Re: [R] gsub() on Matrix

[Tony Plate]

Many more recent regular expression implementations have ways of indicating 
a match on a word boundary.  It's usually \b.

Here's what you did:
 gsub(x1, i1, x1 + x2 + x10 + xx1)
[1] i1 + x2 + i10 + xi1
The following worked for me to just change x1 to i1, while leaving 
alone any larger word that contains x1:

 gsub(\\bx1\\b, i1, x1 + x2 + x10 + xx1)
[1] i1 + x2 + x10 + xx1

Note that the backslash must be escaped itself to get past the R lexical 
analyser, which is independent of the regexp processor.  What the regexp 
processor sees is just a single backslash.

For more on this, look for perl documentation of regular expressions.  Be 
aware that to use full perl regexps, you must supply the perl=T argument to 
gsub().  Also note that \b seems to be part of the most basic regular 
expression language in R; it even works with extended=F:

 gsub(\\bx1\\b, i1, x1 + x2 + x10 + xx1, perl=T)
[1] i1 + x2 + x10 + xx1
 gsub(\\bx1\\b, i1, x1 + x2 + x10 + xx1, perl=F)
[1] i1 + x2 + x10 + xx1
 gsub(\\bx1\\b, i1, x1 + x2 + x10 + xx1, perl=F, ext=F)
[1] i1 + x2 + x10 + xx1

(I assumed the fact that you have a matrix of strings is not relevant.)
Hope this helps,
Tony Plate
At Wednesday 09:07 PM 10/27/2004, Kevin Wang wrote:
Hi,
Suppose I've got a matrix, and the first few elements look like
  x1 + x3 + x4 + x5 + x1:x3 + x1:x4
  x1 + x2 + x3 + x5 + x1:x2 + x1:x5
  x1 + x3 + x4 + x5 + x1:x3 + x1:x5
and so on (have got terms from x1 ~ x14).
If I want to replace all the x1 with i7, all x2 with i14, all x3 with i13,
for example.  Is there an easy way?
I tried to put what I want to replace in a vector, like:
 repl = c(i7, i14, i13, d2, i8, i5,
  i6, i3, A, i9, i2,
  i4, i15, i21)
and have another vector, say:
   orig
 [1] x1  x2  x3  x4  x5  x6  x7  x8  x9  x10
[11] x11 x12 x13 x14
Then I tried something like
  gsub(orig, repl, mat)
## mat is the name of my matrix
but it didn't work *_*.it would replace terms like x10 with i70.
(I know it may be an easy question...but I haven't done much regular
expression)
Cheers,
Kevin

Ko-Kang Kevin Wang
PhD Student
Centre for Mathematics and its Applications
Building 27, Room 1004
Mathematical Sciences Institute (MSI)
Australian National University
Canberra, ACT 0200
Australia
Homepage: http://wwwmaths.anu.edu.au/~wangk/
Ph (W): +61-2-6125-2431
Ph (H): +61-2-6125-7407
Ph (M): +61-40-451-8301
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Re: [R] gsub() on Matrix

[Peter Dalgaard]

Tony Plate [EMAIL PROTECTED] writes:

 Many more recent regular expression implementations have ways of
 indicating a match on a word boundary.  It's usually \b.


Another idea is that if what you need is something that is parseable
as a model formula RHS, then you might want to parse first and
substitute later. Something along these lines:

 e - parse(text=x1 + x3 + x4 + x5 + x1:x3 + x1:x4)[[1]]
 repl = lapply(c(i7, i14, i13, d2, i8, i5),as.name)
 names(repl)-paste(x,1:6,sep=)
 eval(substitute(substitute(e,repl),list(e=e)))
i7 + i13 + d2 + i8 + i7:i13 + i7:d2


 At Wednesday 09:07 PM 10/27/2004, Kevin Wang wrote:
 Suppose I've got a matrix, and the first few elements look like
x1 + x3 + x4 + x5 + x1:x3 + x1:x4
x1 + x2 + x3 + x5 + x1:x2 + x1:x5
x1 + x3 + x4 + x5 + x1:x3 + x1:x5
 and so on (have got terms from x1 ~ x14).
 
 If I want to replace all the x1 with i7, all x2 with i14, all x3 with i13,
 for example.  Is there an easy way?
 
 I tried to put what I want to replace in a vector, like:
   repl = c(i7, i14, i13, d2, i8, i5,
i6, i3, A, i9, i2,
i4, i15, i21)
 and have another vector, say:
 orig
   [1] x1  x2  x3  x4  x5  x6  x7  x8  x9  x10
 [11] x11 x12 x13 x14
 
 Then I tried something like
gsub(orig, repl, mat)
 ## mat is the name of my matrix
 
 but it didn't work *_*.it would replace terms like x10 with i70.

-- 
   O__   Peter Dalgaard Blegdamsvej 3  
  c/ /'_ --- Dept. of Biostatistics 2200 Cph. N   
 (*) \(*) -- University of Copenhagen   Denmark  Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907

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