Re: [R] Interpolate? a line
I would like to thank you very much for making this clear. It seems that the solution you suggested is right one as the second attempt does find all the cells that are touched. Now I ll try to find out how much the line gets into one of this cell as every cell affects acts like a weight. The more you are in a cell the higher the weight will be. Best Regards Alex From: David Winsemius dwinsem...@comcast.net Cc: Rhelp list r-help@r-project.org Sent: Fri, September 17, 2010 6:36:51 PM Subject: Re: [R] Interpolate? a line 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 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
Re: [R] Interpolate? a line
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. 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? I would like to thank you in advance for your help best Regards Alex From: David Winsemius dwinsem...@comcast.net 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,]23 [2,]33 [3,]44 [4,]54 [5,]55 [6,]65 [7,]75 [8,]76 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 __ 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. 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. David Winsemius, MD West Hartford, CT [[alternative HTML version deleted]] __ 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.
Re: [R] Interpolate? a line
I think that's because of your definition for touches. If I read it correctly, David's code assumes that a cell is a square 1 unit wide and then followed the normal geometric definition of touches, ie. the line intersects with at least one point on the cell's boundary. You seem to have a different definition: cell centers perhaps ? or cell lower-left corner points ? Whether or not any particular definition is valid depends on your requirements. In other words - it's all up to you :) Michael On 17 September 2010 21:22, Alaios ala...@yahoo.com 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. 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? I would like to thank you in advance for your help best Regards Alex From: David Winsemius dwinsem...@comcast.net 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 __ 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. 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. David Winsemius, MD West Hartford, CT [[alternative HTML version deleted]] __ 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. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read
Re: [R] Interpolate? a line
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,]23 [2,]33 [3,]44 [4,]54 [5,]55 [6,]65 [7,]75 [8,]76 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
[R] Interpolate? a line
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 [[alternative HTML version deleted]] __ 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.
Re: [R] Interpolate? a line
Hello Alex, Here is one way to do it. It works but it's not pretty :) 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 Michael __ 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.
Re: [R] Interpolate? a line
Thank you very much. I am trying to execute it line by line to get some understanding how this works. Could you please explain me what these two lines do? tie - m[,2] == c(-Inf, m[-nrow(m),2]) m - m[ !tie, ] I would like to thank you in advance for your help Best Regards Alex From: Michael Bedward michael.bedw...@gmail.com Cc: Rhelp r-help@r-project.org Sent: Wed, September 15, 2010 12:52:45 PM Subject: Re: [R] Interpolate? a line Hello Alex, Here is one way to do it. It works but it's not pretty :) 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 Michael [[alternative HTML version deleted]] __ 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.
Re: [R] Interpolate? a line
Ah, now that you ask that I realize I was making an assumption assumption about what you want to achieve. Those two lines remove coords where the y value is equal to that of the preceding coord. That may or may not be what you want to do. To visualize what is happening, type plot(m) before doing those two lines (you'll see most points arranged in horizontal pairs), the execute the two lines and do plot(m) again. Michael On 15 September 2010 21:05, Alaios ala...@yahoo.com wrote: Thank you very much. I am trying to execute it line by line to get some understanding how this works. Could you please explain me what these two lines do? tie - m[,2] == c(-Inf, m[-nrow(m),2]) m - m[ !tie, ] __ 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.
Re: [R] Interpolate? a line
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.) -- 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 __ 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. 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.
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,]23 [2,]33 [3,]44 [4,]54 [5,]55 [6,]65 [7,]75 [8,]76 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 __ 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. 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. 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.
