First three items of a list in Haskell -

i new haskell, , struggling bit function here. premise simple enough: run through list, , combine each 3 items next each other function , return list results. problem in nice way.

here i've got:

foo :: [int] -> [int] foo xs     | length xs < 3 = []     | otherwise     = n : foo (tail xs)     n = calc (xs!!0) (xs!!1) (xs!!2)  -- function more complicated. calc :: int -> int -> int -> int calc x y z = x + y - (z * 2)  -- , can use this: foo [1,2,3]     -- [-3] foo [1,2,3,4]   -- [-3,-3] foo [1,1,5,3,3] -- [-8,0,2] 

what don't like, 5th line, containing !!'s. feels i'm thinking wrong way, , there should better way of doing this. i'd like

foo (x:y:z:xs)     -- ... 

but fail when list gets less 3 items. so, i'd have declare other patterns when list has fewer items?

also, in case there function foo (there is, seems there 1 everything), i'm not interested in it. i'm trying grok haskell way of doing things, more expanding repetoire of functions.

edit: in js, i'd n = calc.apply(null, take(3, xs)). wonder if haskell has apply takes array , applies function parameters.

edit 2 -- solution: (based on comment below)

foo (x:y:z:xs) = calc x y z : foo (y:z:xs) foo _          = [] 

last pattern match catch-all, if first "fails" fall through , return empty list.

well, foo (x:y:z:xs) plus “too short clause” wouldn't bad solution. be

foo xs = case splitat 3 xs of            ([x,y,z],xs') -> calc x y z : foo (y:z:xs')            _ -> [] 

or, perhaps nicest,

import data.list (tails)  foo xs = [ calc x y z | (x:y:z:_) <- tails xs ]