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another basic ?

5 messages · jimi adams, Peter Dalgaard, Roger D. Peng +2 more

jimi adams
# 
i am sure that there is a simple solution that i am just overlooking, but 
there is no logical test for =
there is one for != and >,<, >=, <=...
what do i do if i need to check if something IS EQUAL

or is there an easy way, in working with a matrix of size n x 3,  to sort 
the matrix by one of the columns?

specifically what i am trying to do is i have a matrix with individual and 
group labels that is randomly generated, and is by default sorted by the 
individual labels ( column 1) and i want them to be organized into unique 
group labels (comlumn 2)
thanks
jimi
jimi adams
Department of Sociology
The Ohio State University
300 Bricker Hall
190 N. Oval Mall
Columbus, OH 43210-1353
614-688-4261

our mind has a remarkable ability to think of contents as being independent 
of the act of thinking
                                             -georg simmel

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Peter Dalgaard
# 
jimi adams <adams.644 at osu.edu> writes:
Double it: ==
m[order(m[,2]),]
Roger D. Peng
#
On Mon, 25 Feb 2002, jimi adams wrote:

            

      
      
      
    

        
Depends.  To test if two objects are the same, you can use identical().
Or you might use == in other situations.  It's hard to know without more
details.

            

      
      
      
    

        
If m is an n x 3 matrix, to sort by column 1, you could do:

            

      
      
      
    

        
-roger
_______________________________
UCLA Department of Statistics
rpeng at stat.ucla.edu
http://www.stat.ucla.edu/~rpeng

            

      
      
      
    

        
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      Andrew Perrin
    

    

      
      #
    

  


  

      
It's == if I'm understanding your question right.

----------------------------------------------------------------------
Andrew J Perrin - andrew_perrin at unc.edu - http://www.unc.edu/~aperrin
 Assistant Professor of Sociology, U of North Carolina, Chapel Hill
      269 Hamilton Hall, CB#3210, Chapel Hill, NC 27599-3210 USA

        
On Mon, 25 Feb 2002, jimi adams wrote:

        

            

      
      
      
    

        
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      Mark Myatt
    

    

      
      #
    

  


  

      
jimi adams <adams.644 at osu.edu> writes:

            

      
      
      
    

        
==

            

      
      
      
    

        
Use order() with the column (or row) indices:

        my.matrix[order(my.matrix[,column.to.sort.on]), ]

As in:

        > my.matrix <- matrix(c(5:1, 6:8, 12:9), nrow = 4)
        > my.matrix

             [,1] [,2] [,3]
        [1,]    5    1   12
        [2,]    4    6   11
        [3,]    3    7   10
        [4,]    2    8    9

        > my.sorted.matrix <- my.matrix[order(my.matrix[ ,1]), ]
        > my.sorted.matrix

             [,1] [,2] [,3]
        [1,]    2    8    9
        [2,]    3    7   10
        [3,]    4    6   11
        [4,]    5    1   12

Best wishes,

Mark

--
Mark Myatt


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