# Re: [R] vectorizing an iterative process.

From: Christos Hatzis <christos_at_nuverabio.com>
Date: Tue 26 Dec 2006 - 16:58:07 GMT

In your case, the recurrence relationship for x can be solved easily: Notice that

sum{i=1,n}(x[i]-x[i-1]) = x[n] - x[0]

and therefore

x[n] = x[0] + sum{i=1,n}(y[i-1] for n=1, N, with the appropriate initial condition for i=0, (x[0],y[0]).

Thus cumsum on y will give you a direct answer.

-Christos

Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com

-----Original Message-----
From: r-help-bounces@stat.math.ethz.ch
[mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Geoffrey Zhu Sent: Tuesday, December 26, 2006 11:06 AM To: r-help@stat.math.ethz.ch
Subject: Re: [R] vectorizing an iterative process.

I meant x[i] <- x[i-1] + y[i-1] and Y[i] <- y[i-1] + x[i] below.

The mailing list software just keep adding 3D's. Sorry.

-----Original Message-----
From: r-help-bounces@stat.math.ethz.ch
[mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Geoffrey Zhu Sent: Tuesday, December 26, 2006 10:03 AM To: Richard M. Heiberger; r-help@stat.math.ethz.ch Subject: Re: [R] vectorizing an iterative process.

Hi Richard,

3D is automatically generated by the mailing list software, probably because I had ] followed by =3D3D without a space in the original post.

What I meant was to calculate x[i] =3D3D x[i-1] + y[i-1]

For example, if X <- 1:10

Then I want the vector Y to be 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, or in other words Y[i] =3D3D y[i-1] + x[i].

Yes, cumsum does the trick for this. This is what I need. Thanks.

Just curious, do you know how to calculate the more generic x[i] <- f( x[i-1], y[i-1] )?

Thanks,
Geoffrey

-----Original Message-----
From: Richard M. Heiberger [mailto:rmh@temple.edu] Sent: Tuesday, December 26, 2006 9:56 AM To: Geoffrey Zhu; r-help@stat.math.ethz.ch Subject: Re: [R] vectorizing an iterative process.

It looks like

x <- cumsum(y)

What does 3D mean?

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