haskell - 在 haskell 中加入或合并函数

Question

Haskell 中是否有一个函数可以执行 SQL 连接或 R 合并的等效操作？

基本上我有两个元组列表，并想根据它们的键“压缩”它们。我知道每个键只有一个或零值

a = [(1, "hello"), (2, "world")]
b = [(3, "foo"), (1, "bar")]

得到类似的东西

 [ (Just (1, "hello), Just (1, "bar))
 , (Just (2, "world), Nothing)
 , (Nothing         , Just (3, "foo"))
 ]

Answer 1

使用有序集(列表)和Ord key

key = fst

fullOuterJoin xs [] = map (\x -> (Just x, Nothing)) xs
fullOuterJoin [] ys = map (\y -> (Nothing, Just y)) ys
fullOuterJoin xss@(x:xs) yss@(y:ys) =
  if key x == key y
    then (Just x, Just y): fullOuterJoin xs ys
    else
      if key x < key y
        then (Just x, Nothing): fullOuterJoin xs yss
        else (Nothing, Just y): fullOuterJoin xss ys

(复杂度为 O(n+m) 但如果您必须排序，则为 O(n log n + m log m))

例子

setA = [(1, "hello"), (2, "world")]
setB = [(1, "bar"), (3, "foo")]

*Main> fullOuterJoin setA setB
[(Just (1,"hello"),Just (1,"bar")),(Just (2,"world"),Nothing),(Nothing,Just (3, "foo"))]

(显然有 sort 支持

fullOuterJoin' xs ys = fullOuterJoin (sort xs) (sort ys)

正如@Franky 所说，您可以避免 if，例如

fullOuterJoin xs [] = [(Just  x, Nothing) | x <- xs]
fullOuterJoin [] ys = [(Nothing, Just  y) | y <- ys]
fullOuterJoin xss@(x:xs) yss@(y:ys) =
  case (compare `on` key) x y of
    EQ -> (Just  x, Just  y): fullOuterJoin xs  ys
    LT -> (Just  x, Nothing): fullOuterJoin xs  yss
    GT -> (Nothing, Just  y): fullOuterJoin xss ys