Thank you, that is the solution.
And the RRcoeffizient (number of agreements/total number) is this:
(ma%*%t(ma)) / ncol(ma)
Regards, Petra
I don't know the RussellRao coefficient but maybe this will help:
You can compute the number of times y and x are both 1 (that's what your
function f does) by counting 1s in y*x, i.e., sum(y*x), aka t(y)%*%x.
Ordinary matrix multiplication does this rowbycolumn, so if M and N are
two binary matrices, M%*%t(N) will have (i,j) entry equal to the count of 1s
in common to row i of M and row j of N.
Hope I've understood the problem correctly.
Reid Huntsinger
Original Message
From: Petra Steiner [mailto:steinep at unimuenster.de]
Sent: Tuesday, March 26, 2002 2:08 PM
To: rhelp at stat.math.ethz.ch
Subject: [R] comparing row by row in matrix
Hello,
and thanks for the two responses to my questions on binary matrixes, which
showed me that the functions I needed do not exist.
To get a distance matrix with the RussellRaocoefficient, I first have to
compare each row of a binary matrix with each row and count how many
elements are
a. equal
and
b. 1.
by
f < function(x,y) (sum(x == 1& y == x))
Now how can I iterate this over a matrix without a loop? I think apply etc.
will not work in this case.
Thanks for any help.
Regards,
Petra


Petra Steiner
Arbeitsbereich Linguistik
Universitaet Muenster
Huefferstrasse 27
48149 Muenster
Tel: 0251 / 83 39442
petra at marley.unimuenster.de
http://santana.unimuenster.de/~petra/..................................... .
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