Thanks so much Josh. That absolutely does help. I was just stuck on the
function regexpr and didn't consider how handy the other string handling
functions would be. As far as attaching 'i' to 'vector', that was indeed a
typo. As I was going through my code to make it more readable (at least that
was my intent), I forgot to switch the variable 'vector' to 'indices'. It
was the first of our weekly 'beer Fridays' at the office and I wasn't as
sharp an editor as I would have been otherwise. Nonetheless, I wanted to get
the e-mail off before I left.

I'll make sure to put some working code up next time.

-Ben



On Fri, Apr 1, 2011 at 7:08 PM, Joshua Wiley <jwiley.psych@gmail.com> wrote:

> Hi Ben,
>
> I am having some trouble figuring out what it is exactly you want.  A
> workable example would be nice and then (even if just manually) typed
> out what you would like it to return based on a given input.  I almost
> think you just want grep().
>
> grep("test", c("test", "not", "test2", "not"), value = TRUE)
> [1] "test"  "test2"
>
> using your example:
>
> grep("string", d[,"Column.You.Want"], value = TRUE)
>
> On Fri, Apr 1, 2011 at 5:39 PM, Ben Hunter <bjameshunter@gmail.com> wrote:
> > I'm stuck here. The following code works great if I just use it on the
> > command line in a 'for' loop. Once I try to define it as a function as
> seen
> > below, I get bad results. I'm trying to return a vector of column indices
> > based on a short string that is contained in a selection (length of about
> > 70) of long column names that changes from time to time. Is there an
> > existing function that will do this for me? The more I think about this
> > problem, the more I feel there has to be a function out there. I've not
> > found it.
> >
> >
> > ind <- function(col, string, vector){  # this is really the problem. I
> don't
> > feel like I'm declaring these arguments properly.
> >  indices <- vector(mode = 'numeric') # I am not entirely confident that
> > this use is necessary. Is indices <- c() okay?
>
> indices <- c() would sort of work, but vector() is better.  Also, if
> you actually want your function to be returning integers, you should
> instatiate indices as an integer class vector, not numeric.
>
> >  for (i in 1:length(col)){
> >    num <- regexpr(str, col[i])
>
> str() is a function, and as far as I can tell, a variable "str" is not
> defined anywhere in your function or your functions argument.  Did you
> mean "string"?
>
> >    if (num != -1){
> >         indices <- c(vector, i) # I've also had success with indices <-
> > append(indices, i)
>
> why are you combining "vector" with i over and over?  This will give
> you something like:
>
> c("vector", i1, "vector", i2, "vector", i3, etc.) except obviously
> replace i1, i3 with their values and "vector" with its value.  Is that
> what you want?
>
> >  }
> > }
> > indices
> > }
> >
> > ind(d[,'Column.I.want'], 'string', 'output.vector')
> >
> > Am I wrong here? I've read that the last statement in the function is
> what
> > it will return, and what I want is a vector of integers.
>
> yes, if return() is not explicitly specified inside the function, then
> it will return the output of the last statement.
>
> >
> > Thanks,
>
> Thank you for providing code of what you tried.  For future reference,
> it is good to at least provide useable input data and then show us
> what the output you would like is.  For example: "Blah blah blah, I
> have a vector d <- c(1, 2, 3), how can I find the average of this
> (i.e., 2)?" To which you would get the answer: "mean(d)" or some such.
>
> If grep() is not actually what you are after, can you let us know a
> little bit more about what you want?
>
> Hope this helps,
>
> Josh
>
>
> >
> > -Ben
> >
> >        [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
> --
> Joshua Wiley
> Ph.D. Student, Health Psychology
> University of California, Los Angeles
> http://www.joshuawiley.com/
>

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