Computing Moving Average in Haskell - haskell

Question

I'm working on learning Haskell, so I tried to implement a moving average function. Here is my code:
mAverage :: Int-> [Int] -> [Float]
mAverage x a = [fromIntegral k / fromIntegral x | k <- rawAverage]
where
rawAverage = mAverage' x a a
-- First list contains original values; second list contains moving average computations
mAverage' :: Int -> [Int] -> [Int] -> [Int]
mAverage' 1 a b = b
mAverage' x a b = mAverage' (x - 1) a' b'
where
a' = init a
b' = zipWith (+) a' (tail b)
where the user calls mAverage with a length for each average and the list of values (e.g. mAverage 4 [1,2..100]).
However, when I run the code on the input mAverage 4 [1,2..100000], I get that it takes 3.6 seconds in ghci (using :set +s) and uses a gigabyte of memory. This seems very inefficient to me, as the equivalent function takes a fraction of a second in Python. Is there some way that I could make my code more efficient?

Answer 1

If you want to learn something new you can take a look at this nice solution for Moving Average problem. It is written by one of my students so I won't claim authorship. I really like it because it's very short. The only problem here is average function. Such functions are known to be bad. Instead you can use Beautiful folds by Gabriel Gonzalez. And yes, this function takes O(k) time (where k is size of window) for calculating average of window (I find it better because you can face floating point errors if you try to add only new element to window and subtract last). Oh, it also uses State monad :)
{-# LANGUAGE UnicodeSyntax #-}
module MovingAverage where
import Control.Monad (forM)
import Control.Monad.State (evalState, gets, modify)
moving :: Fractional a ⇒ Int → [a] → [a]
moving n _ | n <= 0 = error "non-positive argument"
moving n xs = evalState (forM xs $ \x → modify ((x:) . take (n-1)) >> gets average) []
where
average xs = sum xs / fromIntegral n

Answer 2

Here is a straightforward list-based solution which is idiomatic and fast enough, though requires more memory.
import Data.List (tails)
mavg :: Fractional b => Int -> [b] -> [b]
mavg k lst = take (length lst-k) $ map average $ tails lst
where average = (/ fromIntegral k) . sum . take k
This solution allows to use any function instead of average in a moving window.
The following solution is less universal but it is constant in space and seems to be the fastest one.
import Data.List (scanl')
mavg :: Fractional b => Int -> [b] -> [b]
mavg k lst = map (/ fromIntegral k) $ scanl' (+) (sum h) $ zipWith (-) t lst
where (h, t) = splitAt k lst
Finally, the solution which uses a kind of Okasaki's persistent functional queue, to keep the moving window. It does make sense when dealing with streaming data, like conduits or pipes.
mavg k lst = map average $ scanl' enq ([], take k lst) $ drop k lst
where
average (l,r) = (sum l + sum r) / fromIntegral k
enq (l, []) x = enq ([], reverse l) x
enq (l, (_:r)) x = (x:l, r)
And as it was mentioned in comments to the original post, do not use ghci for profiling. For example, you won't be able to see any benefits of scanl' in ghci.

Answer 3

Here's a solution for you.
The idea is to scan two lists, one where the averaging window starts, and another where it ends. Getting a tail end of a list costs as much as scanning the part we're skipping, and we're not copying anything. (If the windows size was usually quite large, we could calculate the remaining_data along with counting the sum initial_data, in one go.)
We generate a list of partial sums as described in my comment, then divide them by the windows width to get averages.
While slidingAverage calculates averages for biased position (window width to the right), centeredSlidingAverage calculates centered averages, using half window width to the left and to the right.
import Data.List (splitAt, replicate)
slidingAverage :: Int -> [Int] -> [Double] -- window size, source list -> list of averages
slidingAverage w xs = map divide $ initial_sum : slidingSum initial_sum xs remaining_data
where
divide = (\n -> (fromIntegral n) / (fromIntegral w)) -- divides the sums by window size
initial_sum = sum initial_data
(initial_data, remaining_data) = splitAt w xs
centeredSlidingAverage :: Int -> [Int] -> [Double] -- window size, source list -> list of averages
centeredSlidingAverage w xs = slidingAverage w $ left_padding ++ xs ++ right_padding
where
left_padding = replicate half_width 0
right_padding = replicate (w - half_width) 0
half_width = (w `quot` 2) -- quot is integer division
slidingSum :: Int -> [Int] -> [Int] -> [Int] -- window_sum before_window after_window -> list of sums
slidingSum _ _ [] = []
slidingSum window_sum before_window after_window = new_sum : slidingSum new_sum new_before new_after
where
value_to_go = head before_window
new_before = tail before_window
value_to_come = head after_window
new_after = tail after_window
new_sum = window_sum - value_to_go + value_to_come
When I try length $ slidingAverage 10 [1..1000000], it takes less than a second on my MBP. Due to the laziness, centeredSlidingAverage takes about the same time.

Answer 4

One simple way of doing it that also uses O(n) complexity
movingAverage :: (Fractional a) => Int -> [a] -> [a]
movingAverage n _ | n <= 0 = error "non-positive argument"
movingAverage n xs = fmap average $ groupBy n xs
where average xs' = sum xs' / fromIntegral (length xs')
groupBy :: Int -> [a] -> [[a]]
groupBy _ [] = []
groupBy n xs = go [] xs
where
go _ [] = []
go l (x:xs') = (x:t) : go (x:l) xs'
where t = take (n-1) l

Answer 5

Another way is to use STUArray.
import Data.Array.Unboxed
import Data.Array.ST
import Data.STRef
import Control.Monad
import Control.Monad.ST
movingAverage :: [Double] -> IO [Double]
movingAverage vals = stToIO $ do
let end = length vals - 1
myArray <- newArray (1, end) 0 :: ST s (STArray s Int Double)
forM_ [1 .. end] $ \i -> do
let cval = vals !! i
let lval = vals !! (i-1)
writeArray myArray i ((cval + lval)/2)
getElems myArray