Haskell:找到自定义数据类型中的最小值

Question

所以我有一个自定义数据类型,让我们调用它Struct,定义如下:

data Struct = Struct  [SubStruct] deriving (Read, Show)
data SubStruct = SubStruct (Int, Int) deriving (Read, Show)

我需要做的是遍历所有元素Struct并找到基于的最小值fst,然后基于snd.我怎么做？更具体地说,我想得到另一个SubStruct如:

SubStruct (-2,-5),基于代码中的示例.

目前,我开始这样做:

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

data Struct = Struct  [SubStruct] deriving (Read, Show)
data SubStruct = SubStruct (Int, Int) deriving (Read, Show  )

struct s sx = Struct(s:sx)

subStruct :: (Int, Int) -> SubStruct
subStruct (x, y) = SubStruct (x, y)

substructs = Struct $ [subStruct (0,1), subStruct (-2, 3), subStruct (4,-5)]

results xs = sortBy (compare `on` fst) (substructs xs)

但我得到这个错误:

Couldn't match expected type `t -> [(a, b)]'
            with actual type `Struct'
Relevant bindings include
  xs :: t (bound at bbox.hs:15:9)
  results :: t -> [(a, b)] (bound at file.hs:15:1)
The function `substructs' is applied to one argument,
but its type `Struct' has none
In the second argument of `sortBy', namely `(substructs xs)'
In the expression: sortBy (compare `on` fst) (substructs xs)

Answer 1

为什么不使用该unzip功能.如果我们定义一个辅助函数:

unSubStruct :: SubStruct -> (Int, Int)
unSubStruct (SubStruct p) = p

然后返回所需元素的函数可以写为:

getMin :: Struct -> SubStruct
getMin (Struct l) = SubStruct (minimum xs, minimum ys)
  where
    (xs, ys) = unzip $ map unSubStruct l

请注意,这将遍历列表两次.如果您定义minimum适用于对的自己的版本,则可以避免这种情况:

getMin :: Struct -> SubStruct
getMin (Struct l) =
    SubStruct $ foldr1 minPair $ map unSubStruct l
  where
    minPair (x0, y0) (x, y) = (min x0 x, min y0 y)