aplpy recursive function on a list

I have 2 series of data a,b and I would like to calculate a new series which
is z[t] = z[t-1]*a[t] + b[t] , z[1] = b[1].
How can I do that without using a loop?

On 25-01-2012, at 19:09, patzoul wrote:

I have 2 series of data a,b and I would like to calculate a new series which
is z[t] = z[t-1]*a[t] + b[t] , z[1] = b[1].
How can I do that without using a loop?

?filter

I have 2 series of data a,b and I would like to calculate a new series which
is z[t] = z[t-1]*a[t] + b[t] , z[1] = b[1].
How can I do that without using a loop?

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On 1/25/2012 10:09 AM, patzoul wrote:

I have 2 series of data a,b and I would like to calculate a new series which
is z[t] = z[t-1]*a[t] + b[t] , z[1] = b[1].
How can I do that without using a loop?

A combination of Reduce and Map will work.  Map to stitch together the a and b vectors, Reduce to step along them, allowing for accumulation.

a <- c(2,4,3,5)
b <- c(1,3,5,7)

z <- Reduce(function(zm1, coef) {coef[1] * zm1 + coef[2]},
           Map(c, a, b),
           init = b[1], accumulate = TRUE)

z

[1]   1   3  15  50 257

u

On 26-01-2012, at 17:58, Brian Diggs wrote:

On 1/25/2012 10:09 AM, patzoul wrote:

I have 2 series of data a,b and I would like to calculate a new series which
is z[t] = z[t-1]*a[t] + b[t] , z[1] = b[1].
How can I do that without using a loop?

........

I don't think so.

a <- c(2,4,3,5)
b <- c(1,3,5,7)

z <- rep(0,length(a))
z[1] <- b[1]
for( t in 2:length(a) ) { z[t] <- a[t] * z[t-1] + b[t] }
z

gives

[1]   1   7  26 137

and this agrees with a manual calculation.

You get a vector of length 5 as result. It should be of length 4 with your data.
If you change the Reduce expression to this

u <- Reduce(function(zm1, coef) {coef[1] * zm1 + coef[2]},
          Map(c, a[-1], b[-1]),
          init = b[1], accumulate = TRUE)

then you get the correct result.

u

[1]   1   7  26 137

On 26-01-2012, at 19:10, Berend Hasselman wrote:

On 26-01-2012, at 17:58, Brian Diggs wrote:

On 1/25/2012 10:09 AM, patzoul wrote:

I have 2 series of data a,b and I would like to calculate a new series which
is z[t] = z[t-1]*a[t] + b[t] , z[1] = b[1].
How can I do that without using a loop?

........

I don't think so.

a<- c(2,4,3,5)
b<- c(1,3,5,7)

z<- rep(0,length(a))
z[1]<- b[1]
for( t in 2:length(a) ) { z[t]<- a[t] * z[t-1] + b[t] }
z

gives

[1]   1   7  26 137

and this agrees with a manual calculation.

You get a vector of length 5 as result. It should be of length 4 with your data.
If you change the Reduce expression to this

u<- Reduce(function(zm1, coef) {coef[1] * zm1 + coef[2]},
           Map(c, a[-1], b[-1]),
           init = b[1], accumulate = TRUE)

then you get the correct result.

u

[1]   1   7  26 137

And the loop especially if byte compiled with cmpfun  appears to be quite a bit quicker.

Nrep<- 1000

tfrml.loop<- function(a,b) {
     z<- rep(0,length(a))
     z[1]<- b[1]
     for( t in 2:length(a) ) {
         z[t]<- a[t] * z[t-1] + b[t]
     }

     z
}

tfrml.rdce<- function(a,b) {
     u<- Reduce(function(zm1, coef) {coef[1] * zm1 + coef[2]},
                Map(c, a[-1], b[-1]),
                init = b[1], accumulate = TRUE)
     u
}

library(compiler)
tfrml.loop.c<- cmpfun(tfrml.loop)
tfrml.rdce.c<- cmpfun(tfrml.rdce)

z.loop<- tfrml.loop(a,b)
z.rdce<- tfrml.rdce(a,b)
all.equal(z.loop, z.rdce)

library(rbenchmark)

N<- 500
set.seed(1)
a<- runif(N)
b<- runif(N)

benchmark(tfrml.loop(a,b), tfrml.rdce(a,b), tfrml.loop.c(a,b), tfrml.rdce.c(a,b),
                 replications=Nrep, columns=c("test", "replications", "elapsed"))

                 test replications elapsed
1   tfrml.loop(a, b)         1000   2.665
3 tfrml.loop.c(a, b)         1000   0.554
2   tfrml.rdce(a, b)         1000   4.082
4 tfrml.rdce.c(a, b)         1000   3.143

Berend

R-2.14.1 (32-bits), Mac OS X 10.6.8.