Re: [Haskell-cafe] Help to create a function to calculate a n element moving average ??
S. Doaitse Swierstra schrieb: 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 _ _ [] = [] Moving average can be interpreted as convolution (*): [1/n,1/n,1/n,...,1/n] * xs You may drop the first (n-1) values, since they average over initial padding zeros. Convolution is associative and you can decompose the first operand into simpler parts: 1/n · [1,1,1,...] * [1,0,0,,0,0,-1] * xs = 1/n · integrate ([1,0,0,,0,0,-1] * xs) Convolution is commutative, thus you could also write = 1/n · [1,0,0,,0,0,-1] * integrate xs but then integration of xs will yield unbounded values and thus higher rounding errors. This yields: movingAverage :: Int - [Float] - [Float] movingAverage n = drop (n-1) . map (/ fromIntegral n) . scanl1 (+) . (\xs - zipWith (-) xs (replicate n 0 ++ xs)) This should be the same as the implementation above, but maybe a bit nicer. :-) ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Help to create a function to calculate a n element moving average ??
On 29 sep 2010, at 00:58, o...@cs.otago.ac.nz wrote: 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 I very very carefully avoided doing any such thing in my example code. For each output result, my code does two additions and one division. Yours does one addition, one subtraction, and one division, for the required case n = 3. The way I formulated it, each calculation is independent. The way you've formulated it, the error in one calculation accumulates into the next. NOT a good idea. If this an issue then: module MovingAverage where movingAverage :: [Float] - [Float] movingAverage (x:y:l) = movingAverage' x y l where movingAverage' x y (z:zs) = (x+y+z)/3:movingAverage' y z zs movingAverage' _ _ _ = [] movingAverage _ = [] has far fewer pattern matches, Doaitse ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Help to create a function to calculate a n element moving average ??
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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Help to create a function to calculate a n element moving average ??
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
[Haskell-cafe] Help to create a function to calculate a n element moving average ??
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 .. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Help to create a function to calculate a n element moving average ??
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
