R help - Aug 2002 - Counting elements in 2 rows of a matrix

Dear R-users:

Sorry to bother so late with this question, which surely has simple answer.
   I'm working with matrices that contain either "1" or
"0", for example:

   [,1] [,2] [,3] [,4] [,5]
A    1    0    1    0    0
B    0    1    0    1    0
C    1    0    1    0    0
D    1    1    1    0    1

I want to count the number of "1" common to, say, A and B, the number
of
"1" that appear only in A and the number of "1" that appear
only in B.
Please let me know if there's a simple way of doing this without the 
complicated for...next loops that have come across my mind.

Fredy Mej?a
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> x
      [,1] [,2] [,3] [,4] [,5]
[1,]    1    0    1    0    0
[2,]    0    1    0    1    0
[3,]    1    0    1    0    0
[4,]    1    1    1    0    1
 > table(x[1,],x[2,])

     0 1
   0 1 2
   1 2 0
 > as.vector(table(x[1,],x[2,]))
[1] 1 2 2 0
 >prs<-function(x,i,j) as.vector(table(x[i,],x[j,])[2:4])
 > prs(x,1,2)
[1] 2 2 0
 > prs(x,1,3)
[1] 0 0 2

or use

ab.common<-function(x) crossprod(t(x))
ab.unique<-function(x) outer(rowSums(x),rep(1,nrow(x)))-pr11(x)

The first function will give what two rows have in common, the
second will give what they have unique (01 below the
diagonal, 10 above)

 > ab.common(x)
      [,1] [,2] [,3] [,4]
[1,]    2    0    2    2
[2,]    0    2    0    1
[3,]    2    0    2    2
[4,]    2    1    2    4
 > ab.unique(x)
      [,1] [,2] [,3] [,4]
[1,]    0    2    0    0
[2,]    2    0    2    1
[3,]    0    2    0    0
[4,]    2    3    2    0

For ab.unique you can also use
ab.unique <-function(x) matrix(rowSums(x), nrow(x),nrow(x))-ab.common(x)
which may be marginally faster.

On Friday, August 30, 2002, at 10:22 PM, Fredy Mej?a wrote:
> Dear R-users:
>
> Sorry to bother so late with this question, which surely has simple  
> answer.  I'm working with matrices that contain either "1" or
"0", for
> example:
>
>   [,1] [,2] [,3] [,4] [,5]
> A    1    0    1    0    0
> B    0    1    0    1    0
> C    1    0    1    0    0
> D    1    1    1    0    1
>
> I want to count the number of "1" common to, say, A and B, the
number
> of "1" that appear only in A and the number of "1" that
appear only in
> B.  Please let me know if there's a simple way of doing this without  
> the complicated for...next loops that have come across my mind.
>
> Fredy Mej?a
==Jan de Leeuw; Professor and Chair, UCLA Department of Statistics;
US mail: 9432 Boelter Hall, Box 951554, Los Angeles, CA 90095-1554
phone (310)-825-9550;  fax (310)-206-5658;  email: deleeuw at stat.ucla.edu
homepage: http://gifi.stat.ucla.edu
   
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Try either this or crossprod() ...
> x%*%t(x)
     [,1] [,2] [,3] [,4]
[1,]    2    0    2    2
[2,]    0    2    0    1
[3,]    2    0    2    2
[4,]    2    1    2    4

The above is "like" distance matrix. eg B and D have only one
"1" in common.
Notice the symmetry about the diagonal.
The diagonals are not so useful as they tell you how many "1"'s A,
B, C, D
have in them.
  ----- Original Message -----
  From: To: r-help at stat.math.ethz.ch
  Sent: Saturday, August 31, 2002 6:22 AM
  Subject: [R] Counting elements in 2 rows of a matrix


  Dear R-users:

  Sorry to bother so late with this question, which surely has simple
answer. I'm working with matrices that contain either "1" or
"0", for
example:

  [,1] [,2] [,3] [,4] [,5]
  A 1 0 1 0 0
  B 0 1 0 1 0
  C 1 0 1 0 0
  D 1 1 1 0 1

  I want to count the number of "1" common to, say, A and B, the
number of
"1" that appear only in A and the number of "1" that appear
only in B.
Please let me know if there's a simple way of doing this without the
complicated for...next loops that have come across my mind.

  Fredy Mej?a

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r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html
Send "info", "help", or "[un]subscribe"
(in the "body", not the subject !)  To: r-help-request at
stat.math.ethz.ch
_._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._