Part of file name as a variable

Question

I have a list of files with similar names, for example:

• 2002_file
• 2003_file
• 2004_file

I am going to run a simple script:

foreach year in 2002 2003 2004 2005 2006 2007 2008 2009 2011 2012 2013 2014 2015 {  
use "$input\`year'_file.dta ", clear
keep v1 v2 v3
tab v1, gen (v1_dummy)
gen year = 'x'
save "$output\`year'_newfile.dta ", replace
}

However, I would like for one of the files to be created the respective year variable.

How can I add a variable year = 'year' ?

Answer 1

You do not define the local macro x anywhere.

The following should work:

foreach year in 2002 2003 2004 2005 2006 2007 2008 2009 2011 2012 2013 2014 2015 {
    use "$input\`year'_file.dta ", clear
    keep v1 v2 v3
    tab v1, gen (v1_dummy)
    gen year = `year'
    save "$output\`year'_newfile.dta ", replace
}

For a specific year only:

foreach year in 2002 2003 2004 2005 2006 2007 2008 2009 2011 2012 2013 2014 2015 {
    use "$input\`year'_file.dta ", clear
    keep v1 v2 v3
    tab v1, gen (v1_dummy)
    if `year' == 2005 gen year = `year'
    save "$output\`year'_newfile.dta ", replace
}