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7 messages · Anna H. Pryor, Barry Rowlingson, Peter Dalgaard +2 more

Anna H. Pryor
# 
I don't know how to retain my row labels.  I read in a table with row labels, 
ie:  

Price   Floor  Area  Rooms
52.      111.   830    5
54.      128.   710    5
47.4    101   1000   4
...

but then I need to transform the data

raggedRidzeros <- function(x){
	 	  
	y = list(ridzeros(x[[1]]))
	for(i in 2:length(x)){
	  y = c(y,list(ridzeros(x[[i]])))
	}
	y

}

(where ridzeros is a function that I call to get rid of the zeros in the 
vector) 
When I do this, I no longer have row labels.  Is there any way to insert the 
row labels from the original calling function into this function?

Anna
Barry Rowlingson
#
Anna H. Pryor wrote:
I dont see any row labels. I assume you mean column labels!
This can all be done in one line! Here's some data that does have row 
and column labels:

 > tm
         Foo       Bar Baz
Mercury   1 0.6961034   0
Venus     2 0.3137058   0
Earth     3 0.7692529   1
Mars      0 0.2598111   0
Jupiter   1 0.8375288   0
Saturn    0 0.5866152   0

  Now I want to sweep through columns and return a list without the 
zeroes. I do this:

 > nonZero <- apply(tm,2,function(x){x[x!=0]})

  and I get a list:

$Foo
Mercury   Venus   Earth Jupiter
       1       2       3       1

$Bar
   Mercury     Venus     Earth      Mars   Jupiter    Saturn
0.6961034 0.3137058 0.7692529 0.2598111 0.8375288 0.5866152

$Baz
Earth
     1

  Note this preserves column names (as the names of the list elements, 
so I can do nonZero$Foo), and keeps the row names (as names of 
individual elements).

 > nonZero$Bar['Earth']
     Earth
0.7692529

  How it works:

   function(x){x[x!=0]}   is my 'ridzeros' function.

  I use 'apply(tm,2,function(x){x[x!=0]})' to apply the ridzeros 
function to columns (thats the '2')  of the matrix. To do the same by 
rows, use '1'.

  Hardly rocket science :)

Baz
Peter Dalgaard
# 
Barry Rowlingson <B.Rowlingson at lancaster.ac.uk> writes:
...but slightly dangerous if you have a risk of ending up with all
lists elements equally long, because then you get a matrix instead of
a list. If you want to make damn sure that you get a list, I think you
need something like

do.call("c",apply(tm,2,function(x)list(x[x!=0]))
Kelvin Lam
# 
Dear Group,

I have come across with the following error while
running HoltWinters().

optim(c(0.3, 0.1, 0.1), error, method = "L-BFGS-B", 
lower = c(0, 0, 0), upper = c(1, 1, 1))

It only happens to multiplicative seasonality but not
additive.  I read the program but still couldn't
understand it.  

Thanks!
Brian Ripley
# 
That's not an error, but a line of R code.

If you could post an example which generates the error and the actual 
error message, we could try to investigate it.
On Tue, 29 Apr 2003, Kel wrote:

            

      
      
      
    

        

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Kelvin Lam
    

    

      
      #
    

  


  

      
Here's the time series (Jan 1994 - Dec 1998)

            

      
      
      
    

        
Jan   Feb   Mar   Apr   May   Jun   Jul   Aug   Sep  
Oct   Nov   Dec
1994  16.0  17.4  19.2  21.4  24.0  27.0  29.0  29.4 
29.4  29.0  28.2  27.0
1995  28.0  31.8  36.0  40.6  45.6  51.0  53.0  51.0 
48.6  45.8  42.6  39.0
1996  40.0  46.2  52.8  59.8  67.2  75.0  77.0  72.6 
67.8  62.6  57.0  51.0
1997  52.0  60.6  69.6  79.0  88.8  99.0 101.0  94.2 
87.0  79.4  71.4  63.0
1998  64.0  75.0  86.4  98.2 110.4 123.0 125.0 115.8
106.2  96.2  85.8  75.0

            

      
      
      
    

        
Here's the message I get:

Error in optim(c(0.3, 0.1, 0.1), error, method =
"L-BFGS-B", lower = c(0,  : 
L-BFGS-B needs finite values of fn"

Thanks again.

        
--- Prof Brian Ripley <ripley at stats.ox.ac.uk> wrote:

            

      
      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Brian Ripley
    

    

      
      #
    

  


  

      
That's a very short series to expect to fit a seasonal HW model to whilst
choosing the degree of smoothing. and the optimization code is failing,
because the initial guesses are too far away.

The following should give you a clue:

HoltWinters(test5, seasonal = "additive") 

Smoothing parameters:
 alpha:  1 
 beta :  0.8108822 
 gamma:  1 

and using alpha=1 will work with the multiplicative model too.

You can edit the function and choose other starting values than c(0.3,
0.1, 0.1) and thereby get it to converge, but I would have little
confidence in the values found.

        
On Wed, 30 Apr 2003, Kel wrote: