From: Shubha Vishwanath Karanth <shubhak_at_ambaresearch.com>

Date: Wed, 23 Jul 2008 21:01:44 +0530

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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. Received on Wed 23 Jul 2008 - 16:29:26 GMT

Date: Wed, 23 Jul 2008 21:01:44 +0530

Many Thanks Jorge... That was one more way...Is it possible if I can do
this without using rep(1:10,each=8) or the grouping....because I feel
the number 8 here is fixed... If there is some technique of tracking the
position of first 8, then next 8... don't know whether I am clear in
conveying...

Thanks, shubha

From: Jorge Ivan Velez [mailto:jorgeivanvelez_at_gmail.com]
Sent: Wednesday, July 23, 2008 8:59 PM

To: Shubha Vishwanath Karanth

Cc: r-help_at_stat.math.ethz.ch

Subject: Re: [R] sequential sum of a vector...

Dear Shubha,

Try this:

x=1:80

tapply(x,rep(1:10,each=8),sum)

1 2 3 4 5 6 7 8 9 10

36 100 164 228 292 356 420 484 548 612

**HTH,
**
Jorge

On Wed, Jul 23, 2008 at 10:03 AM, Shubha Vishwanath Karanth <shubhak_at_ambaresearch.com> wrote:

c(36,100,164,228,292,356,420,484,548,612)

aggregate(x,list(rep(1:10,each=8)),sum)[-1]

rowsum(x,group=rep(1:10,each=8))

But without grouping, can I achieve the required? Any other ways of doing this?

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