我试图将一些代码从Matlab转换成Python,但我对大量的Matlab语法和功能不熟悉。我已经成功地使用PIL和numpython包进行了一些转换,但是我希望有人能够解释这段代码的某些元素是怎么回事。在
clear all;close all;clc;
% Set gray scale to 0 for color images. Will need more memory
GRAY_SCALE = 1
% The physical mask placed close to the sensor has 4 harmonics, therefore
% we will have 9 angular samples in the light field
nAngles = 9;
cAngles = (nAngles+1)/2;
% The fundamental frequency of the cosine in the mask in pixels
F1Y = 238; F1X = 191; %Cosine Frequency in Pixels from Calibration Image
F12X = floor(F1X/2);
F12Y = floor(F1Y/2);
%PhaseShift due to Mask In-Plane Translation wrt Sensor
phi1 = 300; phi2 = 150;
%read 2D image
disp('Reading Input Image...');
I = double(imread('InputCones.png'));
if(GRAY_SCALE)
%take green channel only
I = I(:,:,2);
end
%make image odd size
I = I(1:end,1:end-1,:);
%find size of image
[m,n,CH] = size(I);
%Compute Spectral Tile Centers, Peak Strengths and Phase
for i = 1:nAngles
for j = 1:nAngles
CentY(i,j) = (m+1)/2 + (i-cAngles)*F1Y;
CentX(i,j) = (n+1)/2 + (j-cAngles)*F1X;
%Mat(i,j) = exp(-sqrt(-1)*((phi1*pi/180)*(i-cAngles) + (phi2*pi/180)*(j-cAngles)));
end
end
Mat = ones(nAngles,nAngles);
% 20 is because we cannot have negative values in the mask. So strenght of
% DC component is 20 times that of harmonics
Mat(cAngles,cAngles) = Mat(cAngles,cAngles) * 20;
% Beginning of 4D light field computation
% do for all color channel
for ch = 1:CH
disp('=================================');
disp(sprintf('Processing channel %d',ch));
% Find FFT of image
disp('Computing FFT of 2D image');
f = fftshift(fft2(I(:,:,ch)));
%If you want to visaulize the FFT of input 2D image (Figure 8 of
%paper), uncomment the next 2 lines
% figure;imshow(log10(abs(f)),[]);colormap gray;
% title('2D FFT of captured image (Figure 8 of paper). Note the spectral replicas');
%Rearrange Tiles of 2D FFT into 4D Planes to obtain FFT of 4D Light-Field
disp('Rearranging 2D FFT into 4D');
for i = 1: nAngles
for j = 1: nAngles
FFT_LF(:,:,i,j) = f( CentY(i,j)-F12Y:CentY(i,j)+F12Y, CentX(i,j)-F12X:CentX(i,j)+F12X)/Mat(i,j);
end
end
clear f
k = sqrt(-1);
for i = 1:nAngles
for j = 1:nAngles
shift = (phi1*pi/180)*(i-cAngles) + (phi2*pi/180)*(j-cAngles);
FFT_LF(:,:,i,j) = FFT_LF(:,:,i,j)*exp(k*shift);
end
end
disp('Computing inverse 4D FFT');
LF = ifftn(ifftshift(FFT_LF)); %Compute Light-Field by 4D Inverse FFT
clear FFT_LF
if(ch==1)
LF_R = LF;
elseif(ch==2)
LF_G = LF;
elseif(ch==3)
LF_B = LF;
end
clear LF
end
clear I
%Now we have 4D light fiel
disp('Light Field Computed. Done...');
disp('==========================================');
% Digital Refocusing Code
% Take a 2D slice of 4D light field
% For refocusing, we only need the FFT of light field, not the light field
disp('Synthesizing Refocused Images by taking 2D slice of 4D Light Field');
if(GRAY_SCALE)
FFT_LF_R = fftshift(fftn(LF_R));
clear LF_R
else
FFT_LF_R = fftshift(fftn(LF_R));
clear LF_R
FFT_LF_G = fftshift(fftn(LF_G));
clear LF_G
FFT_LF_B = fftshift(fftn(LF_B));
clear LF_B
end
% height and width of refocused image
H = size(FFT_LF_R,1);
W = size(FFT_LF_R,2);
count = 0;
for theta = -14:14
count = count + 1;
disp('===============================================');
disp(sprintf('Calculating New ReFocused Image: theta = %d',theta));
if(GRAY_SCALE)
RefocusedImage = Refocus2D(FFT_LF_R,[theta,theta]);
else
RefocusedImage = zeros(H,W,3);
RefocusedImage(:,:,1) = Refocus2D(FFT_LF_R,[theta,theta]);
RefocusedImage(:,:,2) = Refocus2D(FFT_LF_G,[theta,theta]);
RefocusedImage(:,:,3) = Refocus2D(FFT_LF_B,[theta,theta]);
end
str = sprintf('RefocusedImage%03d.png',count);
%Scale RefocusedImage in [0,255]
RefocusedImage = RefocusedImage - min(RefocusedImage(:));
RefocusedImage = 255*RefocusedImage/max(RefocusedImage(:));
%write as png image
clear tt
for ii = 1:CH
tt(:,:,ii) = fliplr(RefocusedImage(:,:,ii)');
end
imwrite(uint8(tt),str);
disp(sprintf('Refocused image written as %s',str));
end
以下是Refocus2d函数:
^{pr2}$如果有人能帮忙,我们将不胜感激。在
提前谢谢
抱歉:下面是对代码的简要说明:
该代码读取光场的图像,根据对全光掩模的先验知识,我们存储了相关的毫微光、掩模的基频和相移,这些都可以用来找到图像的多个光谱副本。在
一旦读入图像并提取出绿色通道,我们对图像执行快速傅立叶变换,并开始从表示光谱副本之一的图像矩阵中提取切片。在
然后对所有光谱副本进行傅里叶逆变换,得到一个光场。在
Refocus2d函数,然后获取4d数据的二维切片,以重新聚焦图像。在
我正在努力解决的具体问题是:
FFT_LF(:,:,i,j) = f( CentY(i,j)-F12Y:CentY(i,j)+F12Y, CentX(i,j)-F12X:CentX(i,j)+F12X)/Mat(i,j);
我们从矩阵f中取一个切片,但是FFT中的数据在哪里?(:,:,i,j)是什么意思?它是多维数组吗?在
大小函数返回什么:
[m,n,p,q] = size(FFTLF);
简单地解释一下如何将其转换为python将是一个很大的帮助。在
感谢大家:)
从这个页面开始怎么样http://www.scipy.org/NumPy_for_Matlab_Users?另外,如果你有一个简短的描述,这将是很好的
你说得对:
FFT_LF(:,:,i,j)
是指多维数组。在本例中,FFT_LF
是一个4维数组,但计算结果是一个二维数组。(:,:,i,j)
告诉MATLAB如何将二维结果放入4d变量中。在实际上,它为每对索引(i,j)存储一个MxN数组。冒号(:)实际上意味着“获取该维度中的每个元素”
[m,n,p,q] = size(FFTLF)
将返回数组中每个维度的长度。因此,如果FFTLF最终是一个5x5x3x2阵列,则可以得到:如果你有可用的MATLAB,输入“help size”可以很好地解释它的作用。对于大多数MATLAB函数也可以这样说:我对它们的文档一直印象深刻。在
希望有帮助
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