blocks|key|1538183|text|提取元素的最简单的解决方案是水平镜像矩阵，然后找到规则对角线：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1538184|out+=+diag(+flip(in,+2),+k+);|code-block|syntax|javascript|1538185|k这里的主对角线为0，对主对角线为正，对于下面的对角线为负值，参见diag。|offset|length|style|CODE|1538186|1538187|要找到指数本身要复杂得多。首先用一个循环来生成候选对，然后找出哪些对是有效的：|1538188|第一对是diagonal.|1538189|Then，第一对是i=1-k，j=n，n是矩阵的宽度，k是您迭代地从j中减去1到i的1，直到j是1。H+214H+115i=1-k>最后删除d16小于1或大于D17的对，矩阵的高度。H+218F+219|unordered-list-item|1538190|1538191|这个过程并不需要一个循环，循环进程可以被矢量化。不过，用这种方式来形容还是比较容易的。|1538192|在代码中，我们可以这样写(未经测试！)：|1538193|[m,n]+=+size(A);
i+=+(1:n)-k;
j+=+n:-1:1;
invalid+=+(i+<+1)+%7C+(i+>+m);
i(invalid)+=+[];
j(invalid)+=+[];|1538194|entityMap|0|LINK|mutability|MUTABLE|url|https://www.mathworks.com/help/matlab/ref/diag.html^0|0|0|0|1|X|4|X|4|0|0|0|0|0|9|5|F|3|J|1|R|1|Y|1|14|1|1A|1|1E|F|1Y|V|0|0|0|17|0|0|K|0|0^^$0|@$1|2|3|4|5|6|7|19|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1B|8|@$I|1C|J|1D|K|L]|$I|1E|J|1F|K|L]]|9|@$I|1G|J|1H|1|1I]]|A|$]]|$1|M|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1K|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1L|8|@]|9|@]|A|$]]|$1|R|3|S|5|T|7|1M|8|@$I|1N|J|1O|K|L]|$I|1P|J|1Q|K|L]|$I|1R|J|1S|K|L]|$I|1T|J|1U|K|L]|$I|1V|J|1W|K|L]|$I|1X|J|1Y|K|L]|$I|1Z|J|20|K|L]|$I|21|J|22|K|L]|$I|23|J|24|K|L]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|25|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|26|8|@$I|27|J|28|K|L]]|9|@]|A|$]]|$1|X|3|Y|5|6|7|29|8|@$I|2A|J|2B|K|L]]|9|@]|A|$]]|$1|Z|3|10|5|D|7|2C|8|@]|9|@]|A|$E|F]]|$1|11|3|-4|5|6|7|2D|8|@]|9|@]|A|$]]]|12|$13|$5|14|15|16|A|$17|18]]]]

The simplest solution to extract the elements is to mirror the matrix horizontally, then find the regular diagonals:
<pre class="lang-matlab prettyprint-override"><code>out = diag( flip(in, 2), k );
</code></pre>
<code>k</code> here is 0 for the main diagonal, positive for diagonals above the main, and negative for those below, see <a href="https://www.mathworks.com/help/matlab/ref/diag.html" rel="nofollow noreferrer"><code>diag</code></a>.
<hr />
To find the indices themselves is a bit more complex. You’d start with a loop to generate candidate pairs, then figure out which pairs are valid:
<ul>
<li>The first pair is <code>i=1-k</code>, <code>j=n</code>, with <code>n</code> the width of the matrix and <code>k</code> the desired diagonal.</li>
<li>Then you iteratively subtract 1 from <code>j</code> and add 1 to <code>i</code>, until <code>j</code> is 1.</li>
<li>Finally you remove the pairs where <code>i</code> is smaller than 1 or larger than <code>m</code>, the height of the matrix.</li>
</ul>
This process doesn’t really require a loop, the loop process can be vectorized. Still, it’s easier to describe this way.
In code we could write that as follows (untested!):
<pre class="lang-matlab prettyprint-override"><code>[m,n] = size(A);
i = (1:n)-k;
j = n:-1:1;
invalid = (i &lt; 1) | (i &gt; m);
i(invalid) = [];
j(invalid) = [];
</code></pre>

blocks|key|1542793|text|下面的脚本是这个问题的解决方案，适用于正方形和矩形二维矩阵。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1542794|clear+all;close+all;clc

A1=randi([-10+10],3,4)

sz1=size(A1)
++++++++
nA2=reshape([1:1:prod(sz1)],sz1)

nd=sum(sz1)+++++%25+amount+diagonals

C1={}+++%25+init+result+cell

for+k=-floor(nd/2):1:floor(nd/2)

++++a2+=+diag(+flip(nA2,+2),+k+)
++++[row,col]=ind2sub(sz1,a2)
++++C1={C1{:}+[row+col]}

end

C1(1)=[]++%25+sought+indexes

for+k=1:1:size(C1,2)

++++C1{k}

end|code-block|syntax|javascript|1542795|解决方案包含在单元格C1中。|offset|length|style|CODE|1542796|在MATLAB中读取单元元素的实际内容，使用花括号索引，如下所示|1542797|+C1{1}
++++=
+++++3+++++4

+C1{2}
++++=
+++++2+++++4
+++++3+++++3

C1{3}
++++=
+++++1+++++4
+++++2+++++3
+++++3+++++2|1542798|使用常规括号进行索引时，MATLAB返回所查找的元素，但在单元格类型中返回。|1542799|C2(2)

+1×1+cell+array
+{2×2+double}|1542800|若要访问解决方案索引的单个对，请使用以下符号|1542801|C1{3}(1,:)
+=
+++1+++++4

C1{3}(2,:)
+=
+++2+++++3

C1{3}(3,:)
+=
+++3+++++2|1542802|可以生成实际变量antidiag1+antidiag2+.带evalin命令|1542803|entityMap^0|0|0|A|2|0|0|0|0|0|0|0|8|9|I|9|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|17|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|19|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|1B|8|@]|9|@]|A|$]]|$1|W|3|X|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|Y|3|Z|5|6|7|1D|8|@$I|1E|J|1F|K|L]|$I|1G|J|1H|K|L]]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|11|$]]

The following script is a solution to this question and works well for square and rectangular 2D matrices
<pre><code>clear all;close all;clc

A1=randi([-10 10],3,4)

sz1=size(A1)
 
nA2=reshape([1:1:prod(sz1)],sz1)

nd=sum(sz1) % amount diagonals

C1={} % init result cell

for k=-floor(nd/2):1:floor(nd/2)

 a2 = diag( flip(nA2, 2), k )
 [row,col]=ind2sub(sz1,a2)
 C1={C1{:} [row col]}

end

C1(1)=[] % sought indexes

for k=1:1:size(C1,2)

 C1{k}

end
</code></pre>
The solution is contained in cell <code>C1</code>
In MATLAB to read the actual contents of cell elements use curly brackets indexing, like this
<pre><code> C1{1}
 =
 3 4

 C1{2}
 =
 2 4
 3 3

C1{3}
 =
 1 4
 2 3
 3 2
</code></pre>
When indexing with conventional parentheses MATLAB returns the sought elements but inside a cell type
<pre><code>C2(2)

 1×1 cell array
 {2×2 double}
</code></pre>
To access individual pairs of solution indexes use the following notation
<pre><code>C1{3}(1,:)
 =
 1 4

C1{3}(2,:)
 =
 2 3

C1{3}(3,:)
 =
 3 2
</code></pre>
One can generate the actual variables <code>antidiag1</code> <code>antidiag2</code> .. with command evalin

I have a <code>m by n</code> matrix in matlab and want to find the indices (row, column) of each antidiagonal elements. For example for a <code>4x3</code> matrix, I would like to have the following indices:
<pre><code>antidiag1 = (1,1)
antidiag2 = (2,1) , (1,2)
antidiag3 = (3,1) , (2,2), (1,3)
antidiag4 = (4,1) , (3,2), (2,3)
antidiag5 = (4,2) , (3,3)
antidiag3 = (4,3) 
</code></pre>
in the above example <code>(m,n)=(row, column)</code>. To show what I mean by anti-diagonal elements see the below matrix in which red lines are anti-diagonals
<a href="https://i.stack.imgur.com/DuwIw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DuwIw.png" alt="enter image description here" /></a>
How to find the indices?

Find the indices of anti-diagonal elements of a matrix in matlab

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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我在matlab中有一个m by n矩阵，希望找到每个反对角线元素的索引(行、列)。例如，对于4x3矩阵，我希望有以下索引：antidiag1 = (1,1)antidiag2 = (2,1) , (1,2)antidiag3 = (3,1) , (2,2), (1,3)antidiag4 = (4,1) , (3,2...

问在matlab中求矩阵的反对角线元素的指标
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问在matlab中求矩阵的反对角线元素的指标EN