Though I'm late to the party I might as well post what I have. Anyways yeah foldr/foldl is the answer when you find yourself iterating over a list with an accumulator. aggregate :: [(Price,Volume)] -> [(Price, Volume)] aggregate = reverse . snd . foldl' proc ((0,0),[]) where proc ((!totalp,!totalv),sofar) (Price !p,Volume !v) = let adjustedprice = round (fromIntegral p * v) in ((adjustedprice+totalp,v+totalv), ((Price $ adjustedprice + totalp, Volume $ v+totalv) : sofar)) Just make sure you use foldl' instead of foldl for performance reasons. And also try to prepend to lists whenever possible and then reverse if you need to. On Tue, May 20, 2014 at 10:12 PM, Dimitri DeFigueiredo < defigueiredo@ucdavis.edu> wrote:
Awesome haskellers,
I am coding up a little function that aggregates "ask orders" in a currency exchange. Another way to look at it, is that the function takes as input a histogram or fdf (in list format) and outputs the cumulative distribution cdf (also in list format). So we are kind of "integrating" the input list.
When given a list of asks in order of increasing price, the function returns a list of points in the graph of the total supply curve.
Here's an example:
asks: returned list:
[ (Price 42, Volume 0.5), [ (Price 21, Volume 0.5), (Price 50, Volume 1 ), (Price 21+50=71, Volume 1.5), (Price 55, Volume 0.2)] (Price 21+50+11=82,Volume 1.7)]
the returned list gives us the total supply curve (price = y-axis, quantity/volume = x-axis, so the order is flipped)
Summarizing
* We're adding up the volumes. The last volume on the list is the total volume available for sale. * We calculate the total amount to be paid to buy the current volume (for each item in the list).
I have written up a simple function to do this:
aggregate :: Num a => [(a,a)] -> [(a,a)] aggregate xs = aggregate' 0 0 xs
aggregate' :: Num a => a -> a -> [(a,a)] -> [(a,a)] aggregate' _ _ [] = [] aggregate' accX accY ((x,y):ls) = let accX' = accX + x * y accY' = accY + y
in (accX',accY') : aggregate' accX' accY' ls
main = print $ aggregate [(42,0.5),(50,1),(55,0.2)]
However, this does not look very good to me and it feels like I'm reinventing the wheel.
Question: Is there a better Haskell way to do this? I'm really anal about making it easy to read.
Thanks!
Dimitri _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners