I was wondering if someone could help me with the performance of this code snippet in Clojure 1.3. I am trying to implement a simple function that takes two vectors and does a sum of products.
So let's say the vectors are X (size 10,000 elements) and B (size 3 elements), and the sum of products are stored in a vector Y, mathematically it looks like this:
Y0 = B0*X2 + B1*X1 + B2*X0
Y1 = B0*X3 + B1*X2 + B2*X1
Y2 = B0*X4 + B1*X3 + B2*X2
and so on ...
For this example, the size of Y will end up being 9997, which corresponds to (10,000 - 3). I've set up the function to accept any size of X and B.
Here's the code: It basically takes (count b)
elements at a time from X, reverses it, maps *
onto B and sums the contents of the resulting sequence to produce an element of Y.
(defn filt [b-vec x-vec]
(loop [n 0 sig x-vec result []]
(if (= n (- (count x-vec) (count b-vec)))
result
(recur (inc n) (rest sig) (conj result (->> sig
(take (count b-vec))
(reverse)
(map * b-vec)
(apply +)))))))
Upon letting X be (vec (range 1 10001))
and B being [1 2 3]
, this function takes approximately 6 seconds to run. I was hoping someone could suggest improvements to the run time, whether it be algorithmic, or perhaps a language detail I might be abusing.
Thanks!
P.S. I have done (set! *warn-on-reflection* true)
but don't get any reflection warning messages.
You are using count many times unnecessary. Below code calculate count one time only
(defn filt [b-vec x-vec]
(let [bc (count b-vec) xc (count x-vec)]
(loop [n 0 sig x-vec result []]
(if (= n (- xc bc))
result
(recur (inc n) (rest sig) (conj result (->> sig
(take bc)
(reverse)
(map * b-vec)
(apply +))))))))
(time (def b (filt [1 2 3] (range 10000))))
=> "Elapsed time: 50.892536 msecs"
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