On Sep 17, 2010, at 7:22 AM, Alaios wrote:
I would like to thank you again for your help.
But it seems that the floor function (ceiling, round too) create
more dots in the matrix that line really "touches".
You said "cells" not "dots". Are you trying to change the problem now?
My concern is rather that it can still miss cells.
unique( floor( cbind( seq(2,62, by=0.1), linefn(seq(2,62,
by=0.1)) ) ) )
You can see that in the picture below
http://yfrog.com/5blineswj
So, how to select the only the cells that the line "touches"?
If you had taken my suggestion of overlaying a grid rather than
plotting dots that fail to represent a "cell" (which I was taking to
be a square of dimension 1 x 1) you would see that my solution was
correct (at least to the point of not missing any cells so defined
that were touched up to a tolerance of 0.01 cell units. If you want to
define "cells" differently, then it's your turn to step up and get
mathematically precise. Calculus still works if you define
neighborhoods as hyperspheres s rather than epsilon by delta hyper-
rectangles.
# Here is the the illustrated sequence of getting to what I am calling
my "final answer",
# even though it could still miss an occasional cell.
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, ]
plot(m) # plots points
lines(c(2,62), c(3, 34)) # overlay line for comparison
#you can add a grid with
abline(v=2:62, h=3:34)
## First attempt at integer values of x
linefn <- function(x) 3+((34-3)/(62-2)) *(x-2)
findInterval(linefn(2:62), 3:34)
# Second attempt at 0.1 intervals
#########
cellidxs <- unique( floor( cbind( seq(2,62, by=0.1), # There will be
many duplicates after rounding down
linefn(seq(2,62, by=0.1)) ) ) ) # the same
function that just gets a y value
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))
# That is the first plot with coarse tolerances
#Third attempt:
# Now calculate a set of cell ids with tolerances that at ten-fold
more numerous
cellid2 <-unique( floor(cbind(seq(2,62, by=0.01), linefn(seq(2,62,
by=0.01) )) ) )
NROW(cellid2) # 91 cells
rect(cellid2[,1], cellid2[,2], cellid2[,1]+1, cellid2[,2]+1,
col="blue")
rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1,
col="red")
lines(2:62, 3+(34-3)/(62-2)*(0:60))
--
Best
David.
I would like to thank you in advance for your help
best Regards
Alex
From: David Winsemius <dwinsem...@comcast.net>
To: Alaios <ala...@yahoo.com>
Cc: Rhelp list <r-help@r-project.org>
Sent: Wed, September 15, 2010 1:55:10 PM
Subject: Re: [R] Interpolate? a line
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
>>
David Winsemius, MD
West Hartford, CT
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