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function returns R object with name based on input

3 messages · Jennifer Brea, David Winsemius

Jennifer Brea
# 
I wanted to ask how I can make a for loop or a function return an R 
object with a unique name based on either some XX of the for loop or 
some input for the function.

For example

if I have a function:

fn<-function(data,year){

which does does some stuff
}

How do I return an object from the function called X.year, such that if 
I run fn(data,1989), the output is an object called X.1989?

In a separate but related process, I'm also trying to subset data by 
year, where there are multiple observations by years, using the subset() 
function.  For example:

data.1946<-subset(data, year==1946)
data.1947<-subset(data, year==1947)
data.1948<-subset(data, year==1948)
data.1949<-subset(data, year==1949)
...

How should I set this up?  I was thinking of writing a for loop, but I 
have never written a for loop that creates objects based on the loop's 
index, for example a loop for(i in 1946:2000) that returns 55 objects 
with the object names based on the index.

Thanks for your help!
David Winsemius
#
On Apr 24, 2009, at 11:56 AM, Jennifer Brea wrote:

            

      
      
      
    

        
Read:
?assign
?paste
#and FAQ 7.21
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f

            

      
      
      
    

        
list.of.subsets <- sapply(1946:200, function(x) subset(data,  
year==x) ) # with no example ... untested

Using data as a dataframe names is poor R programming practice, since  
many functions use data a a parameter name and it is also a function  
name.

            

      
      
      
    

        
David Winsemius, MD
Heritage Laboratories
West Hartford, CT

  
  


  


      

    

    
2 days later

    

      
      

        


  

        

      Jennifer Brea
    

    

      
      #
    

  


  

      
Thanks for your replies.  I ended up using the following:

            

      
      
      
    

        
year          x           y
1 1991  0.5565083 -1.31364232
2 1991  0.1686598 -0.20344656
3 1992 -0.1010090 -0.65681852
4 1992  0.6130324 -0.10788605
5 1993 -0.9061458 -0.64872139
6 1993 -0.4460332  0.07253762
7 1992 -0.3865464 -1.87445996
8 1991  0.9252679  0.14891506

            

      
      
      
    

        
year         x          y
1 1991 0.5565083 -1.3136423
2 1991 0.1686598 -0.2034466
8 1991 0.9252679  0.1489151

            

      
      
      
    

        
year          x          y
3 1992 -0.1010090 -0.6568185
4 1992  0.6130324 -0.1078861
7 1992 -0.3865464 -1.8744600

Notice that split automatically uses a character version of the values 
of the split variable to name its output.

Once you've created the list, you can use sapply or lapply
to process each piece.  Let's say we wanted the regression coefficients 
for the regression of y on x for each year:

            

      
      
      
    

        
1991       1992      1993
(Intercept) -0.6964841 -0.9456066 0.7717261
x            0.4370229  1.5752294 1.5675705

        
David Winsemius wrote: