ehm... this doesn't take very many ideas.
x = runif(n=10e6, min=0, max=1000)
x = round(x)
system.time( {
y = x[-1] - x[-length(x)]
})
I get about 0.5 seconds on my old laptop.
HTH
Peter
On Fri, Jan 27, 2012 at 4:15 PM, Kevin Ummel <kevinum...@gmail.com> wrote:
> Hi everyone,
>
> Speed is the key here.
>
> I need to find the difference between a vector and its one-period lag (i.e.
> the difference between each value and the subsequent one in the vector).
> Let's say the vector contains 10 million random integers between 0 and 1,000.
> The solution vector will have 9,999,999 values, since their is no lag for the
> 1st observation.
>
> In R we have:
>
> #Set up input vector
> x = runif(n=10e6, min=0, max=1000)
> x = round(x)
>
> #Find one-period difference
> y = diff(x)
>
> Question is: How can I get the 'diff(x)' part as fast as absolutely possible?
> I queried some colleagues who work with other languages, and they provided
> equivalent solutions in Python and Clojure that, on their machines, appear to
> be potentially much faster (I've put the code below in case anyone is
> interested). However, they mentioned that the overhead in passing the data
> between languages could kill any improvements. I don't have much experience
> integrating other languages, so I'm hoping the community has some ideas about
> how to approach this particular problem...
>
> Many thanks,
> Kevin
>
> In iPython:
>
> In [3]: import numpy as np
> In [4]: arr = np.random.randint(0, 1000, (10000000,1)).astype("int16")
> In [5]: arr1 = arr[1:].view()
> In [6]: timeit arr2 = arr1 - arr[:-1]
> 10 loops, best of 3: 20.1 ms per loop
>
> In Clojure:
>
> (defn subtract-lag
> [n]
> (let [v (take n (repeatedly rand))]
> (time (dorun (map - v (cons 0 v))))))
>
>
>
>
>
> [[alternative HTML version deleted]]
>
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