R help - Dec 2009 - integer(0) and NA do not equal FALSE

Hi,
   A noobie question:  I'm simply trying to run a conditional statement that
evaluates if a substring is found within a larger string.  I find that if it
IS found, my function returns TRUE (great!), but if not, the condition does
not evaluate to FALSE.

ex):

if( grep("hi", "hop", fixed = TRUE) )
      print('yes, your substring is in your string')
else print('no, your substring is not in your string')

alternatively, I could replace grep with pmatch:

if (pmatch('hi','hop'))
      print('yes, your substring is in your string')
else print('no, your substring is not in your string')


The first example, using grep, returns logical(0).  The second, using
pmatch, returns NA.  Any idea how to convert either of those to FALSE, or
else a different function that would do the trick?

Thanks,
Jon

	[[alternative HTML version deleted]]

try using 'grepl'
> if( grepl("hi", "hop", fixed = TRUE) ){
+      print('yes, your substring is in your string')
+ } else print('no, your substring is not in your string')
[1] "no, your substring is not in your string"
>

On Sat, Dec 19, 2009 at 3:47 PM, Jonathan <jonsleepy@gmail.com> wrote:
> Hi,
>   A noobie question:  I'm simply trying to run a conditional statement
that
> evaluates if a substring is found within a larger string.  I find that if
> it
> IS found, my function returns TRUE (great!), but if not, the condition does
> not evaluate to FALSE.
>
> ex):
>
> if( grep("hi", "hop", fixed = TRUE) )
>      print('yes, your substring is in your string')
> else print('no, your substring is not in your string')
>
> alternatively, I could replace grep with pmatch:
>
> if (pmatch('hi','hop'))
>      print('yes, your substring is in your string')
> else print('no, your substring is not in your string')
>
>
> The first example, using grep, returns logical(0).  The second, using
> pmatch, returns NA.  Any idea how to convert either of those to FALSE, or
> else a different function that would do the trick?
>
> Thanks,
> Jon
>
>        [[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<http://www.r-project.org/posting-guide.html>
> and provide commented, minimal, self-contained, reproducible code.
>


-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

	[[alternative HTML version deleted]]

Hi Jonathan,

grep() returns  a vector giving either the indices of the elements of 
'x' that yielded a match or, if 'value' is 'TRUE', the
matched elements
of 'x' (quoting from the help page, see ?grep).

So you probably want to test whether this vector is empty or not - in 
other words, whether it has length zero or not:

if( length(grep("hi", "hop", fixed = TRUE)) > 0 )
        print('yes, your substring is in your string') else
        print('no, your substring is not in your string')

(And you can remove the "fixed = TRUE" if you are only interested in 
whether the expression matches or not.)

One off-topic point: you want your "else" at the end of the second
line,
not the beginning of the third. R evaluates line by line, and when it 
gets to the end of the second line and doesn't see your "else", it
has
no way of knowing that the "if" is not yet finished. I've found
that
liberal use of curly braces makes life much easier.

HTH,
Stephan



Jonathan schrieb:
> Hi,
>    A noobie question:  I'm simply trying to run a conditional statement
that
> evaluates if a substring is found within a larger string.  I find that if
it
> IS found, my function returns TRUE (great!), but if not, the condition does
> not evaluate to FALSE.
> 
> ex):
> 
> if( grep("hi", "hop", fixed = TRUE) )
>       print('yes, your substring is in your string')
> else print('no, your substring is not in your string')
> 
> alternatively, I could replace grep with pmatch:
> 
> if (pmatch('hi','hop'))
>       print('yes, your substring is in your string')
> else print('no, your substring is not in your string')
> 
> 
> The first example, using grep, returns logical(0).  The second, using
> pmatch, returns NA.  Any idea how to convert either of those to FALSE, or
> else a different function that would do the trick?
> 
> Thanks,
> Jon
> 
> 	[[alternative HTML version deleted]]
> 
> ______________________________________________
> R-help at 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.
>

Hi Jonathan,
  The function "isTRUE" is useful for this sort of thing:
isTRUE(pmatch("hi", "hop")) evaluates to FALSE.
--Gray

On Sat, Dec 19, 2009 at 12:47 PM, Jonathan <jonsleepy at gmail.com>
wrote:
> Hi,
> ? A noobie question: ?I'm simply trying to run a conditional statement
that
> evaluates if a substring is found within a larger string. ?I find that if
it
> IS found, my function returns TRUE (great!), but if not, the condition does
> not evaluate to FALSE.
>
> ex):
>
> if( grep("hi", "hop", fixed = TRUE) )
> ? ? ?print('yes, your substring is in your string')
> else print('no, your substring is not in your string')
>
> alternatively, I could replace grep with pmatch:
>
> if (pmatch('hi','hop'))
> ? ? ?print('yes, your substring is in your string')
> else print('no, your substring is not in your string')
>
>
> The first example, using grep, returns logical(0). ?The second, using
> pmatch, returns NA. ?Any idea how to convert either of those to FALSE, or
> else a different function that would do the trick?
>
> Thanks,
> Jon
>
> ? ? ? ?[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at 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.
>


-- 
Gray Calhoun

Assistant Professor of Economics
Iowa State University