R help - Dec 2007 - number of count of each unique row

Hi,

I have a matrix of duplicate rows. How to output a list the unique rows with
their count? I have used "unique" to have the unique rows, but
can't produce the occurences of each unique row.

Thanks

Louis


             
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Hi Louis,

If I am understanding your question correctly, here is one way:
suppose your matrix is M of dimension n x k.

D <- data.frame(M)  # convert to data frame
ones <- rep(1, n)   # a column of 1s

Now you can count the number of repeats of each unique row.

aggregate( ones, by = as.list(D), FUN = sum)

The output will be a data frame with the unique rows, and a column at
the end labeled "x" with the frequency of each unique row.

Once you get this you can convert to a list, manipulate, etc.  I am
sure that there exist faster/better methods.

Best,
Jay




On Dec 21, 2007 5:03 PM, Louis Martin <louismartinbis at yahoo.fr>
wrote:
> Hi,
>
> I have a matrix of duplicate rows. How to output a list the unique rows
with their count? I have used "unique" to have the unique rows, but
can't produce the occurences of each unique row.
>
> Thanks
>
> Louis
>
>
>
> ---------------------------------
>
>         [[alternative HTML version deleted]]
>
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> R-help at r-project.org mailing list
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>


-- 



***************************************************
G. Jay Kerns, Ph.D.
Assistant Professor / Statistics Coordinator
Department of Mathematics & Statistics
Youngstown State University
Youngstown, OH 44555-0002 USA
Office: 1035 Cushwa Hall
Phone: (330) 941-3310 Office (voice mail)
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E-mail: gkerns at ysu.edu
http://www.cc.ysu.edu/~gjkerns/

Louis Martin wrote:
> Hi,
> 
> I have a matrix of duplicate rows. How to output a list the unique rows
with their count? I have used "unique" to have the unique rows, but
can't produce the occurences of each unique row.
> 
Hi Louis,
If you want the unique rows returned, this might do the job.

unique.rows<-function(x) {
  nrows<-dim(x)[1]
  urows<-1:nrows
  for(i in 1:(nrows-1)) {
   for(j in (i+1):nrows) {
    if(!is.na(urows[j])) if(all(x[i,]==x[j,])) urows[j]<-NA
   }
  }
  return(x[urows[!is.na(urows)],])
}

Jim