Hello! On Wed, Jun 05, 2002 at 07:04:28PM +0100, Andy Fugard wrote: > >===== Original Message From "xoo" <[EMAIL PROTECTED]> ===== > >hi.. i was just wondering if some body could give a simple equation for the > following situation.other than recursion plz..
> >occurrences :: Eq a => a -> [a] -> [a] > >--occurrences xs ys returns the number of times that xs occurs in ys > You may find it easier if you make > occurrences :: Eq a => a -> [a] -> Integer > since it would seem it is to return a number, and not another list! Yep. And not call the parameter for the single 'a' "xs". That's misleading. > Also I would guess the function will have a form something like > occurrences x xs = foldr (countOp x) 0 xs > where countOp :: Eq a => a -> a -> Integer -> Integer > ... > [...] Why not combine filter and length appropriately? Kind regards, Hannah. _______________________________________________ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
