How to create a matrix using Booleans that looks like this:

Question

V_Izepatchy 2022-4-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1686654-how-to-create-a-matrix-using-booleans-that-looks-like-this

编辑： John D'Errico 2022-4-2

I'm trying to teach myself Matlab. I want to create a 5x5 matrix that looks like this: A= [1,1,1,1,1;6,1,1,1,1;7,7,1,1,1;8,8,8,1,1;9,9,9,9,1]

I've tested this so far:

a=@(i,j) (abs(i-j)<=1)+(i+j)+(abs(i-j)>1)*j;;A=create_matrix(a,5,5)

^^This is the closest I've gotten but it's not even close. Any hints appreciated. Thanks

I made the following script...

function [A] = create_matrix(a,n,m)
A = zeros(n,m);
for i=1:n
for j=1:m
A(i,j)=a(i,j);
end
end
end

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Answer 1

Voss 2022-4-1

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https://ww2.mathworks.cn/matlabcentral/answers/1686654-how-to-create-a-matrix-using-booleans-that-looks-like-this#answer_932574

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A = ones(5); % initialize A as a 5-by-5 matrix of all ones
for ii = 1:size(A,1)
    A(ii,1:ii-1) = 4+ii; % fill in each row up to but not including the diagonal with the appropriate value
end
A
A = 5×5
     1     1     1     1     1
     6     1     1     1     1
     7     7     1     1     1
     8     8     8     1     1
     9     9     9     9     1

Of course the first row remains all ones, so the loop could be for ii = 2:size(A,1)

You might also be able to use tril or other functions to do the same.

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Voss 2022-4-1

编辑：Voss 2022-4-1

在 MATLAB Online 中打开

Or did this need to use Booleans (known as logicals in MATLAB lingo) specifically, for some reason?

If so:

A = ones(5);
[m,n] = size(A);
[jj,ii] = meshgrid(1:m,1:n);
A(ii > jj) = ii(ii > jj)+4;
A
A = 5×5
     1     1     1     1     1
     6     1     1     1     1
     7     7     1     1     1
     8     8     8     1     1
     9     9     9     9     1

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Answer 2

John D'Errico 2022-4-2

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1686654-how-to-create-a-matrix-using-booleans-that-looks-like-this#answer_932624

编辑：John D'Errico 2022-4-2

在 MATLAB Online 中打开

There are always a million ways to solve such a problem, however, there is no need to use booleans at all. Best is if you learn to visualize what various tools can give you.

My immediate solution is the simple:

n = 5;
m = 5;
triu(ones(n,m)) + tril((n:2*n-1)' + zeros(1,m),-1)
ans = 5×5
     1     1     1     1     1
     6     1     1     1     1
     7     7     1     1     1
     8     8     8     1     1
     9     9     9     9     1

Note that it should be fairly easy to follow what I did, and that is always important.

Even slightly simpler is this related one:

ones(n,m) + tril((n-1:2*n-2)' + zeros(1,m),-1)
ans = 5×5
   1     1     1     1
   1     1     1     1
   7     1     1     1
   8     8     1     1
   9     9     9     1

But then you should see why this will work, and is simpler yet.

+ tril((n-1:2*n-2)' + zeros(1,m),-1)
ans = 5×5
   1     1     1     1
   1     1     1     1
   7     1     1     1
   8     8     1     1
   9     9     9     1

I'll quit now, before I add 5 more solutions.