comp.lang.idl-pvwave archive - RDF feed
type: channel
description: Messages from Usenet group comp.lang.idl-pvwave, compiled by Paulo Penteado
items: Re: Mapping image into a polar-square coordinate

Re: Mapping image into a polar-square coordinate
type: item
description: I suggest using 2-dimensional interpolation (for instance, "bilinear") to interpolate x,y data to radius and angle. You only need to call it once, so it should be fast. Ciao, Paolo cmejiapr...@gmail.com wrote: > Hi programmers, > > I have an image and I want to map an annulus of it (matrix 981X 981) > onto a rectangular axes whose columns are the angle, and the rows are > the radius to the central pixel. I tried: > > ;data has the image > xx1 = findgen(4096,10)*0. > for i=0,1023 do begin > roll=i*360./4096. > SB=rot(data,-roll,1,490.5,490.5,cubic=-0.5,missing=-1,/pivot ) > xx1[i,*]=SB[50:59,490] > for j=0,9 do xx1[i+2048,j]=SB[930-j,490] > xx1[i+3072,*]=SB[490,50:59] > for j=0,9 do xx1[i+1024,j]=SB[490,930-j] > endfor > > But it takes too long to run, i need something faster. Any advise? > Thanks
creator: pgrigis
date: 2008-07-09T13:29:21-00:00

Re: Mapping image into a polar-square coordinate
type: item
description: seems like my previous post got lost... anyway I am suggesting to use interpolation (i.e. BILINEAR) to convert from cartesian to polar coordinates. That should be pretty fast. Ciao, Paolo cmejiapr...@gmail.com wrote: > Hi programmers, > > I have an image and I want to map an annulus of it (matrix 981X 981) > onto a rectangular axes whose columns are the angle, and the rows are > the radius to the central pixel. I tried: > > ;data has the image > xx1 = findgen(4096,10)*0. > for i=0,1023 do begin > roll=i*360./4096. > SB=rot(data,-roll,1,490.5,490.5,cubic=-0.5,missing=-1,/pivot ) > xx1[i,*]=SB[50:59,490] > for j=0,9 do xx1[i+2048,j]=SB[930-j,490] > xx1[i+3072,*]=SB[490,50:59] > for j=0,9 do xx1[i+1024,j]=SB[490,930-j] > endfor > > But it takes too long to run, i need something faster. Any advise? > Thanks
creator: pgrigis
date: 2008-07-09T14:47:05-00:00

Re: Mapping image into a polar-square coordinate
type: item
description: On Jul 9, 7:47 am, pgri...@gmail.com wrote: > seems like my previous post got lost... > anyway I am suggesting to use interpolation (i.e. BILINEAR) > to convert from cartesian to polar coordinates. > That should be pretty fast. > > Ciao, > Paolo > > cmejiapr...@gmail.com wrote: >> Hi programmers, > >> I have an image and I want to map an annulus of it (matrix 981X 981) >> onto a rectangular axes whose columns are the angle, and the rows are >> the radius to the central pixel. I tried: > >> ;data has the image >> xx1 = findgen(4096,10)*0. >> for i=0,1023 do begin >> roll=i*360./4096. >> SB=rot(data,-roll,1,490.5,490.5,cubic=-0.5,missing=-1,/pivot ) >> xx1[i,*]=SB[50:59,490] >> for j=0,9 do xx1[i+2048,j]=SB[930-j,490] >> xx1[i+3072,*]=SB[490,50:59] >> for j=0,9 do xx1[i+1024,j]=SB[490,930-j] >> endfor > >> But it takes too long to run, i need something faster. Any advise? >> Thanks yeah, but i dont know how to extract a rectangular matrix which rows are radius and columns are angles
creator: Camilo Mejia
date: 2008-07-09T19:01:12-00:00

Re: Mapping image into a polar-square coordinate
type: item
description: > yeah, but i dont know how to extract a rectangular matrix which rows > are radius and columns are angles Paolo's suggestion of bilinear is a good one. The best thing to do is construct a polar coordinate system and then transform that into a rectangular system that is equivalent to your pixel indices. Suppose there is a rectangular coordinate system, centered on the middle pixel of your 981 x 981 data. Then if we want to extract the annulus which is between 100 and 200 pixels from the center, we could do something like this. ----- image: 981 x 981 (same as your ``data'' array) new_image: 4096 x 10 ;; first construct the equivalent polar coordinates min_r = 100.0 max_r = 200.0 ;; this is theta = [0, 2*pi) new_th = rebin(dindgen(4096) / 4096d * (2d * !dpi), 4096, 10) ;; this is r = [r_min, r_max] new_r = rebin(transpose((max_r - min_r) * dindgen(10) / 9d + min_r), 4096, 10) ;; now convert to rectangular coordinates ;; and shift such that the origin lies not at the center ;; but at image[0,0] new_x = new_r * cos(new_th) + 490.0 new_y = new_r * sin(new_th) + 490.0 ;; new_x and new_y are fractional pixel coordinates ;; use bilinear to extract the values new_img = bilinear(image, new_x ,new_y) ----- Hope that helps, Josiah
date: 2008-07-09T19:43:58-00:00

Re: Mapping image into a polar-square coordinate
type: item
description: On Jul 9, 12:43 pm, "jsch...@gmail.com" <jsch...@gmail.com> wrote: >> yeah, but i dont know how to extract a rectangular matrix which rows >> are radius and columns are angles > > Paolo's suggestion of bilinear is a good one. > > The best thing to do is construct a polar coordinate system and then > transform that into a rectangular system that is equivalent to your > pixel indices. > > Suppose there is a rectangular coordinate system, centered on the > middle pixel of your 981 x 981 data. Then if we want to extract the > annulus which is between 100 and 200 pixels from the center, we could > do something like this. > > ----- > > image: 981 x 981 (same as your ``data'' array) > new_image: 4096 x 10 > > ;; first construct the equivalent polar coordinates > > min_r = 100.0 > max_r = 200.0 > > ;; this is theta = [0, 2*pi) > new_th = rebin(dindgen(4096) / 4096d * (2d * !dpi), 4096, 10) > > ;; this is r = [r_min, r_max] > new_r = rebin(transpose((max_r - min_r) * dindgen(10) / 9d + min_r), > 4096, 10) > > ;; now convert to rectangular coordinates > ;; and shift such that the origin lies not at the center > ;; but at image[0,0] > > new_x = new_r * cos(new_th) + 490.0 > new_y = new_r * sin(new_th) + 490.0 > > ;; new_x and new_y are fractional pixel coordinates > ;; use bilinear to extract the values > > new_img = bilinear(image, new_x ,new_y) > > ----- > > Hope that helps, > Josiah Thanks a lot Josiah and Paolo, it works awesome Camilo
creator: Camilo Mejia
date: 2008-07-09T20:00:21-00:00