Shifting pixels in an image without loops

Matt. (view profile)

on 17 Sep 2019 at 17:27
Latest activity Answered by Image Analyst

Image Analyst (view profile)

on 18 Sep 2019 at 0:59
Hi, I would like to shift some pixels in an image (Ny,Nx,Nc) given a 2D shift map (Ny,Nx).
I can easily do it with loops but I am having troubles to vectorize it:
% test shift of image with a map
clear all
close all
% example of image (vertical gradient) + horizontal red stripe
img=repmat(uint8((1:1980)'*255/1980),[1,2880,3]);
img(800:900,:,1)=255;
figure
imagesc(img)
title('original image')
% example of shift map (vertical gradient) +vertical stripe
shift_map=repmat((1:1980)'*1/4,[1,2880]);
shift_map(:,1000:1500)=255;
figure
imagesc(shift_map)
title('original shift map')
%% with loops
img_out_1=inf(size(img));
tic
for x=1:size(img,2)
for y=1:size(img,1)
indx_new=round(x-shift_map(y,x));
if 0<indx_new && indx_new<=size(img,2)
img_out_1(y,indx_new,:)=img(y,x,:);
end
end
end
% median filter to smooth the missing values
img_out_1=medfilt1(img_out_1,3,[],2);
toc
figure
imagesc(uint8(img_out_1))
title('final image')
I have managed to remove the loop on x and it is already much faster:
%% with loop on y only
img_out_2=inf(size(img));
tic
Nx=size(shift_map,2);
x=linspace(1,Nx,Nx);
for y=1:size(img,1)
indx_new=round(x-shift_map(y,:));
ind=find(0<indx_new & indx_new<Nx);
img_out_2(y,indx_new(ind),:)=img(y,x(ind),:);
end
img_out_2=medfilt1(img_out_2,3,[],2);
toc
figure
imagesc(uint8(img_out_2))
title('final image 2')
Naively I tried a similar approach to remove the loop on x, but I am getting very confused as my inddices ind_x becomes a matrix. Obviously img_out_3(yy(ind),indx_new(ind),:) fails with Requested 5212348x5212348x3 (75908.1GB) array exceeds maximum array size preference. Creation of arrays greater than this limit may take a long time and cause MATLAB to become unresponsive. See array size limit or preference panel for more information.
img_out_3=inf(size(img));
tic
Nx=size(img,2);
Ny=size(img,1);
Nc=size(img,3);
x=linspace(1,Nx,Nx);
y=linspace(1,Ny,Ny);
[yy,xx]=meshgrid(y,x);
indx_new=round(x-shift_map);
ind=find(0<indx_new & indx_new<Nx);
img_out_3(yy(ind),indx_new(ind),:)=img(yy(ind),xx(ind),:);
img_out_3=medfilt1(img_out_3,3,[],2);
toc
figure
imagesc(uint8(img_out_3))
title('final image 3')
Thanks for the heads up,
Matt

KALYAN ACHARJYA

KALYAN ACHARJYA (view profile)

on 17 Sep 2019 at 18:59
Shifting pixels in an image without loops
Sometimes simple representation of the problemmis much simpler to undestand than actual code.
1. Suppose you have an image,
2. Now you want to shift those pixels to ....??? what happen on present pixels location?
3. When the pixel shift to other location, is it oerwrite on present pixels?
4. Location: source and destination pixels location?
Can you share the detail, it would be much easier to understand?
Matt.

Matt. (view profile)

on 18 Sep 2019 at 0:14
Thanks for the help. I will try to clarify my problem. Te first part of my code is doing what I want but it is very inefficient.
1) so I have an image M(x,y) plus a function that gives the shift in the x direction S(x,y).
2) I want to create an image of M where each pixels are moved in x by the corresponding S(x,y) value.
On the pixel location two things can happen:
-either it is the image of an other pixel and the new value is updated,
-either it is not updated and the value is inf.
3) yes. If two original pixel ends up to the same pixel/location, we can keep the latest value. It is not my biggest concern so far so a solution that keeps the first value is also ok for me.
4)in fact I want to get M(x+S(x,y),y)
Origin is the pixel at posiion (x,y). I want to move it to position (x+S(x,y),y)

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