Avoiding repeated additions: movingAverage :: Int -> [Float] -> [Float] movingAverage n l = runSums (sum . take n $l) l (drop n l) where n' = fromIntegral n runSums sum (h:hs) (t:ts) = sum / n' : runSums (sum-h+t) hs ts runSums _ _ [] = [] Doaitse On 28 sep 2010, at 03:40, Richard O'Keefe wrote:
On 27/09/2010, at 5:20 AM, rgowka1 wrote:
Type signature would be Int -> [Double] -> [(Double,Double)]
Any thoughts or ideas on how to calculate a n-element moving average of a list of Doubles?
Let's say [1..10]::[Double]
what is the function to calculate the average of the 3 elements?
[(1,0),(2,0),(3,2),(4,3)....] :: [(Double,Double)]
moving_average3 (xs0 @ (_ : (xs1 @ (_ : xs2)))) = zipWith3 (\x y z -> (x+y+z)/3) xs0 xs1 xs2
*Main> moving_average3 [1..10] [2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0]
The result is two elements shorter than the original, but that _is_ the definition of moving average after all. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe