Haskell - List Comprehension - get Input-Elements -
i have problem list comprehension, if input list.
in these iii excercises it's not allowed use: map, filter , concat!!!
part i
requirements:
a funktion f1 gets list xs of tripels (a, b, cs) a , b of type int c of type [int]
the function should generate list of pairs (a · b, b + c), c in cs , in generated list should appear pairs in 1st element bigger 2nd 1 - (a · b) > b + c.
example:
f1 [(10,20,[1,10,100]), (4,5,[5,15,25])] should return following list:
[(200,21),(200,30),(200,120),(20,10)] my attempts:
f1 :: int -> int -> [int] -> [(int, int)] f1 b cs = [(a*b, b+c)| c<-cs, (a*b)>(b+c)] it works fine, not lists input.
so tried several ways, unfortunately not right 1 :-(
f1 :: [(int, int, [int])] -> [(int, int)] 1st approach:
f1 xs = [((xs !! 0)*(xs !! 1), (xs !! 1)+c)| c<-(xs !! 2), ((xs !! 0)*(xs !! 1))>((xs !! 1)+c)] 2nd approach:
f1 let (a, b, cs) = xs = [(a*b, b+c)| c<-cs, (a*b)>(b+c)] 3rd approach:
f1 (a b cs) = [(a*b, b+c)| c<-cs, (a*b)>(b+c)] all 3 don't work!
solution dave4420:
f1 :: [(int, int, [int])] -> [(int, int)] f1 xs = [ (a*b, b+c) | (a, b, cs) <- xs, c <- cs, (a*b)>(b+c) ] part ii
requirements:
a function g1 gets list of pairs of same type , generate plain list out of it.
example:
g1 [(1,2),(3,4),(5,6)] returns [1,2,3,4,5,6] my attempt:
g1 :: [(int, int)] -> [int] g1 xs = [a,b | (a,b)<-xs] i compiling error because a,b in output of list comprehension not have correct syntax.
however can return or b or e.g. a+b:
g1 xs = [a | (a,b)<-xs] or
g1 xs = [a+b | (a,b)<-xs] could please me out of too?
thanks once again
part iii coming...
f1 :: [(int, int, [int])] -> [(int, int)] f1 xs = [ {-todo-} | (a, b, cs) <- xs, c <- cs, {-todo-} ] i have left couple of {-todo-}s fill in.
part ii.
if had write
g1' :: [[int]] -> [int] how that? can modify g1' g1 desire?
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