[R] function returns R object with name based on input

[Jennifer Brea]

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



> df = data.frame(year 
> =c(1991,1991,1992,1992,1993,1993,1992,1991),x=rnorm(8),y=rnorm(8))
> df
  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
> dfs = split(df,df$year)
> dfs[['1991']]
  year         x          y
1 1991 0.5565083 -1.3136423
2 1991 0.1686598 -0.2034466
8 1991 0.9252679  0.1489151
> dfs[['1992']]
  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:

> regs = sapply(dfs,function(d)coef(lm(y~x,data=d)))
> regs
                  1991       1992      1993
(Intercept) -0.6964841 -0.9456066 0.7717261
x            0.4370229  1.5752294 1.5675705







David Winsemius wrote:
>
> On Apr 24, 2009, at 11:56 AM, Jennifer Brea wrote:
>
>> 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?
>
> 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 
>
>>
>> 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)
>> ...
>
> 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.
>
>>
>>
>> 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.
>>
>
> David Winsemius, MD
> Heritage Laboratories
> West Hartford, CT