foldr :: (a -> b -> b) -> b -> [a] -> b
foldr f v [] = v
foldr f v (x:xs) = f x (foldr f v xs)
foldr :: (a -> b -> b) -> b -> [a] -> b
foldr accum initial [] = initial
foldr accum initial (x:xs) = accum x (foldr accum initial xs)
foldr :: (a -> b -> b) -> b -> [a] -> b
-- 어떤 함수 y가 있을 때 f와 v에 대해서 다음 꼴을 만족한다면
y [] = v
y (x:xs) = f x (y xs)
-- y는 foldr로 정의할 수 있습니다.
y = foldr f v
foldr :: (a -> b -> b) -> b -> [a] -> b
-- 어떤 함수 y가 있을 때 f와 v에 대해서 다음 꼴을 만족한다면
y [] = v
y (x:xs) = f x (y xs)
-- y는 foldr로 정의할 수 있습니다.
y = foldr f v
sum :: Num a => [a] -> a
-- v = 0
sum [] = 0
-- f = (+)
sum (x:xs) = x + (sum xs)
foldr :: (a -> b -> b) -> b -> [a] -> b
-- 어떤 함수 y가 있을 때 f와 v에 대해서 다음 꼴을 만족한다면
y [] = v
y (x:xs) = f x (y xs)
-- y는 foldr로 정의할 수 있습니다.
y = foldr f v
product :: Num a => [a] -> a
-- v = 1
product [] = 1
-- f = (*)
product (x:xs) = x * (product xs)
foldr :: (a -> b -> b) -> b -> [a] -> b
foldr f v [] = v
foldr f v (x:xs) = f x (foldr f v xs)
foldl :: (b -> a -> b) -> b -> [a] -> b
foldl f v [] = v
foldl f v (x:xs) = foldl (f v x) xs
foldr :: (a -> b -> b) -> b -> [a] -> b
foldr f v [] = v
foldr f v (x:xs) = f x (foldr f v xs)
foldr (+) 0 [1,2,3]
1 + (foldr (+) 0 [2,3])
1 + (2 + (foldr (+) 0 [3]))
1 + (2 + (3 + (foldr (+) 0 [])))
1 + (2 + (3 + 0))
1 + (2 + 3)
1 + 5
6
foldl :: (b -> a -> b) -> b -> [a] -> b
foldl f v [] = v
foldl f v (x:xs) = foldl (f v x) xs
foldl (+) 0 [1,2,3]
foldl (+) (1 + 0) [2,3]
foldl (+) (2 + (1 + 0)) [3]
foldl (+) (3 + (2 + (1 + 0))) []
(3 + (2 + (1 + 0)))
(3 + (2 + 1))
(3 + 3)
6
foldr :: (a -> b -> b) -> b -> [a] -> b
fstArg a b = a
foldr fstArg 3 [1,2,3]
-> fstArg 1 (foldr fstArg 3 [2,3])
-> 1
foldl' :: (b -> a -> b) -> b -> [a] -> b
foldl' (+) 0 [1,2,3]
foldl' (+) 1 [2,3]
foldl' (+) 3 [3]
foldl' (+) 6 []
6
map :: (a -> b) -> [a] -> [b]
map f [] = []
map f (x:xs) = f x : map f xs
map f (x:xs) = (:) (f x) (map f xs)
map f (x:xs) = ((:) . f) x (map f xs)
map = foldr ((:) . f) []
filter :: (a -> Bool) -> [a] -> [b]
filter p [] = []
filter p (x:xs) = if p x then x:(filter p xs) else (filter p xs)
-- 위 함수를 foldr 형식으로 바꿔보겠습니다.
filter p (x:xs) = (++) (if p x then [x] else []) (filter p xs)
filter p (x:xs) = ((++) . (\x -> if p x then [x] else [])) x (filter p xs)
-- 따라서
filter p = foldr ((++) . (\x -> if p x then [x] else [])) []
y = foldr f v
y [] = v
y (x:xs) = x `f` (y xs)
dropWhile :: (a -> Bool) -> ([a] -> [a])
dropWhile p [] = []
dropWhile p (x:xs) = if p x then dropWhile p xs else x : xs
dropWhile p (x:xs) = (\y ys -> if p y then ys else y : xs) x (dropWhile p xs)
y = foldr f v
y [] = v
y (x:xs) = x `f` (y xs)
dropWhile' :: (a -> Bool) -> ([a] -> ([a], [a]))
dropWhile' p xs = (dropWhile p xs, xs)
dropWhile' p [] = ([], [])
dropWhile' p (x:xs) = if p x then (dropWhile p xs, x:xs) else (x:xs, x:xs)
dropWhile' p (x:xs) =
let (zs, _) = dropWhile' p xs
in if p x then (zs, x:xs) else (x:xs, x:xs)
-- zs = dropWhile p xs
dropWhile' p (x:xs) =
(\y (zs, ys) -> if p y then (zs, y:ys) else (y:ys, y:ys)) x (dropWhile' p xs)
compose :: [x -> x] -> (x -> x)
compose = foldr (.) id
-- 예시)
-- compose [ (+1), (+2), (+3) ]
-- -> (+1) . ( (+2) . ( (+3) . id) )
-- -> (+1) . ( (+2) . (+3) )
-- -> (+1) . (+5)
-- -> (+6)
suml :: [Int] -> Int
suml xs = suml' xs 0
where
suml' [] n = n
suml' (x:xs) n = suml' xs (n + x)
suml' [] = v
suml' (x:xs) = f x (suml' xs)
v 찾기
suml' [] n = v n
-- -> suml' [] n = n = v n
-- -> v = id
suml :: [Int] -> Int
suml xs = suml' xs 0
where
suml' [] n = n
suml' (x:xs) n = suml' xs (n + x)
f 찾기
suml' (x:xs) n = f x (suml' xs) n
-> suml' xs (n + x) = f x (suml' xs) n
-- suml' xs를 y로 치환
-> y (n + x) = f x y n
-> f = \x y n -> y (n + x)
-- 인자를 2개로 맞춥니다.
-> f = \x y -> (\n -> y (n + x))
suml' = foldr (\x y -> (\n -> y (n + x))) id
suml xs = foldr (\x y -> (\n -> y (n + x))) id xs 0