[R] regex challenge

[arun]

Slightly modified, also seems to work.

gsubfn( "([[:alpha:]][[:alnum:]]*)((?=\\s*[-+~*)])|$)",function(x,...) paste0(toupper(x),'z'), test, perl=TRUE )
#[1] "Y1z + Y2z ~ Az*(Bz + Cz) + Dz + Fz * (h == 3) + (sex == 'male')*Iz"
A.K.



----- Original Message -----
From: Greg Snow <538280 at gmail.com>
To: Frank Harrell <f.harrell at vanderbilt.edu>
Cc: RHELP <R-help at stat.math.ethz.ch>
Sent: Thursday, August 15, 2013 5:07 PM
Subject: Re: [R] regex challenge

Here is a first stab:

library(gsubfn)

test <- "y1 + y2 ~ a*(b + c) + d + f * (h == 3) + (sex == 'male')*i"

gsubfn( "([a-zA-Z][a-zA-Z0-9]*)((?=\\s*[-+~)*])|\\s*$)",
function(x,...) paste0(toupper(x),'z'), test, perl=TRUE )



On Wed, Aug 14, 2013 at 9:13 PM, Frank Harrell <f.harrell at vanderbilt.edu>wrote:

> I would like to be able to use gsub or gsubfn to process a formula and to
> translate the variables but to ignore expressions in the formula. Supposing
> that the R formula has already been transformed into a character string and
> that the transformation is to convert variable names to upper case and to
> append z to the names, an example would be to convert y1 + y2 ~ a*(b + c) +
> d + f * (h == 3) + (sex == 'male')*i to Y1z + Y2z ~ Az*(Bz + Cz) + Dz + Fz
> * (h == 3) + (sex == 'male')*Iz.  Any expression that is not just a simple
> variable name would be left alone.
>
> Does anyone want to try their hand at creating a regex that would
> accomplish this?
>
> Thanks
> Frank
> --
> Frank E Harrell Jr Professor and Chairman      School of Medicine
>                    Department of Biostatistics Vanderbilt University
>
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-- 
Gregory (Greg) L. Snow Ph.D.
538280 at gmail.com

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