[R] Need very fast application of 'diff' - ideas?

[Kevin Ummel]

Thanks. I've played around with pure R solutions. The fastest re-write of diff (for the 1 lag case) I can seem to find is this:

diff2 = function(x) {
  y = c(x,NA) - c(NA,x)
  y[2:length(x)]
}

#Compiling via 'cmpfun' doesn't seem to help (or hurt):
require(compiler)
diff2 = cmpfun(diff2)

But that only gets ~10% improvement over default 'diff' on my machine. Still too slow for my particular application.

I'm inclined towards Michael's suggestion of inline+Rcpp (or some other use of C under the hood).

Could someone show me how to go about doing that?

Thanks!
Kevin

On Jan 28, 2012, at 9:14 AM, Peter Langfelder wrote:

> 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 <kevinummel at 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))))))
>> 
>> 
>> 
>> 
>> 
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>> 
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