generating a vector of y_t = \sum_{i = 1}^t (alpha^i * x_{t - i + 1})
You can write a function to do so and then there are all sorts of options of what to do with it - mostly they degenerate into putting it in a personal package or into your .Rprofile for convenient use. If you want to use compiled code, that's slightly trickier but good facilities exist to do so from C, C++, and Fortran. Any follow-up, after reading the R extensions manual and the documentation to ?package.skeleton, ?STARTUP, and ?function, probably deserves its own thread. Michael
On Nov 27, 2011, at 12:34 PM, "Yen, Steven T" <syen at utk.edu> wrote:
Hi Is there a way to link a frequently-used sub-routine (in binary form or preferably in ascii form) in a program, similar to the following in Gauss? Thanks. #include c:\programs\mylib1.g; -- Steven T. Yen, Professor of Agricultural Economics The University of Tennessee http://web.utk.edu/~syen/
________________________________________
From: r-help-bounces at r-project.org [r-help-bounces at r-project.org] on behalf of David Winsemius [dwinsemius at comcast.net]
Sent: Sunday, November 27, 2011 11:46 AM
To: R. Michael Weylandt <michael.weylandt at gmail.com>
Cc: r-help at r-project.org
Subject: Re: [R] 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.
______________________________________________
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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
______________________________________________
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.