I'm trying to implement Fabian Timm's eye-tracking algorithm [http://www.inb.uni-luebeck.de/publikationen/pdfs/TiBa11b.pdf] (found here: [http://thume.ca/projects/2012/11/04/simple-accurate-eye-center-tracking-in-opencv/]) in numpy and OpenCV and I've hit a snag. I think I've vectorized my implementation decently enough, but it's still not fast enough to run in real time and it doesn't detect pupils with as much accuracy as I had hoped. This is my first time using numpy, so I'm not sure what I've done wrong.
def find_pupil(eye):
eye_len = np.arange(eye.shape[0])
xx,yy = np.meshgrid(eye_len,eye_len) #coordinates
XX,YY = np.meshgrid(xx.ravel(),yy.ravel()) #all distance vectors
Dx,Dy = [YY-XX, YY-XX] #y2-y1, x2-x1 -- simpler this way because YY = XXT
Dlen = np.sqrt(Dx**2+Dy**2)
Dx,Dy = [Dx/Dlen, Dy/Dlen] #normalized
Gx,Gy = np.gradient(eye)
Gmagn = np.sqrt(Gx**2+Gy**2)
Gx,Gy = [Gx/Gmagn,Gy/Gmagn] #normalized
GX,GY = np.meshgrid(Gx.ravel(),Gy.ravel())
X = (GX*Dx+GY*Dy)**2
eye = cv2.bitwise_not(cv2.GaussianBlur(eye,(5,5),0.005*eye.shape[1])) #inverting and blurring eye for use as w
eyem = np.repeat(eye.ravel()[np.newaxis,:],eye.size,0)
C = (np.nansum(eyem*X, axis=0)/eye.size).reshape(eye.shape)
return np.unravel_index(C.argmax(), C.shape)
and the rest of the code:
def find_eyes(face):
left_x, left_y = [int(floor(0.5 * face.shape[0])), int(floor(0.2 * face.shape[1]))]
right_x, right_y = [int(floor(0.1 * face.shape[0])), int(floor(0.2 * face.shape[1]))]
area = int(floor(0.2 * face.shape[0]))
left_eye = (left_x, left_y, area, area)
right_eye = (right_x, right_y, area, area)
return [left_eye,right_eye]
faceCascade = cv2.CascadeClassifier("haarcascade_frontalface_default.xml")
video_capture = cv2.VideoCapture(0)
while True:
# Capture frame-by-frame
ret, frame = video_capture.read()
gray = cv2.cvtColor(frame, cv2.COLOR_BGR2GRAY)
faces = faceCascade.detectMultiScale(
gray,
scaleFactor=1.1,
minNeighbors=5,
minSize=(30, 30),
flags=cv2.CASCADE_SCALE_IMAGE
)
# Draw a rectangle around the faces
for (x, y, w, h) in faces:
cv2.rectangle(frame, (x, y), (x+w, y+h), (0, 255, 0), 2)
roi_gray = gray[y:y+h, x:x+w]
roi_color = frame[y:y+h, x:x+w]
eyes = find_eyes(roi_gray)
for (ex,ey,ew,eh) in eyes:
eye_gray = roi_gray[ey:ey+eh,ex:ex+ew]
eye_color = roi_color[ey:ey+eh,ex:ex+ew]
cv2.rectangle(roi_color,(ex,ey),(ex+ew,ey+eh),(255,0,0),2)
px,py = find_pupil(eye_gray)
cv2.rectangle(eye_color,(px,py),(px+1,py+1),(255,0,0),2)
# Display the resulting frame
cv2.imshow('Video', frame)
if cv2.waitKey(1) & 0xFF == ord('q'):
break
# When everything is done, release the capture
video_capture.release()
cv2.destroyAllWindows()
You can perform many of those operations that save replicated elements and then perform some mathematical opertaions by directly performing the mathematical operatrions after creating singleton dimensions that would allow NumPy broadcasting. Thus, there would be two benefits - On the fly operations to save workspace memory and performance boost. Also, at the end, we can replace the nansum calculation with a simplified version. Thus, with all of that philosophy in mind, here's one modified approach -
def find_pupil_v2(face, x, y, w, h):
eye = face[x:x+w,y:y+h]
eye_len = np.arange(eye.shape[0])
N = eye_len.size**2
eye_len_diff = eye_len[:,None] - eye_len
Dlen = np.sqrt(2*((eye_len_diff)**2))
Dxy0 = eye_len_diff/Dlen
Gx0,Gy0 = np.gradient(eye)
Gmagn = np.sqrt(Gx0**2+Gy0**2)
Gx,Gy = [Gx0/Gmagn,Gy0/Gmagn] #normalized
B0 = Gy[:,:,None]*Dxy0[:,None,:]
C0 = Gx[:,None,:]*Dxy0
X = ((C0.transpose(1,0,2)[:,None,:,:]+B0[:,:,None,:]).reshape(N,N))**2
eye1 = cv2.bitwise_not(cv2.GaussianBlur(eye,(5,5),0.005*eye.shape[1]))
C = (np.nansum(X,0)*eye1.ravel()/eye1.size).reshape(eye1.shape)
return np.unravel_index(C.argmax(), C.shape)
There's one repeat still left in it at Dxy. It might be possible to avoid that step and Dxy0 could be fed directly into the step that uses Dxy to give us X, but I haven't worked through it. Everything's converted to broadcasting based!
Runtime test and output verification -
In [539]: # Inputs with random elements
...: face = np.random.randint(0,10,(256,256)).astype('uint8')
...: x = 40
...: y = 60
...: w = 64
...: h = 64
...:
In [540]: find_pupil(face,x,y,w,h)
Out[540]: (32, 63)
In [541]: find_pupil_v2(face,x,y,w,h)
Out[541]: (32, 63)
In [542]: %timeit find_pupil(face,x,y,w,h)
1 loops, best of 3: 4.15 s per loop
In [543]: %timeit find_pupil_v2(face,x,y,w,h)
1 loops, best of 3: 529 ms per loop
It seems we are getting close to 8x speedup!
Related
I have an image and I would like to blur it in one specific direction and distance using Matlab.
I found out there is a filter called fspecial('motion',len,theta).
Here there is an example:
I = imread('cameraman.tif');
imshow(I);
H = fspecial('motion',20,45);
MotionBlur = imfilter(I,H,'replicate');
imshow(MotionBlur);
However the blurred picture is blurred in 2 directions! In this case 225 and 45 degrees.
What should it do in order to blur it just in a specific direction (e.g. 45) and not both?
I think you want what's called a "comet" kernel. I'm not sure what kernel is used for the "motion" blur, but I'd guess that it's symmetrical based on the image you provided.
Here is some code to play with that applies the comet kernel in one direction. You'll have to change things around if you want an arbitrary angle. You can see from the output that it's smearing in one direction, since there is a black band on only one side (due to the lack of pixels there).
L = 5; % kernel width
sigma=0.2; % kernel smoothness
I = imread('cameraman.tif');
x = -L:1.0:L;
[X,Y] = meshgrid(x,x);
H1 = exp((-sigma.*X.^2)+(-sigma.*Y.^2));
kernel = H1/sum((H1(:)));
Hflag = double((X>0));
comet_kernel = Hflag.*H1;
comet_kernel=comet_kernel/sum(comet_kernel(:));
smearedImage = conv2(double(I),comet_kernel,'same');
imshow(smearedImage,[]);
Updated code: This will apply an arbitrary rotation to the comet kernel. Note also the difference between sigma in the previous example and sx and sy here, which control the length and width parameters of the kernel, as suggested by Andras in the comments.
L = 5; % kernel width
sx=3;
sy=10;
theta=0;
I = imread('cameraman.tif');
x = -L:1.0:L;
[X,Y] = meshgrid(x,x);
rX = X.*cos(theta)-Y.*sin(theta);
rY = X.*sin(theta)+Y.*cos(theta);
H1 = exp(-((rX./sx).^2)-((rY./sy).^2));
Hflag = double((0.*rX+rY)>0);
H1 = H1.*Hflag;
comet_kernel = H1/sum((H1(:)))
smearedImage = conv2(double(I),comet_kernel,'same');
imshow(smearedImage,[]);
Based on Anger Density's answer I wrote this code that solves my problem completely:
L = 10; % kernel width
sx=0.1;
sy=100;
THETA = ([0,45,90,135,180,225,270,320,360])*pi/180;
for i=1:length(THETA)
theta=(THETA(i)+pi)*-1;
I = imread('cameraman.tif');
x = -L:1.0:L;
[X,Y] = meshgrid(x,x);
rX = X.*cos(theta)-Y.*sin(theta);
rY = X.*sin(theta)+Y.*cos(theta);
H1 = exp(-((rX./sx).^2)-((rY./sy).^2));
Hflag = double((0.*rX+rY)>0);
H1 = H1.*Hflag;
comet_kernel = H1/sum((H1(:)));
smearedImage = conv2(double(I),comet_kernel,'same');
% Fix edges
smearedImage(:,[1:L, end-L:end]) = I(:,[1:L, end-L:end]); % Left/Right edge
smearedImage([1:L, end-L:end], :) = I([1:L, end-L:end], :); % Top/bottom edge
% Keep only inner blur
smearedImage(L:end-L,L:end-L) = min(smearedImage(L:end-L,L:end-L),double(I(L:end-L,L:end-L)));
figure
imshow(smearedImage,[]);
title(num2str(THETA(i)*180/pi))
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
end
I track the motion of an object from a video file in MATLAB and save the locations from each frame in a numberOfFrames x 2 array.
I know the theoretical path or intended path. When recording the movie the camera is at some unknown angle in space. Therefore, the path is skewed. The only information I have is the scaling between the pixels and millimeters by using the object diameter.
Now I would like to rotate the intended path, and move it around until it is overlaid on the tracked motion path.
I start with my theoretical path (Pth) then rotate it in 3-dimensions to "Pthr". After that I loop over each point in "M". And for each point in "M", I look for the closest point from "Pthr". Then, I repeat for the next point in "M". This probably has a problem of choosing the same point in "Pthr" for multiple points in "M".
I noticed this is sensitive to my initial guess and it gives terrible results.
Also, M is not a perfect path, since it is experimental measurements it is no where near perfect. measured vs. theoretical unrotated path
% M = [Mx,My], is location in x and y, from motion tracking.
% scale = 20; % pixels/mm, using the size of object
% I build the theoretical path (Pth) goes from to (0,0,0) to (0,3,0) to
% (3,3,0) to (3,0,0) to be approximately the same length as M
Pthup = linspace(0,3,num)';
Pthdwn = linspace(3,0,num)';
Pth0 = zeros(size(Pthup));
Pth3 = 3*ones(size(Pthup));
% Pth is approximately same length as M
Pth = scale*[Pth0 Pthup Pth0;Pthup Pth3 Pth0;Pth3 Pthdwn Pth0];
% using fmincon in matlab to minimize the sum of the square
lb = [145 0 -45 min(min(M)) min(min(M))]; %upper bound
ub = [180 90 45 max(max(M)) max(max(M))]; %lower bound
coro = [180 0 0 mean(Mx) mean(My)]; %initial guess
% initial guess (theta(x),theta(y),theta(z), shift in x, shift in y)
cnt = 0; er = 1;
while (abs(er)>0.1)
[const,fval] = fmincon(#(cor)findOrientation(cor,Pth,M),coro,[],[],[],[],lb,ub);
er = sum(const-coro);
coro = const;
cnt = 1+cnt;
if (cnt>50)
cnt = cnt;
break
end
end
%% function findOrientation keeps rotating Pth until it is closest to M
function [Eo] = findOrientation(cor,Pth,M)
% cor = [angle of rotations, center coordinate];(degrees, non-dimensiolaized in pixels)
% M = measured coordinates from movie in pixel
% coor: is output of the form [x-coordiante,y-coordinate, absolute distance from Center(i,:)]
% F = sum of least square, sum(coor(:,3))
%% Rotation of theoretical path about z,y,x and shifting in it in xy
thx = cor(1);
thy = cor(2);
thz = cor(3);
xy = cor([4:5]);
% T = [cosd(thn) -sind(thn);
% sind(thn) cosd(thn)]; %rotation matrix in 3D
Tz = [cosd(thz) -sind(thz) 0;
sind(thz) cosd(thz) 0;0 0 1]; %rotation matrix
Ty = [cosd(thy) 0 -sind(thy);0 1 0;
sind(thy) 0 cosd(thy)]; %rotation matrix
Tx = [1 0 0;0 cosd(thx) -sind(thx);
0 sind(thx) cosd(thx)]; %rotation matrix
Pthr = zeros(size(Pth));
for i = 1:size(Pth,1)
xp = Tz*Pth(i,:)';
xp = Ty*xp;
xp = Tx*xp;
Pthr(i,:) = xp.';
end
Pthr = Pthr(:,[1,2]); % omit third value because it is 2D
Pthr = Pthr + [cor([4:5])];
rin = sqrt(Pthr(:,1).^2+Pthr(:,2).^2); %theoretical radius
Centern = sqrt(M(:,1).^2 + M(:,2).^2);%measured radius
for i = 1:size(M,1) %loop over each point in tracked motion
sub = Pthr-M(i,:); %subtracting M(i,:) from all Pthr
for j = 1:length(sub)
dist(j,1) = norm(sub(j,:));% distance from M(i,:) to all ri
end
%index is based on the min absolute distance between Pthr and M(i,:). It chooses the closest Pthr to a specific M(i,:)
[mn, index] = min(dist);
erri = abs(rin(index)-Centern(i))./rin(index);
coor(i,:) = erri;
end
Eo = sum(coor);
I have solved a system of two second-order differential equations using an implementation of Euler's method in Julia. The below code shows how Euler's method has been called to solve the system in question.
θ1 = 1.1900518004210798; θ2 = 0.3807445263738167
f(t, y) = [y[2], -2(y[1] - θ1) - 4y[2] + 0.5sin(3pi*t),
y[4], -2(y[3] - θ2) - 4(y[4] + abs(y[2])) + 0.5sin(3pi*t)]
y0 = [pi/2, 0, pi/6, 0]; t0 = 0; tFinal = 50; h = 0.001
res = euler(f, y0, t0, tFinal, h)
The result, res, is a vector of four numbers
1.18798735437173
-0.0458294959470722
0.31530569612003573
-0.049213402534541074
The first entry is the angle that the bottom line segment forms with the x-axis while the third entry is the angle that the two line segments form with one another (see below figure).
To create this plot I called plot_robotarm([res[1], res[3]]) which is implemented according to the below code.
function plot_robotarm(thetav)
# Plots a robotarm with angles thetav
R = 1;
xv=zeros(length(thetav)+1)
yv=zeros(length(thetav)+1)
for i in 1:length(thetav)
xv[i+1]=xv[i]+R*cos(thetav[i])
yv[i+1]=yv[i]+R*sin(thetav[i])
end
# Plot with colors
opts = (:circle, 10, 1., :blue, stroke(7, 1., :red))
plt = plot(xv, yv,
marker = opts,
c = :red,
w = 5,
legend = false,
xlims = (0, 2.0),
ylims = (0, 2.0))
display(plt)
end
How can I create an animation that visualizes how consecutive iterations of Euler's method make the robot arm (i.e. the two line segments) move toward the final point at t = 50? (I do not need to plot every iteration, just enough so that it makes for an animation that captures the movement of the robot arm.)
You can use ffmpeg and Luxor.jl's animation features to make an animated GIF. The frame function needs to be modified to reflect graphical display of each step in your program. See the docs for Luxor for more.
using Luxor
using Colors
using BoundaryValueDiffEq
# constants for differential equations and movie
const g = 9.81
const L = 1.0 # pendulum length in meters
const bobd = 0.10 # pendulum bob diameter in meters
const framerate = 50.0 # intended frame rate/sec
const t0 = 0.0 # start time (s)
const tf = 2.3 # end simulation time (s)
const dtframe = 1.0/framerate # time increment per frame
const tspan = LinRange(t0, tf, Int(floor(tf*framerate))) # array of time points in animation
const bgcolor = "black" # gif background
const leaderhue = (0.80, 0.70, 0.20) # gif swing arm hue light gold
const hslcolors = [HSL(col) for col in (distinguishable_colors(
Int(floor(tf*framerate)+3),[RGB(1,1,1)])[2:end])]
const giffilename = "pendulum.gif" # output file
# differential equations copied from docs of DifferentialEquations.jl
simplependulum!(du, u, p, t) = (θ=u[1]; dθ=u[2]; du[1]=dθ; du[2]=-(g/L)*sin(θ))
bc1!(residual, u, p, t) = (residual[1] = u[div(end,2)][1] + pi/2; residual[2] = u[end][1] - pi/2)
bvp1 = TwoPointBVProblem(simplependulum!, bc1!, [pi/2,pi/2], (tspan[1],tspan[end]))
sol2 = solve(bvp1, GeneralMIRK4(), dt=dtframe)
# movie making background
backdrop(scene, framenumber) = background(bgcolor)
function frame(scene, framenumber)
u1, u2 = sol2.u[framenumber]
y, x = L*cos(u1), L*sin(u1)
sethue(leaderhue)
poly([Point(-4.0, 0.0), Point(4.0, 0.0),
Point(160.0x,160.0y)], :fill)
sethue(Colors.HSV(framenumber*4.0, 1, 1))
circle(Point(160.0x,160.0y), 160bobd, :fill)
text(string("frame $framenumber of $(scene.framerange.stop)"),
Point(0.0, -190.0),
halign=:center)
end
muv = Movie(400, 400, "Pendulum Demo", 1:length(tspan))
animate(muv, [Scene(muv, backdrop),
Scene(muv, frame, easingfunction=easeinoutcubic)],
creategif=true, pathname=giffilename)
I am using matlab's wavelet fractional Brownian motion function in order to generate 1D point-like data of a diffusive particle in the physical regimes: sub-diffusion, super-diffusion and normal diffusion.
The problem I encounter with is that the time normalization/variance is weird.
For example for Hurst parameter equals 0.5 (regular Brownian motion) I get standard deviation which isn't unity (1):
>> std(diff(wfbm(0.5,1e6)))
ans =
0.3955
Due to the above, I am not sure how to re-normalize all the 3 trajectories I create for the 3 diffusion cases (sub, super, normal).
I generated trajectories for N pointlike particles of length M:
M=500;
N=200;
nd = zeros(M,N);
sub = zeros(M,N);
sup = zeros(M,N);
Hsub = 0.25;
Hsup = 0.75;
for j=1:N
nd(:,j) = wfbm(0.5, M, 15, 'db10');
sub(:,j) = wfbm(Hsub,M, 10, 'db10');
sup(:,j) = wfbm(Hsup,M, 10, 'db10');
end
Here is how function is implemented in matlab and generates the signal, however I am not sure how to modify it to have a proper brownian motion:
tmp = conv(randn(1,len+nbmax),ckbeta);
tmp = cumsum(tmp);
CA = wkeep(tmp,len,'c');
for j=0:nblev-1
CD = 2^(j/2)*4^(-s)*2^(-j*s)*randn(1,len);
len = 2*len-nbmax;
CA = idwt(CA,CD,fs1,gs1,len);
end
fBm = wkeep(CA,L,'c');
fBm = fBm-fBm(1);
I was trying to understand it from the paper which says it's possible to control the variance of fBm:
This is citation 7 from the snapshot above.
I have a section of code that calculates the percent of pixels in a binary grid with value == 1 using a 50x50 sliding window:
f = #(x) numel(x(x==1))/numel(x);
I2 = nlfilter(buffer,[50 50],f);
I have heard that imfilter is a more efficient way to make focal calculations and, as such, hope to do some benchmarking. What is the imfilter() equivalent of the above nlfilter() function?
The complete code with sample data is attached
% Generate a grid of 0's to begin with.
m = zeros(400, 400, 'uint8');
% Generate 100 random "trees".
numRandom = 100;
linearIndices = randi(numel(m), 1, numRandom);
% Assign a radius value of 1-12 to each tree
m(linearIndices) = randi(12, [numel(linearIndices) 1]);
buffer = false(size(m));
for radius =1:12 % update to actual range
im_r = m==radius;
se = strel('disk',radius);
im_rb = imfilter(im_r, double(se.getnhood()));
buffer = buffer | im_rb;
end
% The imfilter approach
% The nlfilter approach
f = #(x) numel(x(x==1))/numel(x);
I2 = nlfilter(buffer,[50 50],f);
imshowpair(buffer,I2, 'montage')
For binary images (only 0s and 1s), what you have done is simple summation in a sliding window. Thus, the average filter of imfilter can be adopted here as:
h = fspecial( 'average', 50 );
I2 = imfilter( double( buffer ), h );