2 Haskell的数据列表比较

我在haskell中有两个数据列表，每个列表包含一个id属性。我想比较两个列表，并返回第一个列表中的所有数据对象，而不是第二个列表中的所有数据对象

以下是一个例子：

---

您可以使用

数据列表中的
相交对象
和
\\

使用
intersectBy
可以创建公共元素列表，然后使用
\\
删除列表中的公共元素

期望结果的数学术语是一个集合在另一个集合中的相对补充，所以我将这样调用主函数

注意：我为书中定义了Eq实例，其中equality表示id的相等，但这只是为了elem的利益。如果该定义不符合您的需要，则将删除中的elem use替换为检查ID的内容。

导入数据。列表仅用于intersectBy函数

import Data.List

type IDLibrary = Int
type Location = String
type Author = String

data Book = Book IDLibrary Author deriving Show
data Library = LibraryBook IDLibrary Location deriving Show

instance Eq Book where
  (==) (Book id _) (Book id' _) = id == id'

listB :: [Book]
listB = [Book 1 "Anthony C", Book 3 "David Weber"]
listL :: [Library]
listL = [LibraryBook 1 "Alaska", LibraryBook 2 "Majorka"]

relativeComplement :: [Book] -> [Library] -> [Book]
relativeComplement xs ys = xs `remove` common xs ys

remove :: [Book] -> [Book] -> [Book]
remove xs ys = filter (not . (`elem` ys)) xs

common :: [Book] -> [Library] -> [Book]
common xs ys = intersectBy sameID xs (intoBooks ys)

sameID :: Book -> Book -> Bool
sameID (Book id1 _) (Book id2 _) = id1 == id2

intoBooks :: [Library] -> [Book]
intoBooks = map turnIntoBook
          where turnIntoBook (LibraryBook id _) = Book id ""

输出：

[第三册“大卫·韦伯”]

PS决不是说我的代码质量很高，我也在学习Haskell。
你能添加一些代码，使这些列表包含的内容更加明确吗？当然可以！type idLibrary=Int type location=String type author=String data Book=Book idLibrary author派生Show data Library=Book idLibrary location派生Show Book1=（Book 1“Anthony C”）Library 1=（Library 2“Alaska”）listB:：[Book]listB=[Book1]listL:[Library]listL=[Library 1]这是无效的haskellcode@EraseMe您应该能够编辑问题（我认为某些类型的问题也需要大写字母才能有效），您可以提供一个示例吗？
import Data.List

type IDLibrary = Int
type Location = String
type Author = String

data Book = Book IDLibrary Author deriving Show
data Library = LibraryBook IDLibrary Location deriving Show

instance Eq Book where
  (==) (Book id _) (Book id' _) = id == id'

listB :: [Book]
listB = [Book 1 "Anthony C", Book 3 "David Weber"]
listL :: [Library]
listL = [LibraryBook 1 "Alaska", LibraryBook 2 "Majorka"]

relativeComplement :: [Book] -> [Library] -> [Book]
relativeComplement xs ys = xs `remove` common xs ys

remove :: [Book] -> [Book] -> [Book]
remove xs ys = filter (not . (`elem` ys)) xs

common :: [Book] -> [Library] -> [Book]
common xs ys = intersectBy sameID xs (intoBooks ys)

sameID :: Book -> Book -> Bool
sameID (Book id1 _) (Book id2 _) = id1 == id2

intoBooks :: [Library] -> [Book]
intoBooks = map turnIntoBook
          where turnIntoBook (LibraryBook id _) = Book id ""