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

Petr PIKAL petr.pikal at precheza.cz
Tue Aug 18 15:55:05 CEST 2009

```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-bounces at 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,]    1    0    0
> [2,]    0    0    1
> [3,]    1    0    1
> [4,]    0    1    0
> [5,]    1    1    0
>
> 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 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.
>
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> and provide commented, minimal, self-contained, reproducible code.

```

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