[R] Interpolate? a line

[David Winsemius]

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 <dwinsemius at comcast.net>
> To: Alaios <alaios at yahoo.com>
> Cc: Rhelp list <r-help at 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