[R] for loop using index values

[Frede Aakmann Tøgersen]

As David says it will be good to see how the structure of the list subset is. Now I'm going to guess that the i'th element of subset is a dataframe since you are using nrow(subset[[i]]).

Now you're doing 

select<-which(nrow(subset[[i]])>=100) ## added a missing closing )

so now select holds the number of rows of subset[[i]] only.

Perhaps it should be something like

select <- which(lapply(subset, nrow) >= 100).

Using some dummy data:

> tmp <- list(matrix(0,2,3),matrix(0,9,3),matrix(0,8,3),matrix(0,3,3),matrix(0,6,3))
> which((lapply(tmp, nrow) > 6))
[1] 2 3

 
By the way there is a function called subset so you could run into trouble later on when calling your list for subset.

Yours sincerely / Med venlig hilsen


Frede Aakmann Tøgersen
Specialist, M.Sc., Ph.D.
Plant Performance & Modeling

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> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
> On Behalf Of David Winsemius
> Sent: 15. maj 2014 06:49
> To: David Gwenzi
> Cc: r-help at r-project.org
> Subject: Re: [R] for loop using index values
> 
> 
> On May 14, 2014, at 9:23 PM, David Gwenzi wrote:
> 
> > Hi all
> >
> > I have point data along a transect and I want to divide the transect
> > into
> > small blocks of 10m length each. I have named the blocks as a list i.e
> > subset[[i]]. Now the issue is I want to process only those blocks
> > that have
> > at least 100 data points and keep the original index values of those
> > subsets. How do I set the for loop. I have tried the following but
> > it is
> > still processing everything
> >
> > select<-which(nrow(subset[[i]])>=100
> > for (i in c(select)){
> > ..............
> > }
> > Thank you in advance,
> 
> So can you provide str( subset[[1]] )?
> 
> 
> >
> > David Gwenzi
> > Graduate Degree Program in Ecology
> > Natural Resources Ecology Lab
> > Colorado State University
> >
> > 	[[alternative HTML version deleted]]
> 
> And this is a plain text mailing list as described in the Posting Guide.
> 
> --
> 
> David Winsemius, MD
> Alameda, CA, USA
> 
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