On Thu, Jan 26, 2012 at 08:29:22AM -0800, yan wrote:
> what if I don't need to store the combination results, I just want to get the
> combination result for a given row.
> e.g. for the 5 elements divided into 3 groups , and if I give another
> parameter which is the row number of the results, in p
On Thu, Jan 26, 2012 at 08:29:22AM -0800, yan wrote:
> what if I don't need to store the combination results, I just want to get the
> combination result for a given row.
> e.g. for the 5 elements divided into 3 groups , and if I give another
> parameter which is the row number of the results, in p
what if I don't need to store the combination results, I just want to get the
combination result for a given row.
e.g. for the 5 elements divided into 3 groups , and if I give another
parameter which is the row number of the results, in petr's example, say if
I set row number to 1, then I get 1,2,3
Of David Winsemius
> Sent: Wednesday, January 25, 2012 10:23 AM
> To: yan
> Cc: r-help@r-project.org
> Subject: Re: [R] function for grouping
>
>
> On Jan 25, 2012, at 12:10 PM, yan wrote:
>
> > thanks petr,
> > what if I got 200 elements, so I have to write exp
On Wed, Jan 25, 2012 at 09:10:13AM -0800, yan wrote:
> thanks petr,
> what if I got 200 elements, so I have to write expand.grid(x1=1, x2=1:2,
> x3=1:3, x4=1:3, x5=1:3x200=1:3))?
Hi.
As others told you, the problem is not in reformulating
the command in a concise way, but in enormous memory
a
No, you have not. Take a look at
expand.grid(rep(list(1:5), 2))
HTH,
Jorge.-
On Wed, Jan 25, 2012 at 12:10 PM, yan <> wrote:
> thanks petr,
> what if I got 200 elements, so I have to write expand.grid(x1=1, x2=1:2,
> x3=1:3, x4=1:3, x5=1:3x200=1:3))?
>
> Many thanks
>
> yan
>
> --
> View
You will run out of memory. 2*3^198 is about 5e95 rows and 4GB of RAM can
probably only hold about 5e8 rows.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#.
On Jan 25, 2012, at 12:10 PM, yan wrote:
thanks petr,
what if I got 200 elements, so I have to write expand.grid(x1=1,
x2=1:2,
x3=1:3, x4=1:3, x5=1:3x200=1:3))?
Perhaps same thing that will happen when those monks finish the Towers
of Hanoi?
2*3^198
[1] 5.902533e+94
--
David Wins
thanks petr,
what if I got 200 elements, so I have to write expand.grid(x1=1, x2=1:2,
x3=1:3, x4=1:3, x5=1:3x200=1:3))?
Many thanks
yan
--
View this message in context:
http://r.789695.n4.nabble.com/function-for-grouping-tp4324436p4327812.html
Sent from the R help mailing list archive at Na
On Tue, Jan 24, 2012 at 07:05:37PM +0100, Petr Savicky wrote:
> On Tue, Jan 24, 2012 at 05:19:42PM +, yan jiao wrote:
> > Dear All,
> >
> > I'm wondering if there is a R function could give me all the
> > combinations of the grouping/cluster result, given the number of the groups.
> > e.g.
>
On Tue, Jan 24, 2012 at 05:19:42PM +, yan jiao wrote:
> Dear All,
>
> I'm wondering if there is a R function could give me all the
> combinations of the grouping/cluster result, given the number of the groups.
> e.g.
> 3 objects: x1 x2 x3, number of groups is 2
> so the result will be
> group
Dear All,
I'm wondering if there is a R function could give me all the
combinations of the grouping/cluster result, given the number of the groups.
e.g.
3 objects: x1 x2 x3, number of groups is 2
so the result will be
group1:x1,x2; group2: x3
group1: x1;group2: x2,x3
group1: x1,x3;group2: x2
m
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