Many Thanks¡¡¡

I will try this night, I have read this I think could help me.

I´m conscient the question was badly formulated now, I will try to explain
better next time¡¡¡

On a side note: apply always accesses the function you use at least once. If
the input is a dataframe without any rows but with defined variables, it
sends "FALSE" as an argument to the function. If the dataframe is completely
empty, it sends a logical(0) to the function.

> x <- data.frame(a=numeric(0))
> str(x)
'data.frame':   0 obs. of  1 variable:
 $ a: num

> y <- apply(x,MARGIN=1,FUN=function(x){print(x)})
[1] FALSE

> x <- data.frame()

> str(x)
'data.frame':   0 obs. of  0 variables

> y <- apply(x,MARGIN=1,FUN=function(x){print(x)})
logical(0)




2011/7/12 Sarah Goslee <sarah.goslee@gmail.com>

> Hi,
>
> You don't provide us with a reproducible example, so I can't provide you
> with
> actual code. But two approaches come to mind:
> 1. Create da2 with one row and n columns, then change the appropriate
> elements,
> if any, based on your conditions.
> 2. Do the conditional parts, then check to see whether da2 is empty.
> If it is, then
> replace the empty data frame with a data frame of one row and n columns.
>
> Sarah
>
> On Tue, Jul 12, 2011 at 3:51 AM, Trying To learn again
> <tryingtolearnagain@gmail.com> wrote:
> > Hi all,
> >
> > I first create a matrix/data frame called "d2" if another matrix
> > accomplishes some restrictions "dacc2"
> >
> > da2<-da1[colSums(dacc2)>9,]
> > da2<-da2[(da2[,13]=24),]
> > write.csv(da2, file =paste('hggi', i,'.csv',sep = ''))
> >
> > The thing is if finally da2 cannot get/passs the filters, it cannot
> writte a
> > csv because there is no any true condition.
> >
> > How can I create anyway a csv with zeros of one row and "n" columns
> (being n
> > the number of columns of da2?
> >
> > I need a loop?
>
> Rarely.
>
>
> --
> Sarah Goslee
> http://www.functionaldiversity.org
>

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