On Fri, 11 Jun 2010, Marcin Gomulka wrote:
AFAIK a script run through source() does not have any legit way to learn
about it's own location.
You mean like 'test.R' being able to figure out that that is what it is
called when it is invoked from source()?
cat("print(eval(sys.calls()[[1]][[2]]))",file='test.R')
source('test.R')
[1] "test.R"
setwd('projects')
source('../test.R')
[1] "../test.R"
source(list.files('..','test.R$',full=TRUE))
[1] "../test.R"
See
?sys.calls
HTH,
Chuck
I need this to make sure that the script will find its datafiles after I
move the whole directory. (The datafiles are in the same directory.)
Here is a hack I invented to work around it:
print(getwd())
source_pathname = get("ofile",envir = parent.frame())
source_dirname = dirname(source_pathname )
setwd(source_dirname)
print(getwd())
Question: Is there a better, cleaner way?
Thanks,
mrgomel
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