使用列表列表筛选列表列表

Question

我有两个列表，一个包括所有记录，例如[['eggs', 'milk', 'butter'], ['ham', 'spam', 'milk'], ['cereal', 'skittles']]和一个包含规则[['milk', 'eggs'], ['milk','ham']]。

我试图通过list_of_rules过滤记录，然而，我想捕获[['eggs', 'milk', 'butter'], ['ham', 'spam', 'milk']]，尽管它不完全匹配[['milk', 'eggs'], ['milk','ham']]顺序和额外的项目明智。

records = [['eggs', 'milk', 'butter'], ['ham', 'spam', 'milk'], ['cereal', 'skittles']]

list_of_rules = [['milk', 'eggs'], ['milk','ham']]

# this list comprehension only filters for exact matches

results = [[x for x in L if x in records] for L in list_of_rules] 


# expected output

print(results)
>>[['eggs', 'milk', 'butter'], ['ham', 'spam', 'milk']]

任何和所有的建议都非常感谢。

Answer 1

您可以使用此列表理解：

records = [['eggs', 'milk', 'butter'], ['ham', 'spam', 'milk'], ['cereal', 'skittles']]

list_of_rules = [['milk', 'eggs'], ['milk','ham']]

results = [L for L in records if any(set(R).issubset(L) for R in list_of_rules)]

print(results) # => [['eggs', 'milk', 'butter'], ['ham', 'spam', 'milk']]

它循环每个记录列表L，并检查是否至少存在一个规则列表R (使用内置函数any)，以便R包含在L中(使用set方法issubset)。

Answer 2

您可以使用规则的list of sets，并要求任何带有内部列表的规则交叉点‘与set相同(即集合中的所有项也都存在于内部列表中)：

records = [['eggs', 'milk', 'butter'], ['ham', 'spam', 'milk'], ['cereal', 'skittles']]

list_of_rules = [{'milk', 'eggs'}, {'milk','ham'}]

# this list comprehension only filters for exact matches

# take the full inner list if all things in any rule are in this inner list
results = [ x for x in records if any( p.intersection(x) == p for p in list_of_rules)  
print(results)

输出：

[['eggs', 'milk', 'butter'], ['ham', 'spam', 'milk']]