Save 360 arrays in a cell
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
mikel lasa
am 15 Apr. 2021
Beantwortet: mikel lasa
am 15 Apr. 2021
Hello,
My programs loads 360 images (laser stripe) and from each image gets the 0 crossing of all rows. My issue is that after processing all images, im not able to save the 0 crossing arrays into a cell so I can post process them. Here my code:
clc; clear all; close all;
%% CARGAR LAS IMAGENES (load images)
for k = 1:360
jpgFilename = sprintf('%d.jpg', k);
fullFileName = fullfile('C:\MASTER\S2\percepcion\laser360\imagenes', jpgFilename);
if exist(fullFileName, 'file')
imageData = imread(fullFileName );
imageData=imrotate(imageData,90);
% se binariza la imagen (binarization)
img_o=rgb2gray(imageData);
mask=(img_o(:,:) > 50);
imageData = bsxfun(@times, img_o, cast(mask, 'like', img_o));
% figure()
% imshow(imageData)
% title('Binarizada')
%
[rows, columns, numberOfColorChannels] = size(imageData);
img_peaks = NaN(rows, 1);
img_puntos = zeros(rows, columns);
Xp = [];
Yp = [];
Xn=[];
Yn=[];
m=[];
n=[];
zerocross=[];
coefficients=[];
lineaslaser=cell(k,1);
%se busca la linea del 0 crossing
for icross=1:rows
img_puntos(icross,:)=imageData(icross,:);
%filtrado
filtrado=sgolayfilt(double(img_puntos(icross,:)),3,33 );
for ks=1:length(filtrado)
if filtrado(ks)<0
filtrado(ks)=0;
end
end
%derivada (gradient)
derivada=gradient(filtrado);
%maximo y minimo
[Yp(icross),Xp(icross)]=max(derivada);
[Yn(icross),Xn(icross)]=min(derivada);
%coeficientes de la recta (line coefs)
m(icross)= (Yp(icross)-Yn(icross))/(Xp(icross)-Xn(icross));
n(icross)= (Yp(icross)-m(icross)*Xp(icross));
% 0 crossing
zerocross(icross)= -n(icross)/m(icross);
img_peaks(icross) = zerocross(icross);
end
axis=1:1920;
img_peaks((1:10),1)=NaN;
img_peaks((1500:1920),1)=NaN;
% save 360 lines in a cell
lineaslaser{k}=img_peaks(k);
% figure()
% plot(axis,img_peaks)
% title('plot perfil de la pieza (0 crossing)')
else
warningMessage = sprintf('Warning: image file does not exist:\n%s', fullFileName);
uiwait(warndlg(warningMessage));
end
end
3 Kommentare
DGM
am 15 Apr. 2021
Okay. If you move the preallocation outside the main loop and set that line to
lineaslaser{k}=img_peaks;
does it help?
Akzeptierte Antwort
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Image Processing Toolbox finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!