[R] Autofilling a large matrix in R

[jim holtman]

To avoid FAQ 7.31, you probably should use:

seq(0, 10000) / 10000

On Fri, Oct 12, 2012 at 11:12 AM, Mark Lamias <mlamias at yahoo.com> wrote:
> If you are after all the possible percentages with two decimal places, why don't you use this:
>
> seq(from=0, to=100, by=.01)/100
>
> I'm not really sure what you are trying to do in terms of rows and columns, however.  Can you be a bit more specific on what each row/column is?
> Are you trying to group the numbers so that all the entries in a row add up to 100% and then, once it does, split the following entries onto the next row until they add up to 100%, etc.?
> Thanks.
>
>
>
> ________________________________
>  From: wwreith <reith_william at bah.com>
> To: r-help at r-project.org
> Sent: Friday, October 12, 2012 10:30 AM
> Subject: [R] Autofilling a large matrix in R
>
> I wish to create a matrix of all possible percentages with two decimal place
> percision. I then want each row  to sum to 100%. I started with the code
> below with the intent to then subset the data based on the row sum. This
> works great for 2 or 3 columns, but if I try 4 or more columns the number of
> rows become to large. I would like to find a way to break it down into some
> kind of for loop, so that I can remove the rows that don't sum to 100%
> inside the for loop rather than outside it. My first thought was to take
> list from 1:10, 11:20, etc. but that does not get all of the points.
>
> g<-as.matrix(expand.grid(rep(list(1:100), times=3)))
>
> Any thoughts how to split this into pieces?
>
>
>
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> ______________________________________________
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>



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.