blocks|key|1117573|text|您在每次循环迭代中创建一个大小为3x3的新v，即覆盖它。您可以轻松地添加第三维，并将所有单独的矩阵存储在那里：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1117574|n+=+3;++%25+Amount+of+rows+of+the+sub-matrices
m+=+3;++%25+Amount+of+columns+of+the+sub-matrices
v+=+zeros(size(M,1)./n,size(M,2)./m,(size(M,1)*size(M,2))./(n*m));+%25+Initialise+v

kk+=+1;++%25+Initialise+page+counter
for+ii+=+1:n:size(M,1)
++++for+jj+=+1:m:size(M,2)
++++++++v(:,:,kk)+=+M(ii:ii%2B2,+jj:jj%2B2);+%25+Save+on+page+kk
++++++++kk+=+kk%2B1;++%25+Increase+page+counter
++++end
end|code-block|syntax|javascript|1117575|其他一些注意事项：|1117576|1117577|为了泛化我在任何地方都使用size(M)的代码，这样您就可以轻松地将M扩展到更大的尺寸。|unordered-list-item|1117578|1117579|1117580|不要使用i或j作为索引/变量名，as+they+denote+the+imaginary+unit。使用这些工具很容易导致难以调试的错误。|1117581|1117582|1117583|ii:ii%2B2已是数组，添加方括号是多余的|1117584|1117585|1117586|初始化v，即在进入循环之前告诉MATLAB它将有多大。否则，MATLAB将不得不在每次迭代中增加其大小，这是非常缓慢的。这称为preallocation.|1117587|1117588|1117589|n和m显然需要是整数，但不仅如此。您需要整数个子矩阵来填充每个维度，即size(M,1)/n和size(M,2)/m都需要是整数。如果不是这样，则此代码将在初始化v.|1117590|时出错|1117591|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/q/14790740/5211833|1|https://mathworks.com/help/matlab/matlab_prog/preallocating-arrays.html^0|L|1|0|0|0|0|D|7|Y|1|0|0|0|4|1|6|1|G|X|0|0|0|0|0|7|0|0|0|3|1|1R|D|1|0|0|0|0|1|2|1|Z|B|1B|B|29|1|0|0^^$0|@$1|2|3|4|5|6|7|1I|8|@$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1L|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1M|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1N|8|@]|D|@]|E|$]]|$1|N|3|O|5|P|7|1O|8|@$9|1P|A|1Q|B|C]|$9|1R|A|1S|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1T|8|@]|D|@]|E|$]]|$1|R|3|-4|5|6|7|1U|8|@]|D|@]|E|$]]|$1|S|3|T|5|P|7|1V|8|@$9|1W|A|1X|B|C]|$9|1Y|A|1Z|B|C]]|D|@$9|20|A|21|1|22]]|E|$]]|$1|U|3|-4|5|6|7|23|8|@]|D|@]|E|$]]|$1|V|3|-4|5|6|7|24|8|@]|D|@]|E|$]]|$1|W|3|X|5|P|7|25|8|@$9|26|A|27|B|C]]|D|@]|E|$]]|$1|Y|3|-4|5|6|7|28|8|@]|D|@]|E|$]]|$1|Z|3|-4|5|6|7|29|8|@]|D|@]|E|$]]|$1|10|3|11|5|P|7|2A|8|@$9|2B|A|2C|B|C]]|D|@$9|2D|A|2E|1|2F]]|E|$]]|$1|12|3|-4|5|6|7|2G|8|@]|D|@]|E|$]]|$1|13|3|-4|5|6|7|2H|8|@]|D|@]|E|$]]|$1|14|3|15|5|P|7|2I|8|@$9|2J|A|2K|B|C]|$9|2L|A|2M|B|C]|$9|2N|A|2O|B|C]|$9|2P|A|2Q|B|C]|$9|2R|A|2S|B|C]]|D|@]|E|$]]|$1|16|3|17|5|6|7|2T|8|@]|D|@]|E|$]]|$1|18|3|-4|5|6|7|2U|8|@]|D|@]|E|$]]]|19|$1A|$5|1B|1C|1D|E|$1E|1F]]|1G|$5|1B|1C|1D|E|$1E|1H]]]]

You make a new <code>v</code> of size 3x3 every loop iteration, i.e. you overwrite it. What you can easily do is to add a third dimension, and store all your separate matrices along there:
<pre><code>n = 3; % Amount of rows of the sub-matrices
m = 3; % Amount of columns of the sub-matrices
v = zeros(size(M,1)./n,size(M,2)./m,(size(M,1)*size(M,2))./(n*m)); % Initialise v

kk = 1; % Initialise page counter
for ii = 1:n:size(M,1)
 for jj = 1:m:size(M,2)
 v(:,:,kk) = M(ii:ii+2, jj:jj+2); % Save on page kk
 kk = kk+1; % Increase page counter
 end
end
</code></pre>
A few other notes:
<ul>
<li>In order to generalise the code I used <code>size(M)</code> everywhere, such that you can easily extend <code>M</code> to larger sizes.
</li>
<li>Don't use <code>i</code> or <code>j</code> as indices/variables names, <a href="https://stackoverflow.com/q/14790740/5211833">as they denote the imaginary unit</a>. Using those can easily lead to hard to debug errors.
</li>
<li><code>ii:ii+2</code> already is an array, adding square brackets is superfluous
</li>
<li>Initialise <code>v</code>, i.e. tell MATLAB how large it will be before going into the loop. Otherwise, MATLAB will have to increase its size every iteration,which is very slow. This is called <a href="https://mathworks.com/help/matlab/matlab_prog/preallocating-arrays.html" rel="nofollow noreferrer">preallocation</a>.
</li>
<li><code>n</code> and <code>m</code> obviously need to be integers, but not only that. You need an integer amount of sub-matrices to fill each dimension, i.e. <code>size(M,1)/n</code> and <code>size(M,2)/m</code> both need to be integers. If this is not the case, this code will error out on the initialisation of <code>v</code>.
</li>
</ul>

blocks|key|1117601|text|您也可以使用mat2cell函数。您需要声明input_size和chunk_size。然后，您需要在每个维度中声明块的数量(sc变量)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1117602|例如：|1117603|M+=+[0+++0+++0+++0+++0+++0+++0+++7+++6
+++++0+++2+++0+++9+++0+++0+++0+++0+++0
+++++0+++3+++8+++0+++5+++4+++1+++0+++0
+++++9+++0+++0+++5+++0+++0+++0+++3+++0
+++++0+++0+++0+++0+++1+++8+++0+++6+++7
+++++4+++0+++0+++0+++0+++0+++2+++0+++0
+++++0+++1+++0+++8+++0+++0+++0+++5+++0
+++++0+++0+++0+++0+++0+++7+++0+++0+++0
+++++6+++0+++2+++0+++3+++0+++8+++9+++4];

%25+Z+=+zeros(size(M,+1),+size(M,+2));
input_size+=+size(M);
chunk_size+=+[3+3];
sc+=+input_size+./+chunk_size;

C+=+mat2cell(M,+chunk_size(1)*ones(sc(1),+1),...
++++++++++++++++chunk_size(2)*ones(sc(2),+1));
celldisp(C')|code-block|syntax|javascript|1117604|输出：|1117605|ans{1,1}+=
+
+++++0+++++0+++++0
+++++0+++++2+++++0
+++++0+++++3+++++8

+
+
ans{2,1}+=
+
+++++0+++++0+++++0
+++++9+++++0+++++0
+++++0+++++5+++++4

+
+
ans{3,1}+=
+
+++++0+++++7+++++6
+++++0+++++0+++++0
+++++1+++++0+++++0

+
+
ans{1,2}+=
+
+++++9+++++0+++++0
+++++0+++++0+++++0
+++++4+++++0+++++0

+
+
ans{2,2}+=
+
+++++5+++++0+++++0
+++++0+++++1+++++8
+++++0+++++0+++++0

+
+
ans{3,2}+=
+
+++++0+++++3+++++0
+++++0+++++6+++++7
+++++2+++++0+++++0

+
+
ans{1,3}+=
+
+++++0+++++1+++++0
+++++0+++++0+++++0
+++++6+++++0+++++2

+
+
ans{2,3}+=
+
+++++8+++++0+++++0
+++++0+++++0+++++7
+++++0+++++3+++++0

+
+
ans{3,3}+=
+
+++++0+++++5+++++0
+++++0+++++0+++++0
+++++8+++++9+++++4|1117606|entityMap^0|6|8|M|A|X|A|1R|2|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|13|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|R|$]]

You can also use <code>mat2cell</code> function. You need to declare the <code>input_size</code> and the <code>chunk_size</code>. Then you need to declare the number of chunks in each dimension (<code>sc</code> variable).
For example:
<pre><code>M = [0 0 0 0 0 0 0 7 6
 0 2 0 9 0 0 0 0 0
 0 3 8 0 5 4 1 0 0
 9 0 0 5 0 0 0 3 0
 0 0 0 0 1 8 0 6 7
 4 0 0 0 0 0 2 0 0
 0 1 0 8 0 0 0 5 0
 0 0 0 0 0 7 0 0 0
 6 0 2 0 3 0 8 9 4];

% Z = zeros(size(M, 1), size(M, 2));
input_size = size(M);
chunk_size = [3 3];
sc = input_size ./ chunk_size;

C = mat2cell(M, chunk_size(1)*ones(sc(1), 1),...
 chunk_size(2)*ones(sc(2), 1));
celldisp(C')
</code></pre>
Output:
<pre><code> 
ans{1,1} =
 
 0 0 0
 0 2 0
 0 3 8

 
 
ans{2,1} =
 
 0 0 0
 9 0 0
 0 5 4

 
 
ans{3,1} =
 
 0 7 6
 0 0 0
 1 0 0

 
 
ans{1,2} =
 
 9 0 0
 0 0 0
 4 0 0

 
 
ans{2,2} =
 
 5 0 0
 0 1 8
 0 0 0

 
 
ans{3,2} =
 
 0 3 0
 0 6 7
 2 0 0

 
 
ans{1,3} =
 
 0 1 0
 0 0 0
 6 0 2

 
 
ans{2,3} =
 
 8 0 0
 0 0 7
 0 3 0

 
 
ans{3,3} =
 
 0 5 0
 0 0 0
 8 9 4
</code></pre>

blocks|key|1117631|text|仅使用线性索引和隐式扩展：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1117632|%25+++Random+example+matrix
M+++=+randi(9,[9,12])

%25+++Block+size
n+++=+3;

%25+++Indexing
s+++=+size(M,1)
ind+=+[1:n].'%2B[0:n-1]*s%2Breshape(floor((0:n:(numel(M)/n)-1)/s)*n*s%2Bmod(1:n:numel(M)/n,s)-1,1,1,[])

%25+++Split+each+block+into+a+third+dimension.
B+++=+M(ind)|code-block|syntax|javascript|1117633|其中：|1117634|[1:n].'%2B[0:n-1]*s+=

+++++1+++10+++19
+++++2+++11+++20
+++++3+++12+++21

%25+The+linear+index+of+the+first+block|1117635|和：|1117636|reshape(floor((0:n:(numel(M)/n)-1)/s)*n*s%2Bmod(1:n:numel(M)/n,s)-1,1,1,[])+=+

++++ans(:,:,1)+=+0
++++ans(:,:,2)+=+3
++++ans(:,:,3)+=+6
++++ans(:,:,4)+=+27
++++ans(:,:,5)+=+30
++++ans(:,:,6)+=+33
++++ans(:,:,7)+=+54
++++...

%25+The+shifting+number+for+each+block|1117637|请注意，矩阵M的每个维度的长度必须能被n整除|offset|length|style|CODE|1117638|entityMap^0|0|0|0|0|0|0|6|1|J|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@$Q|13|R|14|S|T]|$Q|15|R|16|S|T]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|V|$]]

Using only linear indexing and implicit expansion:
<pre><code>% Random example matrix
M = randi(9,[9,12])

% Block size
n = 3;

% Indexing
s = size(M,1)
ind = [1:n].'+[0:n-1]*s+reshape(floor((0:n:(numel(M)/n)-1)/s)*n*s+mod(1:n:numel(M)/n,s)-1,1,1,[])

% Split each block into a third dimension.
B = M(ind)
</code></pre>
Where:
<pre><code>[1:n].'+[0:n-1]*s =

 1 10 19
 2 11 20
 3 12 21

% The linear index of the first block
</code></pre>
And:
<pre><code>reshape(floor((0:n:(numel(M)/n)-1)/s)*n*s+mod(1:n:numel(M)/n,s)-1,1,1,[]) = 

 ans(:,:,1) = 0
 ans(:,:,2) = 3
 ans(:,:,3) = 6
 ans(:,:,4) = 27
 ans(:,:,5) = 30
 ans(:,:,6) = 33
 ans(:,:,7) = 54
 ...

% The shifting number for each block
</code></pre>
Noticed that the length of each dimension of the matrix <code>M</code> has to be divisible by <code>n</code>

blocks|key|1154569|text|您可以创建一个平铺索引矩阵，然后使用它来获得3*3个数组|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1154570|idx+=+repelem(+reshape(1:9,3,3),+3,+3+);
out+=+arrayfun(+@(x)+reshape(A(idx==x),3,3),+1:9,+'uni',+0+);|code-block|syntax|javascript|1154571|解释：|1154572|idx是一个平铺数组，如下所示：|offset|length|style|CODE|1154573|idx+=+
+++++1+++++1+++++1+++++4+++++4+++++4+++++7+++++7+++++7
+++++1+++++1+++++1+++++4+++++4+++++4+++++7+++++7+++++7
+++++1+++++1+++++1+++++4+++++4+++++4+++++7+++++7+++++7
+++++2+++++2+++++2+++++5+++++5+++++5+++++8+++++8+++++8
+++++2+++++2+++++2+++++5+++++5+++++5+++++8+++++8+++++8
+++++2+++++2+++++2+++++5+++++5+++++5+++++8+++++8+++++8
+++++3+++++3+++++3+++++6+++++6+++++6+++++9+++++9+++++9
+++++3+++++3+++++3+++++6+++++6+++++6+++++9+++++9+++++9
+++++3+++++3+++++3+++++6+++++6+++++6+++++9+++++9+++++9|1154574|然后，arrayfun行遍历值1+..+9，并提取idx与循环的索引值匹配的矩阵。我们必须在这里使用最终的reshape，因为逻辑索引将数组转换为1D。|1154575|输出是一个9x1单元数组，其中每个单元都是一个子矩阵。|1154576|entityMap^0|0|0|0|0|3|0|0|3|8|F|6|P|3|1H|7|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Z|8|@$K|10|L|11|M|N]]|9|@]|A|$]]|$1|O|3|P|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|13|8|@$K|14|L|15|M|N]|$K|16|L|17|M|N]|$K|18|L|19|M|N]|$K|1A|L|1B|M|N]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1C|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|V|$]]

You can create a tiled index matrix, then use this to get the 3*3 arrays
<pre><code>idx = repelem( reshape(1:9,3,3), 3, 3 );
out = arrayfun( @(x) reshape(A(idx==x),3,3), 1:9, 'uni', 0 );
</code></pre>
Explanation:
<code>idx</code> is a tiled array as shown:
<pre><code>idx = 
 1 1 1 4 4 4 7 7 7
 1 1 1 4 4 4 7 7 7
 1 1 1 4 4 4 7 7 7
 2 2 2 5 5 5 8 8 8
 2 2 2 5 5 5 8 8 8
 2 2 2 5 5 5 8 8 8
 3 3 3 6 6 6 9 9 9
 3 3 3 6 6 6 9 9 9
 3 3 3 6 6 6 9 9 9
</code></pre>
Then the <code>arrayfun</code> line loops through the values <code>1 .. 9</code>, and extracts the matrix where <code>idx</code> matches the indexing value of the loop. We have to use a final <code>reshape</code> here because the logical index turns the array 1D.
The output is a 9 by 1 cell array, where each cell is a submatrix.

I'm trying to decompose a 9x9 matrix into 9 3x3 matrices. I've already used the command <code>reshape</code>, but the matrices it returns are the ones as transforming the columns of the 9x9 matrix to 3x3 matrices, but that isn't what I need. This is 9x9 the matrix M
<pre><code>0 0 0 0 0 0 0 7 6
0 2 0 9 0 0 0 0 0
0 3 8 0 5 4 1 0 0
9 0 0 5 0 0 0 3 0
0 0 0 0 1 8 0 6 7
4 0 0 0 0 0 2 0 0
0 1 0 8 0 0 0 5 0
0 0 0 0 0 7 0 0 0
6 0 2 0 3 0 8 9 4
</code></pre>
And I need it in form
<pre><code>0 0 0 0 0 0 0 7 6
0 2 0 9 0 0 0 0 0
0 3 8 0 5 4 1 0 0 etc...
</code></pre>
This code generates the matrices exactly as I need them, but only saving the last one:
<pre><code>for i=[1 4 7]
 for j=[1 4 7]
 v(:,:)=M([i:i+2],[j:j+2])
 end
end
</code></pre>

How to decompose a 9x9 matrix to 3x3 matrices?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试将一个9x9矩阵分解成9个3x3矩阵。我已经使用了命令reshape，但它返回的矩阵是将9x9矩阵的列转换为3x3矩阵时返回的矩阵，但这不是我需要的。这是9x9矩阵M0   0   0   0   0   0   0   7   60   2   0   9   0   0   0   0   00   3 ...

问如何将一个9x9矩阵分解成3x3矩阵？
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问如何将一个9x9矩阵分解成3x3矩阵？EN