How to make column names/ dataframe names a string with a variable in it (R)

Question

this.year<-2014 
x<-this.year-1 
y<-this.year-2
x.s<-x-2000 
y.s<-y-2000
tpop_y.s<-acs.fetch(endyear=x,span=1,geography=mystates,variable="B01003_001", col.names="DM_TPOP_x.s")  
tpop_x.s<-acs.fetch(endyear=y,span=1,geography=mystates,variable="B01003_001",col.names="DM_TPOP_y.s")

I'm using the package acs to pull out data from the American community survey to update an infographics website. I hope to run the code every year by inputing the current year in this.year and having the code update data for the past 2 years, x and y .

If this.year is 2015, x is 2014, xs is 14, y is 2013, ys is 13. The end result I want is (for y) the data frame name tpop_13 with the column name DM_TPOP_13 . (for x) the data frame name tpop_14 with the column name DM_TPOP_14 .

The code is pulling the desired data correctly, but this code returns (for y) the data frame name tpop_y.s with the column name DM_TPOP_y.s . (for x) the data frame name tpop_x.s with the column name DM_TPOP_x.s . I tried searching for similar questions and found this one: How to print R variables in middle of String

I tried applying the answer by using the quotes \\"',xs,'\\" to solve my problem, but it doesn't work. The code returns (for x) the column name DM_TPOP_...xs.. . I understand that R does not evaluate any expression within the quotes - it just prints the string you defined. But how can you get around this problem so that there can be a variable in a string?

Help would be greatly appreciated!

Answer 1

You can use functions like paste , paste0 , and sprintf to construct strings from string constants and variables. There is also functionality in the gsubfn package to do Perl like string interpolation.

Answer 2

Here is how you can work: store the data in the list kk , so that first element , kk[[1]] or kk[["tpop_13"]] , gives the data for 2013 and second element, kk[[2]] or kk[["tpop_12"]] gives the data for 2012, with the name of each element as you suggested.

wa=geo.make(state="WA")
kk<-lapply(c(x,y),function(i){
  acs.fetch(endyear=i,span=1,geography=wa,variable="B01003_001", col.names=paste0("DM_TPOP_",i-2000)) 
   })


names(kk)<-paste0("tpop_",c(x-2000,y-2000))
   kk 
        $tpop_13
    ACS DATA: 
     2013 ;
      Estimates w/90% confidence intervals;
      for different intervals, see confint()
               DM_TPOP_13   
    Washington 6971406 +/- 0

    $tpop_12
    ACS DATA: 
     2012 ;
      Estimates w/90% confidence intervals;
      for different intervals, see confint()
               DM_TPOP_12   
    Washington 6897012 +/- 0