Question

我有一个嵌套列表：

regions = [[1,2,3],[3,4],[1,3,4],[1,2,3,5]]

我想删除此嵌套列表中包含在另一个列表中的每个列表，即[1,3,4]中包含的[3,4]和[1,2,3中包含的[1,2,3] ，3,5]，因此结果为：

result = [[1,3,4],[1,2,3,5]]

到目前为止，我正在做

regions_remove = []
for i,reg_i in enumerate(regions):
    for j,reg_j in enumerate(regions):
        if j != i and list(set(reg_i)-set(reg_j)) == []:
regions_remove.append(reg_i)
regions = [list(item) for item in set(tuple(row) for row in regions) -
               set(tuple(row) for row in regions_remove)]

我得到了：regions = [[1, 2, 3, 5], [1, 3, 4]]，这是一个解决方案，但是我想知道什么是最有效的pythonic解决方案？

（很抱歉以前没有发布我的完整代码，对此我是新手。

Answer 1

我绝对忽略了一条更简单的路线，但是这种方法有效

列表理解

from itertools import product

l = [[1,2,3],[3,4],[1,3,4],[1,2,3,5]]
bad = [i for i in l for j in l if i != j if tuple(i) in product(j, repeat = len(i))]
final = [i for i in l if i not in bad]

详细说明

from itertools import product
l = [[1,2,3],[3,4],[1,3,4],[1,2,3,5]]
bad = []
for i in l:
    for j in l:
        if i != j:
            if tuple(i) in product(j, repeat = len(i)):
                bad.append(i)

final = [i for i in l if i not in bad]
print(final)

[[1, 3, 4], [1, 2, 3, 5]]

Answer 2

这是一个具有列表理解和sudo npm install <package>函数的解决方案：

all()

返回：

nested_list = [[1,2,3],[3,4],[1,3,4],[1,2,3,5],[2,5]]
result = list(nested_list)      #makes a copy of the initial list
for l1 in nested_list:          #list in nested_list
    rest = list(result)         #makes a copy of the current result list
    rest.remove(l1)             #the list l1 will be compared to every other list (so except itself)
    for l2 in rest:             #list to compare
        if all([elt in l2 for elt in l1]): result.remove(l1)
                                #if all the elements of l1 are in l2 (then all() gives True), it is removed

更多帮助

all（）内置函数：https://docs.python.org/2/library/functions.html#all

复制列表：https://docs.python.org/2/library/functions.html#func-list

列表理解：https://www.pythonforbeginners.com/basics/list-comprehensions-in-python

如果列表包含在同一嵌套列表Python的另一个列表中，则将其删除

2 个答案: