[R] regexpr and parsing question

[Kimpel, Mark William]

The main problem I am trying to solve it this:

I am importing a tab delimited file whose first line contains only one
column, which is a descriptor of the form "col_1 col_2 col_3", i.e. the
colnames are not tab delineated but are separated by whitespace. I would
like to parse this first line and make such that it becomes the colnames
of the rest of the file, which I am reading into R using read.delim().
The file is so huge that I must do this in R.

My first question is this: What is the best way to accomplish what I
want to do?

My other questions revolve around some failed attempts on my part to
solve the problem on my own using regular expressions. I thought that
perhaps I could change the first line to "c("col_1", "col_2", "col_3")
using gsub. I was having trouble figuring out how R uses the backslash
character because I know that sometimes the backslash one would use in
Perl needs to be a double backslash in R.

Here is a sample of what I tried and what I got:

a<-"col_1 col_2 col_3"

> gsub("\\s", " " , a) 

[1] "col_1 col_2 col_3"

> gsub("\\s", "\\s" , a) 

[1] "col_1scol_2scol_3"

As you can see, it looks like R is taking a regular expression for
"pattern", but not taking it for "replacement". Why is this?

Assuming that I did want to solve my original problem with gsub and then
turn the string into an R object, how would I get gsub to return
"c("col_1", "col_2", "col_3") using my original string?

Finally, is there a way to declare a string as a regular expression so
that R sees it the same way other languages, such as Perl do, i.e. make
the backslash be interpreted the same way? For someone who is just
learning regular expressions as I am, it is very frustrating to read
about them in references and then have to translate what I've learned
into R syntax. I was thinking that instead of enclosing the string in
"", one could use THIS.IS.A.REGULAR.EXPRESSION(), similar to the way we
use I() in formulae.

These are a bunch of questions, but obviously I have a lot to learn!

Thanks,

Mark

Mark W. Kimpel MD 

 

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