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list - 根据长度过滤子集?

转载 作者:行者123 更新时间:2023-12-04 10:59:46 25 4
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尝试使用过滤器提取长度为 k 的子集。不知道如何处理它?该列表有 100 个元素 .

subsets :: [a] -> [[a]]
subsets [] = [[]]
subsets (x:xs) = [zs | ys <- subsets xs, zs <- [ys, (x:ys)]]

如果我使用过滤器,这就是我认为的:
filter (length(3)) subsets [1,2,3,4,5]

但我可能错了。如果有不同的方法而不是过滤器?我是haskell的新手,所以不太确定。

最佳答案

当我在过滤中遇到一点困惑时,我会升级并使用 foldr在这种情况下将很简单:

filterLength3 = foldr (\x rs -> if (length x) == 3 then  x : rs else rs) [] 

filterLength3 (subsets [1,2,3,4,5])

output


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

filter应该:
filter ((==3) . length) (subsets [1,2,3,4,5])

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

Edit



想了很多,在chi的帮助下,问 this question我能够解决它:
import Data.List

subsetsOfThree ws = [ [x,y,z] | (x:xs) <- tails ws, (y:ys) <- tails xs, z <- ys ]

一些例子:
  subsetsOfThree [1..3]
=> [[1,2,3]]
subsetsOfThree [1..4]
=> [[1,2,3],[1,2,4],[1,3,4],[2,3,4]]
subsetsOfThree [1..5]
=> [[1,2,3],[1,2,4],[1,2,5],[1,3,4],[1,3,5],[1,4,5],[2,3,4],[2,3,5],[2,4,5],[3,4,5]]
subsetsOfThree [1..10]
=> [[1,2,3],[1,2,4],[1,2,5],[1,2,6],[1,2,7],[1,2,8],[1,2,9],[1,2,10],[1,3,4],[1,3,5],[1,3,6],[1,3,7],[1,3,8],[1,3,9],[1,3,10],[1,4,5],[1,4,6],[1,4,7],[1,4,8],[1,4,9],[1,4,10],[1,5,6],[1,5,7],[1,5,8],[1,5,9],[1,5,10],[1,6,7],[1,6,8],[1,6,9],[1,6,10],[1,7,8],[1,7,9],[1,7,10],[1,8,9],[1,8,10],[1,9,10],[2,3,4],[2,3,5],[2,3,6],[2,3,7],[2,3,8],[2,3,9],[2,3,10],[2,4,5],[2,4,6],[2,4,7],[2,4,8],[2,4,9],[2,4,10],[2,5,6],[2,5,7],[2,5,8],[2,5,9],[2,5,10],[2,6,7],[2,6,8],[2,6,9],[2,6,10],[2,7,8],[2,7,9],[2,7,10],[2,8,9],[2,8,10],[2,9,10],[3,4,5],[3,4,6],[3,4,7],[3,4,8],[3,4,9],[3,4,10],[3,5,6],[3,5,7],[3,5,8],[3,5,9],[3,5,10],[3,6,7],[3,6,8],[3,6,9],[3,6,10],[3,7,8],[3,7,9],[3,7,10],[3,8,9],[3,8,10],[3,9,10],[4,5,6],[4,5,7],[4,5,8],[4,5,9],[4,5,10],[4,6,7],[4,6,8],[4,6,9],[4,6,10],[4,7,8],[4,7,9],[4,7,10],[4,8,9],[4,8,10],[4,9,10],[5,6,7],[5,6,8],[5,6,9],[5,6,10],[5,7,8],[5,7,9],[5,7,10],[5,8,9],[5,8,10],[5,9,10],[6,7,8],[6,7,9],[6,7,10],[6,8,9],[6,8,10],[6,9,10],[7,8,9],[7,8,10],[7,9,10],[8,9,10]]

现在你可以把你的怪物变成一个小木偶:
  length $ subsetsOfThree [1..10]
=> 120
length $ subsetsOfThree [1..20]
=> 1140
length $ subsetsOfThree [1..50]
=> 19600
length $ subsetsOfThree [1..100]
=> 161700
length $ subsetsOfThree [1..500]
=> 20708500

关于list - 根据长度过滤子集?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58262463/

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