6

使用套装。

the_list = map(set, the_list)
filters = set(filters)

fl = []
for sub in the_list:
    sub = sub.difference(filters)
    if sub:
        fl.append(list(sub))

Answer 7

这与您拥有的非常接近

    the_list = [['blue'], ['blue', 'red', 'black'], ['green', 'yellow'], 
    ['yellow', 'green'], ['orange'], ['white', 'gray']]
    filters = ['blue', 'white']

    filtered_list = []

    for sublist in the_list:
        sub_filtered_list=[]
        for item in sublist:
           if item in filters:
                continue
            else:
                sub_filtered_list.extend([item])
        if sub_filtered_list==[]:
            continue
        else:
            filtered_list.append(sub_filtered_list)

    print(filtered_list)

Answer 8

filtered_list=[];
for sublist in the_list:
    if len(list(set(sublist).intersection(filters_exclude)))>0:
        break;
    filtered_list.append(sublist);

set（sublist）.intersection（filters_exclude）返回两个列表的交集。 List（）将该集转换为List。 Len（）返回List的长度。

过滤嵌套列表

8 个答案: