List 按筛选器列表筛选元素列表

我在想，如果我有一个元素列表

[1，2，…n]

，还有一个过滤器列表，比如

[（>3）(
以及过滤器列表，如
[（>3），（Bool
），因此您应该使用：

filters :: [a] -> [a -> Bool] -> [a]
我们可以进一步将签名推广到：

filters :: Foldable f => [a] -> f (a -> Bool) -> [a]
filters xs fs = filter (\x -> all ($ x) fs) xs
filters:：Foldable f=>[a]->f（a->Bool）->[a]
filters xs fs=filter（\x->all（$x）fs）xs
一个接一个地链接过滤器

filters :: [a] -> [a -> Bool] -> [a]
filters xs ps  =  foldr (\p r -> filter p r) xs ps

或者，更短一些

filters :: [a] -> [a -> Bool] -> [a]
filters  =  foldr filter

或者我们可以使用
foldl'（翻转过滤器）
，它将列表按从左到右的顺序推送到过滤器中。
最近的现有技术（带有指向更多现有技术的指针）：
filters :: Foldable f => [a] -> f (a -> Bool) -> [a]
filters xs fs = filter (\x -> all ($ x) fs) xs
filters :: [a] -> [a -> Bool] -> [a]
filters xs ps  =  foldr (\p r -> filter p r) xs ps

filters :: [a] -> [a -> Bool] -> [a]
filters  =  foldr filter