Matlab转Python转换

Question

我正在尝试把一些代码从Matlab转换成Python，但对Matlab的语法和功能不太熟悉。我已经用PIL和Numpy这两个Python库完成了一部分转换，但希望有人能帮我解释一下这段代码中的一些元素。

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

这是Refocus2d函数：

function IOut = Refocus2D(FFTLF,theta)


[m,n,p,q] = size(FFTLF);
Theta1 = theta(1);
Theta2 = theta(2);

cTem = floor(size(FFTLF)/2)  +   1;


% find the coordinates of 2D slice
[XX,YY] = meshgrid(1:n,1:m);
cc = (XX - cTem(2))/size(FFTLF,2);
cc = Theta2*cc + cTem(4);
dd = (YY - cTem(1))/size(FFTLF,1);
dd = Theta1*dd + cTem(3);

% Resample 4D light field along the 2D slice
v = interpn(FFTLF,YY,XX,dd,cc,'cubic');

%set nan values to zero
idx = find(isnan(v)==1);
disp(sprintf('Number of Nans in sampling = %d',size(idx,1)))
v(isnan(v)) =   0;

% take inverse 2D FFT to get the image
IOut = real(ifft2(ifftshift(v)));

如果有人能帮忙，我将非常感激。

提前谢谢大家

抱歉：这里简要描述一下这段代码的功能：

这段代码读取一张光场图像，并根据已有的全光学掩模知识，存储相关的nAngles和掩模的基本频率以及相位偏移，这些信息用于找到图像的多个光谱副本。

在读取图像并提取绿色通道后，我们对图像进行快速傅里叶变换（FFT），然后开始从图像矩阵中提取代表某个光谱副本的切片。

接着，我们对所有光谱副本进行逆傅里叶变换，以生成光场。

Refocus2d函数然后从4维数据中提取一个二维切片，以重建一个重新聚焦的图像。

我具体遇到的困难是：

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中提取一个切片，但在FFT_LF中这个数据在哪里？(:,:,i,j)是什么意思？这是一个多维数组吗？

还有size函数返回的是什么：

[m,n,p,q] = size(FFTLF);

如果能简单解释一下这如何转化为Python，那将非常有帮助。

谢谢大家的帮助 :)

numpy matlab image processing fft multidimensional arrays fourier transform image reconstruction optical imaging

Matlab转Python转换

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