iterate along a ray: linear algebra?
I am in the process of implementing an image processing algorithm that requires following rays extending outwards from a starting point and calculating the intensity derivative at each point. The idea is to find the point where the difference in intensity goes beyond a particular threshold. Specifically I'm examining an image of an eye to find the pupil, and the edge of the pupil is a sharp change in intensity. How does one iterate along a line in a 2d matrix, and is there a better way to do this? Is this a problem that linear algebra can help with? Thanks Stephen Emslie
Hi Stephen, I don't know much about image analysis, but in the scipy tutorial (7.2Filtering), there is an example of an image filter that highlights the edges of an image. If I guess correctly, it finds the smoothing spline of the image and then computes the derivative of the spline. A high derivative means that the image intensity shifts rapidly, so maybe that could help you. David 2006/6/30, stephen emslie <stephenemslie@gmail.com>:
I am in the process of implementing an image processing algorithm that requires following rays extending outwards from a starting point and calculating the intensity derivative at each point. The idea is to find the point where the difference in intensity goes beyond a particular threshold.
Specifically I'm examining an image of an eye to find the pupil, and the edge of the pupil is a sharp change in intensity.
How does one iterate along a line in a 2d matrix, and is there a better way to do this? Is this a problem that linear algebra can help with?
Thanks Stephen Emslie
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I've done something similar a few years ago (numarray,numeric). I started roughly at the middle and did 64 points from a reference point (xc,yc). This point together with a point at the edge of the image (xp,yp) also defined a reference angle (a0). (ysize,xsize) is the shape of the intensity image. I used the following code to calculate points of interest: na=64 xr,yr=xsizexc,ysizeyc a0=arctan2(ypyr,xpxc) if a0<0: a0+=2*pi ac=arctan2([yr,yr,yc,yc],[xr,xc,xc,xr]) if numarray: ac[ac<0]+=2*pi else: ac=choose(ac<0,(ac,ac+2*pi)) a1,a2,a3,a4=ac rmaxfn={ 0: lambda a: a<=a1 and xr/cos(a0.0*pi) or yr/cos(0.5*pia), 1: lambda a: a<=a2 and yr/cos(a0.5*pi) or xc/cos(1.0*pia), 2: lambda a: a<=a3 and xc/cos(a1.0*pi) or yc/cos(1.5*pia), 3: lambda a: a<=a4 and yc/cos(a1.5*pi) or xr/cos(2.0*pia) } angles=arange(a0,a0+2*pi,2*pi/na) if numarray: angles[angles>=2*pi]=2*pi else: angles=choose(angles>=2*pi,(angles,angles2*pi)) nr=int(ceil(sqrt(max(yc,yr)**2+max(xc,xr)**2))) crmax=array([int(floor(rmaxfn[floor(a*2/pi)](a))) for a in angles]) cr=outerproduct(ones(na),arange(float(nr))) ca=outerproduct(angles,ones(nr)) x=cr*cos(ca)+xc y=cr*sin(ca)+yc After this I did cubic spline interpolation in the image with these points and did something useful. I don't know how relevant this is to you and it doesn't use the linear algebra package but it might give you some hint. If you find out a nifty way to do your rays please post on this thread. Sidenote  Watch my explicit float argument to arange and even putting in pi there in one case. There's a discussion on this list that floats in arange are troublesome On 6/30/06, stephen emslie <stephenemslie@gmail.com> wrote:
I am in the process of implementing an image processing algorithm that requires following rays extending outwards from a starting point and calculating the intensity derivative at each point. The idea is to find the point where the difference in intensity goes beyond a particular threshold.
Specifically I'm examining an image of an eye to find the pupil, and the edge of the pupil is a sharp change in intensity.
How does one iterate along a line in a 2d matrix, and is there a better way to do this? Is this a problem that linear algebra can help with?
Thanks Stephen Emslie
Using Tomcat but need to do more? Need to support web services, security? Get stuff done quickly with preintegrated technology to make your job easier Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo http://sel.asus.falkag.net/sel?cmd=lnk&kid=120709&bid=263057&dat=121642
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I've found some matlab code that seems to do the same sort of thing. Interestingly enough it just uses trigonomotry to do find the x,y positions in the matrix that correspond to the ray at a particular angle. I had origionally discarded this idea because I thought there must be a more efficient way to do it, but perhaps not. I still have a lot to learn about numpy :) Stephen (sorry for the double post to you Torgil) On 7/3/06, Torgil Svensson <torgil.svensson@gmail.com> wrote:
I've done something similar a few years ago (numarray,numeric). I started roughly at the middle and did 64 points from a reference point (xc,yc). This point together with a point at the edge of the image (xp,yp) also defined a reference angle (a0). (ysize,xsize) is the shape of the intensity image.
I used the following code to calculate points of interest:
na=64 xr,yr=xsizexc,ysizeyc a0=arctan2(ypyr,xpxc) if a0<0: a0+=2*pi ac=arctan2([yr,yr,yc,yc],[xr,xc,xc,xr]) if numarray: ac[ac<0]+=2*pi else: ac=choose(ac<0,(ac,ac+2*pi)) a1,a2,a3,a4=ac rmaxfn={ 0: lambda a: a<=a1 and xr/cos(a0.0*pi) or yr/cos(0.5*pia), 1: lambda a: a<=a2 and yr/cos(a0.5*pi) or xc/cos(1.0*pia), 2: lambda a: a<=a3 and xc/cos(a1.0*pi) or yc/cos(1.5*pia), 3: lambda a: a<=a4 and yc/cos(a1.5*pi) or xr/cos(2.0*pia) } angles=arange(a0,a0+2*pi,2*pi/na) if numarray: angles[angles>=2*pi]=2*pi else: angles=choose(angles>=2*pi,(angles,angles2*pi)) nr=int(ceil(sqrt(max(yc,yr)**2+max(xc,xr)**2))) crmax=array([int(floor(rmaxfn[floor(a*2/pi)](a))) for a in angles]) cr=outerproduct(ones(na),arange(float(nr))) ca=outerproduct(angles,ones(nr)) x=cr*cos(ca)+xc y=cr*sin(ca)+yc
After this I did cubic spline interpolation in the image with these points and did something useful. I don't know how relevant this is to you and it doesn't use the linear algebra package but it might give you some hint.
If you find out a nifty way to do your rays please post on this thread.
Sidenote  Watch my explicit float argument to arange and even putting in pi there in one case. There's a discussion on this list that floats in arange are troublesome
I am in the process of implementing an image processing algorithm that requires following rays extending outwards from a starting point and calculating the intensity derivative at each point. The idea is to find
point where the difference in intensity goes beyond a particular
On 6/30/06, stephen emslie <stephenemslie@gmail.com> wrote: the threshold.
Specifically I'm examining an image of an eye to find the pupil, and the edge of the pupil is a sharp change in intensity.
How does one iterate along a line in a 2d matrix, and is there a better
way
to do this? Is this a problem that linear algebra can help with?
Thanks Stephen Emslie
Using Tomcat but need to do more? Need to support web services, security? Get stuff done quickly with preintegrated technology to make your job easier Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo http://sel.asus.falkag.net/sel?cmd=lnk&kid=120709&bid=263057&dat=121642
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participants (3)

David Huard

stephen emslie

Torgil Svensson