blocks|key|3373583|text|生成对角线为零，非对角线为零的矩阵.用A的转置替换非对角线元素(因为MATLAB是列的主要部分)。再转一次，得到正确的顺序。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3373584|B+=+double(~eye(N));++%25Converting+to+double+since+we+want+to+replace+with+double+entries
B(find(B))+=+A.';+++++%25Replacing+the+entries
B+=+B.';++++++++++++++%25Transposing+again+to+get+the+matrix+in+the+correct+order|code-block|syntax|javascript|3373585|编辑：|BOLD|3373586|作为https://stackoverflow.com/questions/47713435/add-a-zero-diagonal-to-a-matrix-in-matlab#comment82387253_47713640与https://stackoverflow.com/users/3978545/wolfie的相同算法，您可以去掉对double的转换和使用find：|3373587|B+=+1-eye(N);
B(logical(B))+=+A.';+
B+=+B.';|3373588|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/47713435/add-a-zero-diagonal-to-a-matrix-in-matlab#comment82387253_47713640|1|https://stackoverflow.com/users/3978545/wolfie^0|J|1|0|0|0|3|0|4S|6|54|4|2|33|0|36|1A|1|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@$9|16|A|17|B|M]]|D|@]|E|$]]|$1|N|3|O|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]|$9|1G|A|1H|1|1I]]|E|$]]|$1|P|3|Q|5|H|7|1J|8|@]|D|@]|E|$I|J]]|$1|R|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|S|$T|$5|U|V|W|E|$X|Y]]|Z|$5|U|V|W|E|$X|10]]]]

Generate a matrix with zeros at diagonal and ones at non-diagonal indices. Replace the non-diagonal elements with the transpose of <code>A</code> (since MATLAB is column major). Transpose again to get the correct order.

<pre><code>B = double(~eye(N)); %Converting to double since we want to replace with double entries
B(find(B)) = A.'; %Replacing the entries
B = B.'; %Transposing again to get the matrix in the correct order
</code></pre>

Edit:

As <a href="https://stackoverflow.com/questions/47713435/add-a-zero-diagonal-to-a-matrix-in-matlab#comment82387253_47713640">suggested</a> by <a href="https://stackoverflow.com/users/3978545/wolfie">Wolfie</a> for the same algorithm, you can get rid of conversion to <code>double</code> and the use of <code>find</code> with:

<pre><code>B = 1-eye(N);
B(logical(B)) = A.'; 
B = B.';
</code></pre>

blocks|key|3373671|text|您可以使用矩阵的上三角部分和下三角部分(triu和tril)来完成此操作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3373672|这是一个1行的解决方案：|3373673|B+=+[tril(A,-1)+zeros(N,+1)]+%2B+[zeros(N,1)+triu(A)];|code-block|syntax|javascript|3373674|编辑:基准测试|BOLD|3373675|这是循环方法、Sardar's+answer中的两个方法和我上面的方法的比较。|3373676|基准代码，使用timeit来计时，并直接解除代码的问答：|3373677|function+benchie()
++++N+=+1e4;+A+=+rand(N,N-1);+%25+Initialise+large+matrix
++++%25+Set+up+anonymous+functions+for+input+to+timeit
++++s1+=+@()+sardar1(A,N);+s2+=+@()+sardar2(A,N);+
++++w+=++@()+wolfie(A,N);+u+=+@()+user3285148(A,N);
++++%25+timings
++++timeit(s1),+timeit(s2),+timeit(w),+timeit(u)
end
function+sardar1(A,+N)+%25+using+eye+as+an+indexing+matrix
++++B=double(~eye(N));+B(find(B))=A.';+B=B.';
end
function+sardar2(A,N)+%25+similar+to+sardar1,+but+avoiding+slow+operations
++++B=1-eye(N);+B(logical(B))=A.';+B=B.';
end
function+wolfie(A,N)+%25+using+triangular+parts+of+the+matrix
++++B+=+[tril(A,-1)+zeros(N,+1)]+%2B+[zeros(N,1)+triu(A)];
end
function+user3285148(A,+N)+%25+original+looping+method
++++B+=+zeros(N,N);+B(1,:)+=+[0+A(1,:)];+B(N,:)+=+[A(N,:)+0];
++++for+j=2:N-1;+B(j,:)=+[A(j,1:j-1)+0+A(j,j:end)];+end
end|3373678|结果：|3373679|Sardar方法1:+2.83秒|unordered-list-item|3373680|Sardar方法2:+1.82秒|3373681|我的方法:+1.45秒|3373682|循环方法:3.80秒(!)|3373683|结论：|3373684|你想要矢量化这一点是有根据的，循环比其他方法慢得多。|3373685|避免大型矩阵的数据转换和find是很重要的，在Sardar方法之间节省了35%25的处理时间。|3373686|通过避免所有索引在一起，您可以节省20%25的处理时间。|3373687|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/47713640/3978545^0|K|4|P|4|0|0|0|0|7|0|7|F|0|0|7|6|0|0|0|0|0|0|0|0|0|C|4|0|0^^$0|@$1|2|3|4|5|6|7|1M|8|@$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1R|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1S|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1T|8|@$9|1U|A|1V|B|O]]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1W|8|@]|D|@$9|1X|A|1Y|1|1Z]]|E|$]]|$1|R|3|S|5|6|7|20|8|@$9|21|A|22|B|C]]|D|@]|E|$]]|$1|T|3|U|5|J|7|23|8|@]|D|@]|E|$K|L]]|$1|V|3|W|5|6|7|24|8|@]|D|@]|E|$]]|$1|X|3|Y|5|Z|7|25|8|@]|D|@]|E|$]]|$1|10|3|11|5|Z|7|26|8|@]|D|@]|E|$]]|$1|12|3|13|5|Z|7|27|8|@]|D|@]|E|$]]|$1|14|3|15|5|Z|7|28|8|@]|D|@]|E|$]]|$1|16|3|17|5|6|7|29|8|@]|D|@]|E|$]]|$1|18|3|19|5|Z|7|2A|8|@]|D|@]|E|$]]|$1|1A|3|1B|5|Z|7|2B|8|@$9|2C|A|2D|B|C]]|D|@]|E|$]]|$1|1C|3|1D|5|Z|7|2E|8|@]|D|@]|E|$]]|$1|1E|3|-4|5|6|7|2F|8|@]|D|@]|E|$]]]|1F|$1G|$5|1H|1I|1J|E|$1K|1L]]]]

You can do this with upper and lower triangular parts of the matrix (<code>triu</code> and <code>tril</code>).

Then it's a 1 line solution:

<pre><code>B = [tril(A,-1) zeros(N, 1)] + [zeros(N,1) triu(A)];
</code></pre>

<hr>

Edit: benchmark

This is a comparison of the loop method, the 2 methods in <a href="https://stackoverflow.com/a/47713640/3978545">Sardar's answer</a>, and my method above.

Benchmark code, using <code>timeit</code> for timing and directly lifting code from question and answers:

<pre><code>function benchie()
 N = 1e4; A = rand(N,N-1); % Initialise large matrix
 % Set up anonymous functions for input to timeit
 s1 = @() sardar1(A,N); s2 = @() sardar2(A,N); 
 w = @() wolfie(A,N); u = @() user3285148(A,N);
 % timings
 timeit(s1), timeit(s2), timeit(w), timeit(u)
end
function sardar1(A, N) % using eye as an indexing matrix
 B=double(~eye(N)); B(find(B))=A.'; B=B.';
end
function sardar2(A,N) % similar to sardar1, but avoiding slow operations
 B=1-eye(N); B(logical(B))=A.'; B=B.';
end
function wolfie(A,N) % using triangular parts of the matrix
 B = [tril(A,-1) zeros(N, 1)] + [zeros(N,1) triu(A)];
end
function user3285148(A, N) % original looping method
 B = zeros(N,N); B(1,:) = [0 A(1,:)]; B(N,:) = [A(N,:) 0];
 for j=2:N-1; B(j,:)= [A(j,1:j-1) 0 A(j,j:end)]; end
end
</code></pre>

Results:

<ul>
<li>Sardar method 1: 2.83 secs</li>
<li>Sardar method 2: 1.82 secs</li>
<li>My method: 1.45 secs</li>
<li>Looping method: 3.80 secs (!)</li>
</ul>

Conclusions: 

<ul>
<li>Your desire to vectorise this was well founded, looping is way slower than other methods.</li>
<li>Avoiding data conversions and <code>find</code> for large matrices is important, saving ~35% processing time between Sardar's methods. </li>
<li>By avoiding indexing all together you can save a further 20% processing time. </li>
</ul>

blocks|key|3373707|text|如果要在矩阵的对角线上插入任何向量，可以使用普通索引。下面的片段给出了所需对角线的索引，给定方阵n的大小(矩阵是n+by+n)，以及对角线k的数目，其中k=0对应于主对角线，k的正数对应于上对角，k的负数对应于下对角线。ixd最终给出了2D索引。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3373708|function+[idx]+=+diagidx(n,k)
%25+n+size+of+square+matrix
%25+k+number+of+diagonal
if+k==0+%25+identity
++++idx+=+[(1:n).'+(1:n).'];+%25+[row+col]
elseif+k>0+%25+Upper+diagonal
++++idx+=+[(1:n-k).'+(1%2Bk:n).'];
elseif+k<0+%25+lower+diagonal
++++idx+=+[(1%2Babs(k):n).'+(1:n-abs(k)).'];
end
end|code-block|syntax|javascript|3373709|用法：|3373710|n=10;
k=3;
A+=+rand(n);
idx+=+diagidx(n,k);

A(idx)+=+1:(n-k);|3373711|​|3373712|entityMap^0|1C|1|1K|1|1P|1|1X|1|24|3|2F|1|2Q|1|32|3|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]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1C|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|R|$]]

If you want to insert any vector on a diagonal of a matrix, one can use plain indexing. The following snippet gives you the indices of the desired diagonal, given the size of the square matrix <code>n</code> (matrix is <code>n</code> by <code>n</code>), and the number of the diagonal <code>k</code>, where <code>k=0</code> corresponds to the main diagonal, positive numbers of <code>k</code> to upper diagonals and negative numbers of <code>k</code> to lower diagonals. <code>ixd</code> finally gives you the 2D indices.

<pre><code>function [idx] = diagidx(n,k)
% n size of square matrix
% k number of diagonal
if k==0 % identity
 idx = [(1:n).' (1:n).']; % [row col]
elseif k&gt;0 % Upper diagonal
 idx = [(1:n-k).' (1+k:n).'];
elseif k&lt;0 % lower diagonal
 idx = [(1+abs(k):n).' (1:n-abs(k)).'];
end
end
</code></pre>

Usage:

<pre><code>n=10;
k=3;
A = rand(n);
idx = diagidx(n,k);

A(idx) = 1:(n-k);
</code></pre>

Suppose I have a matrix <code>A</code> of dimension <code>Nx(N-1)</code> in MATLAB, e.g.

<pre><code>N=5;
A=[1 2 3 4;
 5 6 7 8;
 9 10 11 12;
 13 14 15 16;
 17 18 19 20 ];
</code></pre>

I want to transform <code>A</code> into an <code>NxN</code> matrix <code>B</code>, just by adding a zero diagonal, i.e.,

<pre><code>B=[ 0 1 2 3 4;
 5 0 6 7 8;
 9 10 0 11 12;
 13 14 15 0 16;
 17 18 19 20 0];
</code></pre>

<hr>

This code does what I want:

<pre><code>B_temp = zeros(N,N); 
B_temp(1,:) = [0 A(1,:)];
B_temp(N,:) = [A(N,:) 0];
for j=2:N-1
 B_temp(j,:)= [A(j,1:j-1) 0 A(j,j:end)];
end
B = B_temp; 
</code></pre>

Could you suggest an efficient way to vectorise it?

Add a diagonal of zeros to 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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假设我在MATLAB中有一个维数A的矩阵Nx(N-1)。N=5;A=[1  2  3  4;   5  6  7  8;   9  10 11 12;   13 14 15 16;   17 18 19 20 ];我想将A转换为NxN矩阵B，只需添加一个零对角线，即，B=[ 0  1   2   3   4;    5...

问在MATLAB中将零对角添加到矩阵中
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