[R] Substring replacement in string

[Gabor Grothendieck]

Replace the + (i.e. 1 or more) in the pattern with a * (i.e. 0 or more):

   x <- "pmin(pmax(pmin(a,B),pmin(a,C,d))==Y,pmax(E,e))"

   gsub("(\\b[a-oq-z][a-z0-9]*)", "1-\\U\\1", x, perl = TRUE)

giving:

   [1] "pmin(pmax(pmin(1-A,B),pmin(1-A,C,1-D))==Y,pmax(E,1-E))"

Here is a visualization of the regular expression:

   https://www.debuggex.com/i/5ByOCQS2zIdPEf-f.png


On Sat, Feb 28, 2015 at 8:16 AM, Alrik Thiem <alrik.thiem at gmail.com> wrote:
> Dear Gabor,
>
> Many thanks. Works like a charm, but I can't get it to work with
>
> "pmin(pmax(pmin(a,B),pmin(a,C,d))==Y,pmax(E,e))"
>
> i.e., with strings where there're no integers following the components in the pmin/pmax functions. Could this be generalized to handle both cases?
>
> Best wishes,
> Alrik
>
> -----Ursprüngliche Nachricht-----
> Von: Gabor Grothendieck [mailto:ggrothendieck at gmail.com]
> Gesendet: Samstag, 28. Februar 2015 13:35
> An: Alrik Thiem
> Cc: r-help at r-project.org
> Betreff: Re: [R] Substring replacement in string
>
> On Fri, Feb 27, 2015 at 5:19 PM, Alrik Thiem <alrik.thiem at gmail.com> wrote:
>> I would like to replace all lower-case letters in a string that are not part
>> of certain fixed expressions. For example, I have the string:
>>
>> "pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))"
>>
>> Where I would like to replace all lower-case letters that do not belong to
>> the functions "pmin" and "pmax" by 1 - toupper(...) to get
>>
>> "pmin(pmax(pmin(1 - X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1 - Z1))"
>>
>
> Assuming x is the input string:
>
> gsub("(\\b[a-oq-z][a-z0-9]+)", "1-\\U\\1", x, perl = TRUE)
> ## [1] "pmin(pmax(pmin(1-X1, X2), pmin(X3, X4)) == Y, pmax(Z1, 1-Z1))"
>
>
>
> --
> Statistics & Software Consulting
> GKX Group, GKX Associates Inc.
> tel: 1-877-GKX-GROUP
> email: ggrothendieck at gmail.com
>



-- 
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com