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# How to build a matrix like this

1 ビュー (過去 30 日間)
Rui Zhu 2017 年 8 月 18 日

if we have a number n, we want a matrix like this:
So one simple example is:
What is the best way to build this?

### 回答 (4 件)

Stephen23 2017 年 8 月 18 日

>> N = 4;
>> V = N:-1:1;
>> bsxfun(@min,V,repmat(V(:),1,N))
ans =
4 3 2 1
3 3 2 1
2 2 2 1
1 1 1 1
##### 0 件のコメント表示非表示 -1 件の古いコメント

José-Luis 2017 年 8 月 18 日
n = 4;
result = repmat((n:-1:1),n,1) - tril(cumsum(tril(ones(n)))-1)
##### 0 件のコメント表示非表示 -1 件の古いコメント

Kuifeng 2017 年 8 月 18 日
The following works, not the 'best' way...
n = 5;
A = [n:-1:1]';
for i = 2:n
A(:,i) = A(:,i-1);
A([1:i],i) = n-(i-1);
end
##### 3 件のコメント表示非表示 2 件の古いコメント
Kuifeng 2017 年 8 月 18 日
Thank you Jose-Luis and Stephen, I noticed this red wave below for some time already, but didnt know why. You cleared my doubt. Good to learn.

Robert U 2017 年 8 月 18 日

Hi Rui Zhu,
another possibility:
function [ A ] = SpecMatrice( n )
tic
A = zeros(n);
for ik = 1:n
A(1:ik,1:ik) = A(1:ik,1:ik) + 1;
end
t = toc;
sprintf('MySolution took %.2f µs',t*1e6)
end
On my PC that solution is for low values of n faster than the presented above. Here values for n=4:
Solution 1 took 806.82 µs
Solution 2 took 888.93 µs
Solution 3 took 441.34 µs
MySolution took 109.33 µs
For high values of n solution 2 & 3 gain speed a lot while solution 1 and MySolution get slowlier due to looping.
Kind regards,
Robert

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