升序中的连续子列表 [英] contiguous sublists from an ascending sequence
问题描述
给予
xs = [1,2,3,4,6,7,9,10,11]
我打算回来
[[1,2,3,4],[6,7],[9,10,11]]
我认为我可以做到:
groupBy (\x y -> succ x == y) xs
但这返回:
[[1,2],[3,4],[6,7],[9,10],[11]]
一点点搜索从Haskell Data.List建议中返回了以下内容页面
a little bit of searching returned the following from the Haskell Data.List suggestion page.
groupBy :: (a -> a -> Bool) -> [a] -> [[a]]
groupBy rel [] = []
groupBy rel (x:xs) = (x:ys) : groupBy rel zs
where (ys,zs) = groupByAux x xs
groupByAux x0 (x:xs) | rel x0 x = (x:ys, zs)
where (ys,zs) = groupByAux x xs
groupByAux y xs = ([], xs)
他们给出的例子之一就是我所寻找的:
One of the examples they give is exacly what I am looking for:
groupBy (\a b -> a+1 == b) [1,2,3,4,6]
[[1,2,3,4],[6]]
所以我的问题...是否还有另一种方法,而不是像重新定义groupBy
那样看起来有些戏剧性?
So My question... Is there another approach to this, as opposed to re-defining groupBy
as it seems a little dramatic?
编辑...
我决定按以下方式实施它:
I have decided to implement it as follows:
pattern :: (Enum a, Eq a) => (a -> a) -> [a] -> [[a]]
pattern f = foldr g []
where g a [] = [[a]]
g a xs | f a == head (head xs) = (a : head xs): tail xs
| otherwise = [a]:xs
可以进行以下操作:
*Main Map> pattern succ "thisabcdeisxyz"
["t","hi","s","abcde","i","s","xyz"]
*Main Map> pattern (+ 3) [3,6,9,12,1,2,3,2,5,8,23,24,25]
[[3,6,9,12],[1],[2],[3],[2,5,8],[23],[24],[25]]
或完全像group
一样起作用-并不是有任何原因:
or to function exactly like group
-- not that there is any reason:
*Main Map> let xs = [1,1,1,2,3,4,5,6,6,6,5]
*Main Map> group xs == pattern id xs
True
推荐答案
有很多方法可以做到这一点.一种方法是使用文件夹
There are many ways to do that. One way can be using foldr
f = foldr g []
where g a [] = [[a]]
g a xs@(x:xs') | a+1 == head x = (a : x): xs'
| otherwise = [a]:xs
现在尝试实际操作
*Main> f [1,2,3,4,6,7,9,10,11]
[[1,2,3,4],[6,7],[9,10,11]]
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