On Sep 15, 2010, at 7:24 AM, David Winsemius wrote:
Replacing context:
Hello everyone.
I have created a 100*100 matrix in R.
Let's now say that I have a line that starts from (2,3) point and
ends to the
(62,34) point. In other words this line starts at cell (2,3) and
ends at cell
(62,34).
Is it possible to get by some R function all the matrix's cells
that this line
transverses?
I would like to thank you for your feedback.
Best Regards
Alex
On Sep 15, 2010, at 6:52 AM, Michael Bedward wrote:
Hello Alex,
Here is one way to do it. It works but it's not pretty :)
If you want an alternative, consider that produces the Y cell
indices (since the x cell indices are already 2:62):
> linefn <- function(x) 3+((34-3)/(62-2)) *(x-2)
> findInterval(linefn(2:62), 3:34)
[1] 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11
11 12 12 13 13 14
[28] 14 15 15 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25
25 26 26 27 27 28
[55] 28 29 29 30 30 31 32
# that seems "off" by two
> linefn(62)
[1] 34
> linefn(2)
[1] 3 # but that checks out and I realized those were just indices
for the 3:34 findInterval vector
> (3:34)[findInterval(linefn(2:62), 3:34)]
[1] 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13
13 14 14 15 15 16
[28] 16 17 17 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27
27 28 28 29 29 30
[55] 30 31 31 32 32 33 34
( no rounding and I think the logic is clearer.)
But I also realized it didn't enumerate all the the cells were crossed
either, only indicating which cell was associated with an integer
value of x. Also would have even more serious problems if the slope
were greater than unity. To enumerate the cell indices that were
crossed, try:
unique( floor( cbind( seq(2,62, by=0.1), linefn(seq(2,62,
by=0.1)) ) ) )
[,1] [,2]
[1,] 2 3
[2,] 3 3
[3,] 4 4
[4,] 5 4
[5,] 5 5
[6,] 6 5
[7,] 7 5
[8,] 7 6
snipping interior results
[83,] 58 32
[84,] 59 32
[85,] 60 32
[86,] 60 33
[87,] 61 33
[88,] 62 34
That could probably be passed to rect() to illustrate (and check logic):
rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1,
col="red")
#redraw line :
lines(2:62, 3+(34-3)/(62-2)*(0:60))
--
David.
interp <- approx(c(2, 62), c(3, 34), method="linear", xout=2:62)
m <- matrix(c(interp$x, round(interp$y)), ncol=2)
tie <- m[,2] == c(-Inf, m[-nrow(m),2])
m <- m[ !tie, ]
You might want to examine the result like this...
plot(m) # plots points
lines(c(2,26), c(3, 34)) # overlay line for comparison
you can add a grid with
abline(v=2:62, h=3:34)
Michael
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David Winsemius, MD
West Hartford, CT
______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
