Hi

You have got several suggestions. When I try your code I get only errors,

not list or any other objects.

> x<-unique(Returns.names$date_)

Error in unique(Returns.names$date_) : object 'Returns.names' not found

> n<-1

> while(n<=19) {

+ Returns.period<-lapply(((n-1)*125+1):((n+1)*125), function (i)

+ which(Returns.filter$date_==x[i]))

+ Returns.period<-unlist(Returns.period)

+

+

Returns.period1[n,1:(length(Returns.period))]<-Returns.filter[Returns.period,]

+ n<-n+1

+ }

Error in which(Returns.filter$date_ == x[i]) :

object 'Returns.filter' not found

>

How do you expect to get reasonable help if you can not present what you

did and how result differ from what you want on some fake data which we

can copy to R.

[hidden email] napsal dne 09.06.2010 03:49:37:

>

> Dear R Gurus,

>

> Thanks for any help in advance!

>

> Date.frame: Returns.names

>

> X id ticker date_ adjClose totret RankStk

> 258060 258060 13645T10 CP 2001-06-29 18.125 1877.758

>

> My data frame is in the above format. I would like to filter by period,

per

what is filter??

Help page says

Description

Applies linear filtering to a univariate time series or to each series

separately of a multivariate time series.

> id (every 125 days) each consisting of 250 days, I.e. 1-250, 126-375,

etc.

<snip>

> Thus, I tried to adjust by dynamically assigning each period to a

column.

> However, I get the error: object 'Returns.period1' not found. And R is

not

> like Java where you can simply declare a variable by typing say

> Returns.period1.

Why do you expect that one language will behave the same way as another

language. If R was like Java what would be the reason to have 2 languages?

Try instead to comply posting guide and make a fake data which can be

directly used e.g. by dput, construct a result which is to be achieved and

present a code which can be copied to R and which gives other then

expected result.

After that we can quit guessing and start providing relevant help.

Regards

Petr

>

> I do not know how to create a flexible empty matrix that would allow

this

> code to work.

>

> If anyone knows how to do this without a loop, that would be even

better!

>

> ##Filtering by Period

> n<-1

> while(n<=19) {

> Returns.period<-lapply(((n-1)*125+1):((n+1)*125), function (i)

> which(Returns.filter$date_==x[i]))

> Returns.period<-unlist(Returns.period)

>

>

Returns.period1[n,1:(length(Returns.period))]<-Returns.filter[Returns.period,]

http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code.

______________________________________________

[hidden email] mailing list

https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide

http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.