generating a vector of y_t = \sum_{i = 1}^t (alpha^i * x_{t - i + 1})
On Nov 27, 2011, at 9:25 AM, R. Michael Weylandt wrote:
You also might look at EMA() in the TTR package for a C implementation. (I think it matches your problem but can't promise)
It's pretty close for EMA(1:1000, n=1, ratio=0.5) and 7 times faster. > EMA(1:10, n=1, ratio=0.5) [1] 1.000000 1.500000 2.250000 3.125000 4.062500 5.031250 6.015625 7.007812 8.003906 [10] 9.001953 > loopRec3(1:10, 0.5) [1] 0.500000 1.250000 2.125000 3.062500 4.031250 5.015625 6.007812 7.003906 8.001953 [10] 9.000977 This series converges very quickly to n-1 when the ratio = 0.5 and to n-3 when ratio = 0.25. It's already within 1% at the fifth iteration. There may be a simple analytical approximation that makes the process even more efficient. The asymptotic result appears to be: n-(1-ratio)/ ratio > tail(EMA(1:1000, n=1, ratio=0.5)) [1] 994 995 996 997 998 999 > tail(loopRec3(1:1000, 0.5)) [1] 994 995 996 997 998 999 > EMA(1:1000, n=1, ratio=0.1)[491:500] [1] 482 483 484 485 486 487 488 489 490 491
David.
> Michael
>
> On Nov 27, 2011, at 9:05 AM, Michael Kao <mkao006rmail at gmail.com>
> wrote:
>
>> Hi Florent,
>>
>> That is great, I was working on solving the recurrence equation and
>> this was part of that equation. Now I know how to put everything
>> together, thanks for all the help e everybody!
>>
>> Cheers,
>>
>> On 27/11/2011 2:05 p.m., Florent D. wrote:
>>> You can make things a lot faster by using the recurrence equation
>>>
>>> y[n] = alpha * (y[n-1]+x[n])
>>>
>>> loopRec<- function(x, alpha){
>>> n<- length(x)
>>> y<- numeric(n)
>>> if (n == 0) return(y)
>>> y[1]<- alpha * x[1]
>>> for(i in seq_len(n)[-1]){
>>> y[i]<- alpha * (y[i-1] + x[i])
>>> }
>>> return(y)
>>> }
>>>
>>>
>>>
>>> On Sun, Nov 27, 2011 at 5:17 AM, Michael
>>> Kao<mkao006rmail at gmail.com> wrote:
>>>> Dear Enrico,
>>>>
>>>> Brilliant! Thank you for the improvements, not sure what I was
>>>> thinking using rev. I will test it out to see whether it is fast
>>>> enough for our implementation, but your help has been
>>>> SIGNIFICANT!!!
>>>>
>>>> Thanks heaps!
>>>> Michael
>>>>
>>>> On 27/11/2011 10:43 a.m., Enrico Schumann wrote:
>>>>> Hi Michael
>>>>>
>>>>> here are two variations of your loop, and both seem faster than
>>>>> the original loop (on my computer).
>>>>>
>>>>>
>>>>> require("rbenchmark")
>>>>>
>>>>> ## your function
>>>>> loopRec<- function(x, alpha){
>>>>> n<- length(x)
>>>>> y<- double(n)
>>>>> for(i in 1:n){
>>>>> y[i]<- sum(cumprod(rep(alpha, i)) * rev(x[1:i]))
>>>>> }
>>>>> y
>>>>> }
>>>>> loopRec(c(1, 2, 3), 0.5)
>>>>>
>>>>> loopRec2<- function(x, alpha){
>>>>> n<- length(x)
>>>>> y<- numeric(n)
>>>>> for(i in seq_len(n)){
>>>>> y[i]<- sum(cumprod(rep.int(alpha, i)) * x[i:1])
>>>>> }
>>>>> y
>>>>> }
>>>>> loopRec2(c(1, 2, 3), 0.5)
>>>>>
>>>>> loopRec3<- function(x, alpha){
>>>>> n<- length(x)
>>>>> y<- numeric(n)
>>>>> aa<- cumprod(rep.int(alpha, n))
>>>>> for(i in seq_len(n)){
>>>>> y[i]<- sum(aa[seq_len(i)] * x[i:1])
>>>>> }
>>>>> y
>>>>> }
>>>>> loopRec3(c(1, 2, 3), 0.5)
>>>>>
>>>>>
>>>>> ## Check whether value is correct
>>>>> all.equal(loopRec(1:1000, 0.5), loopRec2(1:1000, 0.5))
>>>>> all.equal(loopRec(1:1000, 0.5), loopRec3(1:1000, 0.5))
>>>>>
>>>>>
>>>>> ## benchmark the functions.
>>>>> benchmark(loopRec(1:1000, 0.5), loopRec2(1:1000, 0.5),
>>>>> loopRec3(1:1000, 0.5),
>>>>> replications = 50, order = "relative")
>>>>>
>>>>>
>>>>> ... I get
>>>>> test replications elapsed relative user.self
>>>>> sys.self
>>>>> 2 loopRec2(1:1000, 0.5) 50 0.77 1.000000
>>>>> 0.76 0.00
>>>>> 3 loopRec3(1:1000, 0.5) 50 0.86 1.116883
>>>>> 0.85 0.00
>>>>> 1 loopRec(1:1000, 0.5) 50 1.84 2.389610
>>>>> 1.79 0.01
>>>>>
>>>>>
>>>>> Regards,
>>>>> Enrico
>>>>>
>>>>>
>>>>> Am 27.11.2011 01:20, schrieb Michael Kao:
>>>>>> Dear R-help,
>>>>>>
>>>>>> I have been trying really hard to generate the following vector
>>>>>> given
>>>>>> the data (x) and parameter (alpha) efficiently.
>>>>>>
>>>>>> Let y be the output list, the aim is to produce the the following
>>>>>> vector(y) with at least half the time used by the loop example
>>>>>> below.
>>>>>>
>>>>>> y[1] = alpha * x[1]
>>>>>> y[2] = alpha^2 * x[1] + alpha * x[2]
>>>>>> y[3] = alpha^3 * x[1] + alpha^2 * x[2] + alpha * x[3]
>>>>>> .....
>>>>>>
>>>>>> below are the methods I have tried and failed miserably, some
>>>>>> are just
>>>>>> totally ridiculous so feel free to have a laugh but would
>>>>>> appreciate if
>>>>>> someone can give me a hint. Otherwise I guess I'll have to give
>>>>>> RCpp a
>>>>>> try.....
>>>>>>
>>>>>>
>>>>>> ## Bench mark the recursion functions
>>>>>> loopRec<- function(x, alpha){
>>>>>> n<- length(x)
>>>>>> y<- double(n)
>>>>>> for(i in 1:n){
>>>>>> y[i]<- sum(cumprod(rep(alpha, i)) * rev(x[1:i]))
>>>>>> }
>>>>>> y
>>>>>> }
>>>>>>
>>>>>> loopRec(c(1, 2, 3), 0.5)
>>>>>>
>>>>>> ## This is a crazy solution, but worth giving it a try.
>>>>>> charRec<- function(x, alpha){
>>>>>> n<- length(x)
>>>>>> exp.mat<- matrix(rep(x, each = n), nc = n, byrow = TRUE)
>>>>>> up.mat<- matrix(eval(parse(text = paste("c(",
>>>>>> paste(paste(paste("rep(0,
>>>>>> ", 0:(n - 1), ")", sep = ""),
>>>>>> paste("cumprod(rep(", alpha, ",", n:1, "))", sep = "") , sep =
>>>>>> ","),
>>>>>> collapse = ","), ")", sep = ""))), nc = n, byrow = TRUE)
>>>>>> colSums(up.mat * exp.mat)
>>>>>> }
>>>>>> vecRec(c(1, 2, 3), 0.5)
>>>>>>
>>>>>> ## Sweep is slow, shouldn't use it.
>>>>>> matRec<- function(x, alpha){
>>>>>> n<- length(x)
>>>>>> exp.mat<- matrix(rep(x, each = n), nc = n, byrow = TRUE)
>>>>>> up.mat<- sweep(matrix(cumprod(rep(alpha, n)), nc = n, nr = n,
>>>>>> byrow = TRUE), 1,
>>>>>> c(1, cumprod(rep(1/alpha, n - 1))), FUN = "*")
>>>>>> up.mat[lower.tri(up.mat)]<- 0
>>>>>> colSums(up.mat * exp.mat)
>>>>>> }
>>>>>> matRec(c(1, 2, 3), 0.5)
>>>>>>
>>>>>> matRec2<- function(x, alpha){
>>>>>> n<- length(x)
>>>>>> exp.mat<- matrix(rep(x, each = n), nc = n, byrow = TRUE)
>>>>>> up.mat1<- matrix(cumprod(rep(alpha, n)), nc = n, nr = n, byrow
>>>>>> = TRUE)
>>>>>> up.mat2<- matrix(c(1, cumprod(rep(1/alpha, n - 1))), nc = n, nr
>>>>>> = n)
>>>>>> up.mat<- up.mat1 * up.mat2
>>>>>> up.mat[lower.tri(up.mat)]<- 0
>>>>>> colSums(up.mat * exp.mat)
>>>>>> }
>>>>>>
>>>>>> matRec2(c(1, 2, 3), 0.5)
>>>>>>
>>>>>> ## Check whether value is correct
>>>>>> all.equal(loopRec(1:1000, 0.5), vecRec(1:1000, 0.5))
>>>>>> all.equal(loopRec(1:1000, 0.5), matRec(1:1000, 0.5))
>>>>>> all.equal(loopRec(1:1000, 0.5), matRec2(1:1000, 0.5))
>>>>>>
>>>>>> ## benchmark the functions.
>>>>>> benchmark(loopRec(1:1000, 0.5), vecRec(1:1000, 0.5),
>>>>>> matRec(1:1000, 0.5),
>>>>>> matRec2(1:1000, 0.5), replications = 50,
>>>>>> order = "relative")
>>>>>>
>>>>>> Thank you very much for your help.
>>>>>>
>>>>>> ______________________________________________
>>>>>> R-help at r-project.org mailing list
>>>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>>>> PLEASE do read the posting guide
>>>>>> http://www.R-project.org/posting-guide.html
>>>>>> and provide commented, minimal, self-contained, reproducible
>>>>>> code.
>>>>>>
>>>> ______________________________________________
>>>> R-help at r-project.org mailing list
>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>>>> and provide commented, minimal, self-contained, reproducible code.
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT