if(grep(...) == 1) was [R] (no subject)

[ripley@stats.ox.ac.uk]

On Thu, 16 Jan 2003, graham lawrence wrote:

> Why does the second grep reject when the first grep accepts?
> Why does regexpr accept what grep rejects?

It's the if() that is different.  grep returns an integer vector of 
matching indices, and numeric(0) == 1 is logical(0), which gives nothing 
to test (and I guess the error message could be improved),
The idiom often is if(length(grep(...))).

regexpr returns something for each match, but I suspect you want to test 
> 0, not ==1.  See the help page.

> >eqtn
> [1] "(1.2*A%*%B)/2"
> >if(grep("A\\%\\*\\%B",eqtn)==1)erind<-1
> >if(grep("A\\%\\*\\%A",eqtn)==1)erind<-1
> Error in if (grep("A\\%\\*\\%A", eqtn) == 1) erind <- 1 :
>         missing value where logical needed
> >if(regexpr("A\\%\\*\\%A",eqtn)==1)erind<-1

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
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