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Functional Programming Lecture 12 - more higher order functions

Functional Programming Lecture 12 - more higher order functions. Recursion over two arguments. The function zip “zips” together two lists, to produce a new list of pairs. zip :: [a] -> [b] -> [(a,b)] zip (x:xs) (y:ys) = (x,y) : zip xs ys zip _ _ = [] or

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Functional Programming Lecture 12 - more higher order functions

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  1. Functional ProgrammingLecture 12 - more higher order functions

  2. Recursion over two arguments The function zip “zips” together two lists, to produce a new list of pairs. zip :: [a] -> [b] -> [(a,b)] zip (x:xs) (y:ys) = (x,y) : zip xs ys zip _ _ = [] or zip (x:xs) (y:ys) = (x,y) : zip xs ys zip [] (y:ys) = [] zip xs [] = [] (Why not zip (x:xs) [] = [] ?) E.g. zip [1,2,3] [‘a’,’b’,’c’] => [(1,’a’), (2,’b’), (3,’c’)] zip [1,2,3] [‘a’,’b’] => [(1,’a’), (2,’b’)]

  3. The function unzip “unzips” a list of pairs to produce a pair of lists. unzip :: [(a,b)] -> ([a],[b]) tricky definition unzip [] = ([],[]) unzip (x,y):zs = (x:xs,y:ys) where (xs,ys) = unzip zs What is unzip [(1,’a’), (2,’b’), (3,’c’)]? => ([1,2,3],[‘a’,’b’,’c’]) What is unzip (zip xs ys) ? => (xs,ys) if xs and ys have same length What is zip (unzip zs)? => type error, unzip has produced a tuple! What is zip (fst (unzip zs)) (snd (unzip zs)) ? => zs

  4. Fold and Foldr Consider computing the sum of the elements of a list [e1, e2, … en] i.e. e1 + (e2 +( … + en )). sumlist :: [Int] -> Int sumlist [x] = x sumlist (x:xs) = x + sumlist xs Now consider computing the maximum of the elements of a list [e1, e2, … en] i.e. e1 max (e2 max ( … max en)). maxlist :: [Int] -> Int maxlist [x] = x maxlist (x:xs) = max x (maxlist xs) In both cases, we are “folding” the binary function + and max through a non-empty list.

  5. Fold Define fold as the function fold :: (a -> a -> a) -> [a] -> a a binary function a list the result fold f [x] = x fold f (x:xs) = f x (fold f xs) Note: although [x] matches the second equation, we will always apply the first equation. Examples fold (||) [False, True, False] => True fold (++) [“Hello “, “world”, “!”] => “Hello world!” fold (*) [1..5] => 120 n.b infix op. Becomes prefix in ‘(‘ ‘)’

  6. But fold has not been defined for an empty list. What would we like to define fold (||) [] and fold (*) [] as ? There is no general answer, we must provide one for each function we give to fold. The function which defines fold for an empty list is foldr.

  7. Foldr Define foldr as the function foldr :: (a -> b -> b) -> b -> [a] -> b a binary function empty value a list the result Note: foldr has a more general type. foldr f st [] = st foldr f st (x:xs) = f x (foldr f st xs) Note: foldr means fold right. Usually the extra argument is the identity element, i.e. the stopping value st is the identity for f. foldr is defined in the standard prelude.

  8. Examples foldr (||) False [False, True, False] => True foldr (&&) True [True, True] => True foldr (&&) False [True, True] => False foldr (*) 1 [1..5] => 120 1 is an identity for * foldr (*) 0 [1..5] => 0 because 0 is a zero for * fac n = foldr (*) 1 [1..n]

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