On Sat, 13 Nov 1999, Jonathan Rougier wrote:
> May I raise the issue of the behaviour of image()? In image() dim(z) is
> supposed to be c(length(x), length(y)). This suggests that the user is
> supplying the midpoints of the rectangles. For all but the outer
> rectangles this is true. The outer rectangles, however, are only
> displayed in half width. So it is not possible to show correctly an image
> plot of equally-spaced abscissa. I would hazard a guess that these kinds
> of plots make up the bulk of calls to image() either directly or after a
> simple transformation (eg starting with boundary values rather than
> midpoints).
>
> Can I suggest that the code for image() be modified to show full
> rectangles for all values on the basis that x and y are midpoints?
> According to my code for R (which is a bit out of date!) the minimum
> change would require 4 lines to be modified in the function do_image of
> plot3d.c. The alternative, which I think would be more transparent, would
> be to call the function do_image with boundary values having previously
> made the appropriate calculations in image() according to whether
> midpoints or boundary values are supplied. Something like this happens in
> S+ (although not very transparently!).
I needed this in hurry this weekend for some fMRI plots. I've altered
image so that if z is m x n, x can be of length m or m+1 and y of length n
or n+1. If x is of length m it supplies the midpoints, if of length m+1
it supplies the boundaries of the cells. (All S compatible.) The internal
call is made with the boundary values. The outer cells are plotted
in full width.
The one change is if m=1 or n=1. I reckon that is an error, but S
and R previously filled the whole width. Does anyone want that behaviour?
At present those cases are an error.
--
Brian D. Ripley, ripley@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272860 (secr)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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