haskell - 查找列表中的最大列表

Question

我试图获取列表中总和最大的列表，然后返回该列表。但是当我用

调用该函数时

max_list [[1,2],[3,6],[10,34,5]] 

它给了我错误:

Exception: a4.hs:65:1-64: Non-exhaustive patterns in function max_list

这是代码:

max_num :: [Int] -> Int
max_num [x] = x
max_num (x:xs) | (max_num xs) > x = maxVal xs
               | otherwise        = x

max_list :: [[Int]] -> [Int]
max_list [[a]] = head(filter (\x -> (sum_int x) == (max_num [[a]]) [[a]])

我的逻辑是这样的:我会的

1. 对子列表中的元素求和
2. 比较该元素以查看它是否等于列表的最大值
3. 过滤掉不等于最大值的值

调用示例:

head (filter (\x -> (sum x) == 11) [[1,3],[4,7],[2,5]])
> [4,7]

因此，在这种情况下，我事先计算了值 11，每个元素的总和为 [4, 11, 7]，它将给出总和等于最大值的值

Answer 1

Data.List 中有一个名为 maximumBy 的函数，其签名为

maximumBy :: (a -> a -> Ordering) -> [a] -> a

和Data.Function具有带有签名的on

on :: (b -> b -> c) -> (a -> b) -> a -> a -> c

应用 compare 函数 (compare::Ord a => a -> a -> Ordering)，我们可以看到这正是您想要的寻找。

import Data.List     (maximumBy)
import Data.Function (on)

{- for clarity:
   compare          :: Ord a          => b -> b -> Ordering
   (compare `on`)   :: Ord b          => (a -> b) -> a  ->  a  -> Ordering
   compare `on` sum :: (Num a, Ord a) =>            [a] -> [a] -> Ordering
   -- well actually [a] is t a for a foldable t, but same diff -}
result = maximumBy (compare `on` sum) [[1,2],[3,6],[10,34,5]] 

<小时/>

要自己实现这一点，您可以编写一个折叠，根据每个值的总和进行比较，递归直到 x 的总和大于其之前的任何值。

myMaximumBySum []     = [] -- degenerate case
myMaximumBySum [x]    = x  -- tautological case
myMaximumBySum (x:xs)
  | sum x > sum (myMaximumBySum xs) = x
  | otherwise                       = myMaximumBySum xs

-- or more naturally:
myMaximumBySum = foldr f []
  where f x acc = if sum x > sum acc then x else acc