[R] vector replacement 1/0 to P/A

2009-08-18 Thread Lana Schaffer
Hi,
Can someone suggest an efficient way to substitute a vector/matrix
which contains 1's and 0's to P's and A's (resp.)?
Thanks,
Lana
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Re: [R] vector replacement 1/0 to P/A

2009-08-18 Thread Chuck Cleland
On 8/17/2009 10:22 AM, Lana Schaffer wrote:
 Hi,
 Can someone suggest an efficient way to substitute a vector/matrix
 which contains 1's and 0's to P's and A's (resp.)?
 Thanks,
 Lana

  Here is one approach:

mymat - matrix(rbinom(15, 1, .5), ncol=3)

mymat
 [,1] [,2] [,3]
[1,]100
[2,]001
[3,]101
[4,]010
[5,]110

mymat[] - sapply(mymat, function(x){ifelse(x == 1, 'P', ifelse(x == 0,
'A', NA))})

mymat
 [,1] [,2] [,3]
[1,] P  A  A
[2,] A  A  P
[3,] P  A  P
[4,] A  P  A
[5,] P  P  A

 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
Chuck Cleland, Ph.D.
NDRI, Inc. (www.ndri.org)
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
fax: (917) 438-0894

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Re: [R] vector replacement 1/0 to P/A

2009-08-18 Thread Duncan Murdoch

On 17/08/2009 10:22 AM, Lana Schaffer wrote:

Hi,
Can someone suggest an efficient way to substitute a vector/matrix
which contains 1's and 0's to P's and A's (resp.)?


x[x == 1] - P
x[x == 0] - A

(I added the quotes around 0 on the second line because the first line 
changed x to a character vector.  This isn't necessary, 0 == 0 comes 
out TRUE, but I think it is clearer.)


Duncan Murdoch

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Re: [R] vector replacement 1/0 to P/A

2009-08-18 Thread jim holtman
Will this do it:

 x - sample(0:1,10,TRUE)
 x
 [1] 1 0 0 1 0 1 0 0 0 0
 ifelse(x == 1, P, A)
 [1] P A A P A P A A A A



On Mon, Aug 17, 2009 at 10:22 AM, Lana Schafferschaf...@scripps.edu wrote:
 Hi,
 Can someone suggest an efficient way to substitute a vector/matrix
 which contains 1's and 0's to P's and A's (resp.)?
 Thanks,
 Lana
 __
 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
 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?

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Re: [R] vector replacement 1/0 to P/A

2009-08-18 Thread Petr PIKAL
Hi

As matrix is vector with dim attribute

 mymat-ifelse(mymat==0, A,P)

should be sufficient. 

Even with data frame it works

mydf-data.frame(mymat)
mydf-ifelse(mydf==0, A,P)
 mydf
 X1  X2  X3 
[1,] P P A
[2,] A A A
[3,] A A P
[4,] A A P
[5,] P P P

Regards
Petr


r-help-boun...@r-project.org napsal dne 18.08.2009 13:58:51:

 On 8/17/2009 10:22 AM, Lana Schaffer wrote:
  Hi,
  Can someone suggest an efficient way to substitute a vector/matrix
  which contains 1's and 0's to P's and A's (resp.)?
  Thanks,
  Lana
 
   Here is one approach:
 
 mymat - matrix(rbinom(15, 1, .5), ncol=3)
 
 mymat
  [,1] [,2] [,3]
 [1,]100
 [2,]001
 [3,]101
 [4,]010
 [5,]110
 
 mymat[] - sapply(mymat, function(x){ifelse(x == 1, 'P', ifelse(x == 0,
 'A', NA))})
 
 mymat
  [,1] [,2] [,3]
 [1,] P  A  A
 [2,] A  A  P
 [3,] P  A  P
 [4,] A  P  A
 [5,] P  P  A
 
  __
  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
  and provide commented, minimal, self-contained, reproducible code. 
 
 -- 
 Chuck Cleland, Ph.D.
 NDRI, Inc. (www.ndri.org)
 71 West 23rd Street, 8th floor
 New York, NY 10010
 tel: (212) 845-4495 (Tu, Th)
 tel: (732) 512-0171 (M, W, F)
 fax: (917) 438-0894
 
 __
 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
 and provide commented, minimal, self-contained, reproducible code.

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.