# [R] comparing row by row in matrix

Petra Steiner steinep at uni-muenster.de
Tue Mar 26 23:18:24 CET 2002

```Thank you, that is the solution.

And the RR-coeffizient (number of agreements/total number) is this:
(ma%*%t(ma)) / ncol(ma)

Regards, Petra

> I don't know the Russell-Rao 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 row-by-column, 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 uni-muenster.de]
> Sent: Tuesday, March 26, 2002 2:08 PM
> To: r-help 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 Russell-Rao-coefficient, 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.uni-muenster.de
> http://santana.uni-muenster.de/~petra/
>
>
>
>
>
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