CODEWITHSUNDEEP
listhaskell
piggyback
piggyback

Reputation: 9264

Count positive values

Here is my function. It checks for positive values, changes them to ones and sums them.

countPositive :: [Integer] -> Integer
countPositive xs = foldr (+) 0 $ map (^0) (filter (>0) xs)

Is there a better strategy to count positive values without using length but just foldr, map and filter?

Upvotes: 0

Views: 1649

Answers (2)

ertes
ertes

Reputation: 101

Foldr doesn't seem right here. You want foldl' instead. This is my solution:

countPos :: (Num a, Ord a) => [a] -> Int
countPos = length . filter (> 0)

As you don't want to use length for some reason you would basically just reinvent it:

countPos xs = sum (1 <$ filter (> 0) xs)

or yet another method:

countPos = foldl' (\x _ -> succ x) 0 . filter (> 0)

There are lots and lots of way to do this. If 100 people answer to this post, chances are you get 100 different ways to do it, but the simplest way is to use filter and length.

Upvotes: 10

huon
huon

Reputation: 102156

Sure, just directly count them with foldr:

countPositive = foldr (\n count -> if n > 0 then count + 1 else count) 0

Or reimplement length with foldr:

countPositive = foldr (const succ) 0 . filter (>0)

Upvotes: 5

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