2010/9/26 rgowka1 <rgow...@gmail.com>: > 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)] > > (1,0) the average is zero as the length is less than 3 > > (2,0) the average is zero as the length is less than 3 > > (3,2) the average is (1+2+3)/3 = 2 > > (4,3) the average is (2+3+4)/3 = 9
movingAverage n xs = map (/n) $ sums n xs sums 1 xs = xs sums n xx@(x:xs) = zipWith (+) xx (sums (n-1) xs) Tests: *Main> movingAverage 1 [1..10] [1.0,2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0] *Main> movingAverage 2 [1..10] [1.5,2.5,3.5,4.5,5.5,6.5,7.5,8.5,9.5] *Main> movingAverage 3 [1..10] [2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0] *Main> movingAverage 4 [1..10] [2.5,3.5,4.5,5.5,6.5,7.5,8.5] Is it right? ;) It is more interesting to create movingAverage in CPS/iteratees style. That way you'll have solid interface with IO world and you can freely alternate between reading moving averages and waiting for another ticket. No magic usafeInterleaveIO, hGetContents, etc, will be required. I did this once for my friend's pet project in Erlang, we jointly developed a whole library of operators over time series - sums, averages, etc. It is simple and fun. ;) _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
