R help - Dec 2011 - Assign name to object for each iteration in a loop.

Hi R-users,

I'm trying to produce decompositions of a multiple time-series, grouped by a
factor (called "area"). I'm modifying the code in the STLperArea
function of
package ndvits, as this function only plots produces stl plots, it does not
return the underlying data. 

I want to extract the trend component of each decomposition
("x$time.series[,trend]), assign a name based on the factor
"area".

My input data look like this:
Area is a factor, with three (but could be many more) levels.
area
1
2
3

Ystart=2000

TS is a timeseries:

  X2000049   X2000065  X2000081  X2000097  X2000113 
1     0.2080      0.2165      0.2149     0.2314      0.2028  
2     0.1578      0.1671      0.1577     0.1593      0.1672   
3     0.1897      0.1948      0.2290     0.2292      0.2067   

Here's the function:

STLpA<-function(TS, area, Ystart, period=23, nSG="5,5", DSG=0)
{
require (RTisean)
for(i in 1:unique(area)){
vi.metric=TS[area==i]
filt.vi<-sav_gol(vi.metric,n=nSG,D=DSG)
vi.sg<-ts(filt.vi[,1], start=Ystart,frequency=period)
stld.tmp<-stl(vi.sg, s.window="periodic", robust=TRUE,
na.action=na.approx)
stld.trend<-stld.temp$time.series[,trend]
}
assign(paste("stld", i , sep= "."), stld.trend)
vi.trend<-ls(pattern= "^stld..$")
return(vi.trend)
}

When I call this function with signal=STLpA(TS,area,Ystart=2000,period=23,
nSG= "5,5", DSG="0"))

I get this error:

Error in cat(list(...), file, sep, fill, labels, append) : 
  argument 1 (type 'list') cannot be handled by 'cat'
In addition: Warning message:
In 1:unique(area) :
  numerical expression has 3 elements: only the first used

I'm guessing this is because I'm assigning names to each temporary
stl.trend
file incorrectly. Can anyone 
improve on my rather poor efforts here?

Many thanks,

Louise




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I think part of your problem is the loop:

you probably mean

for(i in unique(area))

Michael

On Thu, Dec 1, 2011 at 1:13 PM, lglew <l.glew at soton.ac.uk>
wrote:
> Hi R-users,
>
> I'm trying to produce decompositions of a multiple time-series, grouped
by a
> factor (called "area"). I'm modifying the code in the
STLperArea function of
> package ndvits, as this function only plots produces stl plots, it does not
> return the underlying data.
>
> I want to extract the trend component of each decomposition
> ("x$time.series[,trend]), assign a name based on the factor
"area".
>
> My input data look like this:
> Area is a factor, with three (but could be many more) levels.
> area
> 1
> 2
> 3
>
> Ystart=2000
>
> TS is a timeseries:
>
> ?X2000049 ? X2000065 ?X2000081 ?X2000097 ?X2000113
> 1 ? ? 0.2080 ? ? ?0.2165 ? ? ?0.2149 ? ? 0.2314 ? ? ?0.2028
> 2 ? ? 0.1578 ? ? ?0.1671 ? ? ?0.1577 ? ? 0.1593 ? ? ?0.1672
> 3 ? ? 0.1897 ? ? ?0.1948 ? ? ?0.2290 ? ? 0.2292 ? ? ?0.2067
>
> Here's the function:
>
> STLpA<-function(TS, area, Ystart, period=23, nSG="5,5", DSG=0)
> {
> require (RTisean)
> for(i in 1:unique(area)){
> vi.metric=TS[area==i]
> filt.vi<-sav_gol(vi.metric,n=nSG,D=DSG)
> vi.sg<-ts(filt.vi[,1], start=Ystart,frequency=period)
> stld.tmp<-stl(vi.sg, s.window="periodic", robust=TRUE,
na.action=na.approx)
> stld.trend<-stld.temp$time.series[,trend]
> }
> assign(paste("stld", i , sep= "."), stld.trend)
> vi.trend<-ls(pattern= "^stld..$")
> return(vi.trend)
> }
>
> When I call this function with signal=STLpA(TS,area,Ystart=2000,period=23,
> nSG= "5,5", DSG="0"))
>
> I get this error:
>
> Error in cat(list(...), file, sep, fill, labels, append) :
> ?argument 1 (type 'list') cannot be handled by 'cat'
> In addition: Warning message:
> In 1:unique(area) :
> ?numerical expression has 3 elements: only the first used
>
> I'm guessing this is because I'm assigning names to each temporary
stl.trend
> file incorrectly. Can anyone
> improve on my rather poor efforts here?
>
> Many thanks,
>
> Louise
>
>
>
>
> --
> View this message in context:
http://r.789695.n4.nabble.com/Assign-name-to-object-for-each-iteration-in-a-loop-tp4129752p4129752.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

On Dec 1, 2011, at 1:13 PM, lglew wrote:
> Hi R-users,
>
> I'm trying to produce decompositions of a multiple time-series,  
> grouped by a
> factor (called "area"). I'm modifying the code in the
STLperArea
> function of
> package ndvits, as this function only plots produces stl plots, it  
> does not
> return the underlying data.
>
> I want to extract the trend component of each decomposition
> ("x$time.series[,trend]), assign a name based on the factor
"area".
>
> My input data look like this:
> Area is a factor, with three (but could be many more) levels.
> area
> 1
> 2
> 3
>
> Ystart=2000
>
> TS is a timeseries:
>
>  X2000049   X2000065  X2000081  X2000097  X2000113
> 1     0.2080      0.2165      0.2149     0.2314      0.2028
> 2     0.1578      0.1671      0.1577     0.1593      0.1672
> 3     0.1897      0.1948      0.2290     0.2292      0.2067
>
> Here's the function:
>
> STLpA<-function(TS, area, Ystart, period=23, nSG="5,5", DSG=0)
> {
> require (RTisean)
> for(i in 1:unique(area)){
> vi.metric=TS[area==i]
> filt.vi<-sav_gol(vi.metric,n=nSG,D=DSG)
> vi.sg<-ts(filt.vi[,1], start=Ystart,frequency=period)
> stld.tmp<-stl(vi.sg, s.window="periodic", robust=TRUE,  
> na.action=na.approx)
> stld.trend<-stld.temp$time.series[,trend]
The line above will probably fail. It is more likely to succeed with:

stld.trend<-stld.temp$time.series[ ,"trend"]

Unless you have defined a variable named `trend` with a value of  
"trend".

> }
> assign(paste("stld", i , sep= "."), stld.trend)
> vi.trend<-ls(pattern= "^stld..$")
> return(vi.trend)
> }
>
> When I call this function with  
> signal=STLpA(TS,area,Ystart=2000,period=23,
> nSG= "5,5", DSG="0"))
>
> I get this error:
>
> Error in cat(list(...), file, sep, fill, labels, append) :
You might want to use traceback() at this point to see which of hte  
functions you did call is sending lists to `cat`.

>  argument 1 (type 'list') cannot be handled by 'cat'
> In addition: Warning message:
> In 1:unique(area) :
>  numerical expression has 3 elements: only the first used
>
> I'm guessing this is because I'm assigning names to each temporary
> stl.trend
> file incorrectly.
I'm guessing it's not that since I think that step would fail if you  
got to it for the reasons described above.
> Can anyone
> improve on my rather poor efforts here?
>
> Many thanks,
>
> Louise
>
>
>
>
> --
> View this message in context:
http://r.789695.n4.nabble.com/Assign-name-to-object-for-each-iteration-in-a-loop-tp4129752p4129752.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT

On Thu, 01-Dec-2011 at 10:13AM -0800, lglew wrote:

|> Hi R-users,
|> 
|> I'm trying to produce decompositions of a multiple time-series,
grouped by a
|> factor (called "area"). I'm modifying the code in the
STLperArea function of
|> package ndvits, as this function only plots produces stl plots, it does
not
|> return the underlying data. 
|> 
|> I want to extract the trend component of each decomposition
|> ("x$time.series[,trend]), assign a name based on the factor
"area".
|> 
|> My input data look like this:
|> Area is a factor, with three (but could be many more) levels.
|> area
|> 1
|> 2
|> 3
|> 
|> Ystart=2000
|> 
|> TS is a timeseries:
|> 
|>   X2000049   X2000065  X2000081  X2000097  X2000113 
|> 1     0.2080      0.2165      0.2149     0.2314      0.2028  
|> 2     0.1578      0.1671      0.1577     0.1593      0.1672   
|> 3     0.1897      0.1948      0.2290     0.2292      0.2067   
|> 
|> Here's the function:
|> 
|> STLpA<-function(TS, area, Ystart, period=23, nSG="5,5",
DSG=0)
|> {
|> require (RTisean)
|> for(i in 1:unique(area)){
|> vi.metric=TS[area==i]
|> filt.vi<-sav_gol(vi.metric,n=nSG,D=DSG)
|> vi.sg<-ts(filt.vi[,1], start=Ystart,frequency=period)
|> stld.tmp<-stl(vi.sg, s.window="periodic", robust=TRUE,
na.action=na.approx)
|> stld.trend<-stld.temp$time.series[,trend]
|> }
|> assign(paste("stld", i , sep= "."), stld.trend)
|> vi.trend<-ls(pattern= "^stld..$")
|> return(vi.trend)
|> }
|> 
|> When I call this function with signal=STLpA(TS,area,Ystart=2000,period=23,
|> nSG= "5,5", DSG="0"))
|> 
|> I get this error:
|> 
|> Error in cat(list(...), file, sep, fill, labels, append) : 
|>   argument 1 (type 'list') cannot be handled by 'cat'
|> In addition: Warning message:
|> In 1:unique(area) :
|>   numerical expression has 3 elements: only the first used
|> 
|> I'm guessing this is because I'm assigning names to each temporary
stl.trend
|> file incorrectly. Can anyone 
|> improve on my rather poor efforts here?

I would suggest putting your individual trends into a list and
returning that at the end of your function rather than trying to
assing individual objects.  IMHO it's easier to do and more useful.
Something along these lines:

> trends <- list()
> for(i in 1:unique(area)){
...
+ trends[[i]] <- ....

}

Then you'll have an element in the list trends that you do with what
you will.  It creates less clutter also.

HTH



-- 
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 _( Y )_  	         Average minds discuss events 
(:_~*~_:)                  Small minds discuss people  
 (_)-(_)  	                      ..... Eleanor Roosevelt
	  
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