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ordering x's and y's

2 messages · Venables, Bill (CMIS, Cleveland), robin hankin

Venables, Bill (CMIS, Cleveland)
# 
Robin Hankin asks below for a vectorized version of his solution below:
[WNV]  The original question had a single value for both x and y
with variable repititions.  I prefer to stick to this, but it could be
generalized slightly.

	library(gregmisc)
	xy.sets <- function(nx, ny, x = "x", y = "y") {
		tt <- combinations(nx + ny, nx)
		answer <- matrix(y, nrow(tt), nx + ny)
		answer[cbind(as.vector(row(tt)), as.vector(tt))] <- x
		answer
	}

	Checking:
	> xy.sets(2,3)
	      [,1] [,2] [,3] [,4] [,5]
	 [1,] "x"  "x"  "y"  "y"  "y" 
	 [2,] "x"  "y"  "x"  "y"  "y" 
	 [3,] "x"  "y"  "y"  "x"  "y" 
	 [4,] "x"  "y"  "y"  "y"  "x" 
	 [5,] "y"  "x"  "x"  "y"  "y" 
	 [6,] "y"  "x"  "y"  "x"  "y" 
	 [7,] "y"  "x"  "y"  "y"  "x" 
	 [8,] "y"  "y"  "x"  "x"  "y" 
	 [9,] "y"  "y"  "x"  "y"  "x" 
	[10,] "y"  "y"  "y"  "x"  "x" 

	The trick is the old matrix-as-index thing that surprises so many
people.

	Bill Venables.
robin hankin
# 
WNV writes
[rksh] Bill's solution is indeed vectorized and elegant; I spent some
time pondering the "trick" (as a reformed Matlab user, Bill's comment
about "matrix-as-index" techniques being surprising truly strikes
home, as the following code which I wrote years ago attests:

*****warning: Matlab code follows*********

function out=do_r(matrix,p)
%function out=do_r(matrix,p)
%
%example
%
%      a=reshape(1:12,3,4)
%      p=[1 1;1 4; 2 2]
%      do_r(a,p)    %consider 2nd row of p (=1,4); a(1,4)=10 as reqd.
%   
%      see assign
[n m]=size(matrix);
a=reshape(matrix,n*m,1);
places=(p(:,2)-1)*n+p(:,1);
out=a(places);

function out=assign(matrix,p,values);
%does what a(p)=values should do:
%a=reshape(1:12,3,4) 
%p=[1 2;1 3 ;3 4]
%assign(a,p,0)
%assign(a,p,-[99 99 99])
%
%
%see do_r
[n m]=size(matrix);
a=reshape(matrix,1,n*m);
places=p(:,1)+n*(p(:,2)-1);
a(places)=values;
out=reshape(a,n,m);
*****matlab code ends******

).  

I spent a bit of time trying to generalize Bill's xy.sets() but the
best I could come up with was


xy <- function(x, y) {
  nx <- length(x)
  ny <- length(y)
  tt <- combinations(nx + ny, nx)
  placeholder <- matrix(rep(FALSE,nx+ny), nrow(tt), nx + ny,byrow=F)
  answer <- placeholder
  placeholder[cbind(as.vector(row(tt)),   as.vector(tt))] <- TRUE
  answer <- t(answer)
  placeholder <- t(placeholder)
  answer[placeholder== TRUE] <- x
  answer[placeholder==FALSE] <- y 
  t(answer)
 }

then xy(1:3,10:11) works.  xy() has to deal with the transpose of
answer and placeholder in order to put the elements of x and y in
their correct places.  Is there a better way?

Oh yes, I had omitted to point out that do.thing2() could solve the
original problem with something like

R> do.thing2(rep("x",2),rep("y",3))

best 

rksh
[check deleted]
Robin K. S. Hankin, Lecturer,
School of Geography and Environmental Science
Tamaki Campus
Private Bag 92019 Auckland
New Zealand

r.hankin at auckland.ac.nz
tel 0064-9-373-7599 x6820; FAX 0064-9-373-7042

as of: Thu Dec 12 09:00:00 NZDT 2002
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