On Oct 25, 2009, at 1:51 AM, Kang Min wrote:
Hi Milton,
The matrix can be generated using
p = matrix(1:50, nrow=5)
If I just use levelplot(p), it gives me a graph that is vertical. How
can I rotate it so it becomes horizontal?
I cannot do
q = t(p); levelplot(q)
because this is representing a map from a piece of land, transposing
the data will flip the map upside down laterally.
You can reverse the order of the indices. I cannot quite figure out
which reversal will accomplish your goals because your description
seems ambiguous but take a look at the output of htese and you should
be able to apply hte strategy to your situation:
library(lattice)
#as a courtesy to the helpeRs it's nice to reference the package in
which you function resides
p = matrix(1:50, nrow=5)
q2 = p[5:1, ] ; levelplot(q2)
levelplot(p)
q = t(p)
levelplot(q[10:1, ])
-- David
Thanks,
Kang Min
On Oct 25, 12:40 pm, milton ruser <milton.ru...@gmail.com> wrote:
Hi Kang,
Could you send a reproducible sample-code?
Bests
miltinho
On Sat, Oct 24, 2009 at 11:32 PM, Kang Min <ngokang...@gmail.com>
wrote:
Hi all,
I have a matrix with 5 rows and 10 columns, which represent the
grids
on a rectangular map.
I used the code below to plot, but it gives me the map with the 10
columns as y-axis, and the 5 rows as the x-axis, and the (0,0) point
is at the usual bottom left hand corner. My map starts with the
(0,0)
at the top left hand corner.
How can I rotate the map 90 degrees clockwise so the (0,0) starts at
the top left? That means I need the 10 columns to be x-axis. I
cannot
transpose the matrix because the map would be laterally flipped
over.
levelplot(quadsec.mat, colorkey=list(space="bottom"), scales=list
(tick.number=10), aspect=c("iso"))
Thanks, I'm ready to give more information if needed.
Kang Min
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and provide commented, minimal, self-contained, reproducible code.
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David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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