Re: Counting occurrences question

Hello! On Wed, Jun 05, 2002 at 07:04:28PM +0100, Andy Fugard wrote:

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===== Original Message From "xoo" ===== hi.. i was just wondering if some body could give a simple equation for the following situation.other than recursion plz..

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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 ...

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Why not combine filter and length appropriately? Kind regards, Hannah.