R help - Dec 2010 - Matching a pattern of vector of character strings in another vector of character strings

Dear all,
 
My question is illustrated by the following example:
 
I have a matrix M:
 
> M<-
matrix(c("0","0","1","1","0","1","1","0","0","*","1","1","0","1","*"),nrow=3)
> colnames(M)<-
c("2006","2007","2008","2009","2010")
> M
     2006 2007 2008 2009 2010
[1,] "0"  "1"  "1"  "*"  "0" 
[2,] "0"  "0"  "0"  "1"  "1" 
[3,] "1"  "1"  "0"  "1"  "*" 
 
> pattern<- c("0","1")
 
I would like to find, for each row, if it contains exactly the pattern of two
character strings, beginning with a "0" and followed by a
"1", i.e, exactly "0" "1". If it does, at which
year?
E.g. It should return 2006 for row 1, 2008 for row 2 and 2008 for row 3.
 
For as far as I know, the variations of the grep function group cannot search
for a pattern that has 2 or more character strings. I could do it with a loop
but I seek a more efficient way than a loop. How should I do it? Really
appreciated for your help!!!
 
Best regards,
Jing Liu
  		 	   		  
	[[alternative HTML version deleted]]

On Fri, Dec 17, 2010 at 2:34 PM, Jing Liu <quiet_jing0920 at hotmail.com>
wrote:
>> M<-
matrix(c("0","0","1","1","0","1","1","0","0","*","1","1","0","1","*"),nrow=3)
>> colnames(M)<-
c("2006","2007","2008","2009","2010")
>> M
> ? ? 2006 2007 2008 2009 2010
> [1,] "0" ?"1" ?"1" ?"*"
?"0"
> [2,] "0" ?"0" ?"0" ?"1"
?"1"
> [3,] "1" ?"1" ?"0" ?"1"
?"*"
>
>> pattern<- c("0","1")
>
> I would like to find, for each row, if it contains exactly the pattern of
two character strings, beginning with a "0" and followed by a
"1", i.e, exactly "0" "1". If it does, at which
year?
> E.g. It should return 2006 for row 1, 2008 for row 2 and 2008 for row 3.
>
I could only think of this
> apply(M, 1, function(z) grep('01', paste(z, collapse='')))
[1] 1 1 1
> apply(M, 1, function(z) grepl('01', paste(z, collapse='')))
[1] TRUE TRUE TRUE

But it doesn't return the position of the matched string. So this
isn't what you wanted.

Regards
Liviu

> For as far as I know, the variations of the grep function group cannot
search for a pattern that has 2 or more character strings. I could do it with a
loop but I seek a more efficient way than a loop. How should I do it? Really
appreciated for your help!!!
>
> Best regards,
> Jing Liu
>
> ? ? ? ?[[alternative HTML version deleted]]
>
> ______________________________________________
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> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
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>


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On Fri, Dec 17, 2010 at 09:34:57PM +0800, Jing Liu
wrote:
> 
> Dear all,
>  
> My question is illustrated by the following example:
>  
> I have a matrix M:
>  
> > M<-
matrix(c("0","0","1","1","0","1","1","0","0","*","1","1","0","1","*"),nrow=3)
> > colnames(M)<-
c("2006","2007","2008","2009","2010")
> > M
>      2006 2007 2008 2009 2010
> [1,] "0"  "1"  "1"  "*" 
"0"
> [2,] "0"  "0"  "0"  "1" 
"1"
> [3,] "1"  "1"  "0"  "1" 
"*"
>  
> > pattern<- c("0","1")
>  
> I would like to find, for each row, if it contains exactly the pattern of
two character strings, beginning with a "0" and followed by a
"1", i.e, exactly "0" "1". If it does, at which
year?
> E.g. It should return 2006 for row 1, 2008 for row 2 and 2008 for row 3.
If the pattern is always c("0","1"), the number of rows is
large
and the number of years is relatively small, then this may
computed also using matrix calculations. For example

  M <-
matrix(c("0","0","1","1","0","1","1","0","0","*","1","1","0","1","*"),nrow=3)
  colnames(M) <-
c("2006","2007","2008","2009","2010")
  year <- colnames(M)
  status <- rep(NA, times=nrow(M))
  for (i in seq(length(year) - 1)) {
      status[M[, i] == "0" & M[, i+1] == "1"] <-
year[i]
  }
  status # [1] "2006" "2008" "2008"

Petr Savicky.

On Dec 17, 2010, at 8:34 AM, Jing Liu wrote:
>
> Dear all,
>
> My question is illustrated by the following example:
>
> I have a matrix M:
>
>> M<-  
>> matrix 
>> (c 
>>
("0","0","1","1","0","1","1","0","0","*","1","1","0","1","*"),nrow=3)
>> colnames(M)<-
c("2006","2007","2008","2009","2010")
>> M
>     2006 2007 2008 2009 2010
> [1,] "0"  "1"  "1"  "*" 
"0"
> [2,] "0"  "0"  "0"  "1" 
"1"
> [3,] "1"  "1"  "0"  "1" 
"*"
>
>> pattern<- c("0","1")
>
> I would like to find, for each row, if it contains exactly the  
> pattern of two character strings, beginning with a "0" and
followed
> by a "1", i.e, exactly "0" "1". If it does,
at which year?
> E.g. It should return 2006 for row 1, 2008 for row 2 and 2008 for  
> row 3.
>
> For as far as I know, the variations of the grep function group  
> cannot search for a pattern that has 2 or more character strings. I  
> could do it with a loop but I seek a more efficient way than a loop.  
> How should I do it? Really appreciated for your help!!!
You can just paste() each row with collapse="._" and now can use grep-
ish functions as you were hoping to use.

 > m2 <- apply(M, 1, paste, collapse="_")
 > colnames(M)[(regexpr("0_1", m2)+1)/2]  # assuming number of  
characters per element are all 1
[1] "2006" "2008" "2008"

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West Hartford, CT