N维的小波变换 [英] Wavelet Transform for N dimensions

问题描述

我遇到了这个惊人的回应应用MATLAB的 idwt2 好几次，我自己去了解它。但是，我无法使用与RGB图像一起使用相同的方法。所以，我有3个问题。

I came across this amazing response Applying MATLAB's idwt2 several times which I executed to understand it myself. However, I am unable to get how to use the same with work with an RGB image. So, I have 3 Questions.

1. 如何将代码应用于RGB图像，只有转换后的图像显示在输出与行和列的高频和低频分量一起，是否可以将所有组件的融合视为单个图像？我知道我必须把猫操作员，但我不明白如何去做。

1. How would the code be applied to an RGB image with only the transformed image displayed in the output that is along with the high and low frequency components along row and column,is it possible to view the fusion of all the components as a single image? I am aware that I have to put cat operator, but I cant understand how to go about it.

其次，我也得到一个迷宫图像！我很困惑，因为我似乎无法遵循原因。我还在声明如何生成此图像的语句中添加了相同的代码。

Secondly, I am also getting a mazed image! I am perplexed since I cannot seem to follow the reason. I have also attached the same code with the statement showing how this image has been generated.

3.术语 db1 在函数签名中 dwt 暗示？

3.What does the term db1 in the function signature of dwt imply?

代码：

    load woman;             % Load image data
%startImage=imread('pic_rgb.jpg');  % IF I WANT TO WORK WITH RGB IMAGE
    nLevel = 3;             % Number of decompositions
    nColors = size(map,1);  % Number of colors in colormap
    cA = cell(1,nLevel);    % Approximation coefficients
    cH = cell(1,nLevel);    % Horizontal detail coefficients
    cV = cell(1,nLevel);    % Vertical detail coefficients
    cD = cell(1,nLevel);    % Diagonal detail coefficients
    startImage = X;
    for iLevel = 1:nLevel,
      [cA{iLevel},cH{iLevel},cV{iLevel},cD{iLevel}] = dwt2(startImage,'db1');



     startImage = cA{iLevel};
    end

    figure;colormap(map);
    imagesc(dwt2(startImage,'db1')); %THIS GIVES THE MAZED IMAGE INSTEAD OF THE TRANSFORMED IMAGE
    figure;
    tiledImage = wcodemat(cA{nLevel},nColors);
    for iLevel = nLevel:-1:1,
     tiledImage = [tiledImage                   wcodemat(cH{iLevel},nColors); ...
                    wcodemat(cV{iLevel},nColors) wcodemat(cD{iLevel},nColors)];

    end
    figure;

    imshow(tiledImage,map);

    %reconstruct
    fullRecon = cA{nLevel};
    for iLevel = nLevel:-1:1,
      fullRecon = idwt2(fullRecon,cH{iLevel},cV{iLevel},cD{iLevel},'db1');
    end
    partialRecon = cA{nLevel};
    for iLevel = nLevel:-1:1,
      partialRecon = idwt2(partialRecon,[],[],[],'db1');
    end
    figure;
    imshow([X fullRecon; partialRecon zeros(size(X))],map,...
           'InitialMagnification',50);

推荐答案

我对其他问题的回答是索引图像，因此需要进行一些更改才能使代码正常工作 RGB图像。

The sample image used in my answer to that other question was an indexed image, so there are a few changes that need to be made to get that code working for an RGB image.

我将首先解决你传递给'db1'参数的问题。 mathworks.com/help/toolbox/wavelet/ref/dwt2.htmlrel =nofollow noreferrer> DWT2 。这指定了用于分解的小波类型（在本例中为 Daubechies wavelet ） 。有关可用小波的更多信息，请参阅 WFILTERS <函数的文档。 / a>和 WAVEINFO 。

I'll first address your question about the 'db1' argument passed to DWT2. This specifies the type of wavelet to use for the decomposition (in this case, a Daubechies wavelet). More information about available wavelets can be found in the documentation for the functions WFILTERS and WAVEINFO.

我将通过向您展示如何修改我的其他答案中的代码来处理RGB图像来解决前两个问题。我将使用示例'peppers.png'图像。您首先要加载图像并定义每个颜色分量所具有的值的数量。由于样本图像是无符号8位整数类型（最常见的情况）， nColors 将为256：

I'll address your first two questions by showing you how to modify the code from my other answer to work for an RGB image. I'll use the sample 'peppers.png' image. You'll first want to load your image and define the number of values each color component has. Since the sample image is an unsigned 8-bit integer type (the most common situation), nColors will be 256:

X = imread('peppers.png');  %# Load sample image
nColors = 256;              %# Number of values per color component

如果您的图像是较大的无符号整数类型（例如<$ c） $ c>'uint16'），查找颜色值数量的一般方法是使用函数 INTMAX 如下：

If your images are larger unsigned integer types (e.g. 'uint16'), a general way to find the number of color values is to use the function INTMAX like so:

nColors = double(intmax(class(X)))+1;

对于随后的代码，图像类型'uint8'。

For the ensuing code, an image type of 'uint8' is assumed.

应用分解与索引图像的情况没有什么不同。系数矩阵将简单地是M×N×3矩阵而不是M-by-N矩阵：

Applying the decompositions is no different than in the indexed image case. The coefficient matrices will simply be M-by-N-by-3 matrices instead of M-by-N matrices:

nLevel = 3;             %# Number of decompositions
cA = cell(1,nLevel);    %# Approximation coefficient storage
cH = cell(1,nLevel);    %# Horizontal detail coefficient storage
cV = cell(1,nLevel);    %# Vertical detail coefficient storage
cD = cell(1,nLevel);    %# Diagonal detail coefficient storage
startImage = X;
for iLevel = 1:nLevel,  %# Apply nLevel decompositions
  [cA{iLevel},cH{iLevel},cV{iLevel},cD{iLevel}] = dwt2(startImage,'db1');
  startImage = cA{iLevel};
end

创建平铺图像以显示水平，垂直和对角线的代码由于我们现在正在使用3-D矩阵并且必须使用 CAT 函数而不是连接运算符 [] ：

The code to create the tiled image to show the horizontal, vertical, and diagonal components for each decomposition will change due to the fact that we are now working with 3-D matrices and must use the CAT function instead of the concatenation operator []:

tiledImage = wcodemat(cA{nLevel},nColors);
for iLevel = nLevel:-1:1
  tiledImage = cat(1,cat(2,tiledImage,...
                           wcodemat(cH{iLevel},nColors)),...
                     cat(2,wcodemat(cV{iLevel},nColors),...
                           wcodemat(cD{iLevel},nColors)));
end
figure;
imshow(uint8(tiledImage-1));  %# Convert to unsigned 8-bit integer to display

这将显示以下图像显示水平（每个分解步骤的右上角，垂直（左下角）和对角线（右下角）组件，以及缩小的图像（左上角）：

This will give the following image showing the horizontal (top right), vertical (bottom left), and diagonal (bottom right) components for each decomposition step, along with the reduced image (top left):

重建步骤与其他步骤保持不变回答。只需要修改显示最终图像的代码：

The reconstruction steps are unchanged from the other answer. Only the code for displaying the final images needs to be modified:

fullRecon = cA{nLevel};
for iLevel = nLevel:-1:1,
  fullRecon = idwt2(fullRecon,cH{iLevel},cV{iLevel},cD{iLevel},'db1');
end
partialRecon = cA{nLevel};
for iLevel = nLevel:-1:1,
  partialRecon = idwt2(partialRecon,[],[],[],'db1');
end
figure;
tiledImage = cat(1,cat(2,X,uint8(fullRecon)),...
                   cat(2,uint8(partialRecon),zeros(size(X),'uint8')));
imshow(tiledImage,'InitialMagnification',50);

您将获得一张显示原始RGB图像的图像（左上角），完全重建的图像使用所有存储的细节系数矩阵（右上），使用没有存储的细节系数矩阵的部分重建图像（左下）：

And you will get an image showing the original RGB image (top left), the fully-reconstructed image using all of the stored detail coefficient matrices (top right), and the partially-reconstructed image using none of the stored detail coefficient matrices (bottom left):

这篇关于N维的小波变换的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！