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how to replace elements in top third, middle third, and bottom third of matix
조회 수: 74 (최근 30일)
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This question is soft-locked: new answers that are equivalent to already posted answers may be deleted without prior notice.
My task is the following:
Write a function called trio that takes two positive integer inputs n and m. The function returns a 3n-by-m matrix called T. The top third of T (an n by m submatrix) is all 1s, the middle third is all 2-s while the bottom third is all 3-s. See example below:
M = trio(2,4)
M =
1 1 1 1
1 1 1 1
2 2 2 2
2 2 2 2
3 3 3 3
3 3 3 3
This is the code that I wrote, but it only works for T = trio (4,3). I want my code to work for any input of n,m.
function T = trio (n, m)
T = randi (10, (3 * n) , m);
T ( 1:n , :) = 1;
T ( (n+1):(end-(n-1)) , :) = 2;
T ( (n+3):end, :) = 3;
end
How is it possible to call out only top third, middle third, and bottom third of any matrix?
Thank you in advance.
댓글 수: 6
mahmoud khaled
2020년 8월 11일
function T=trio(n,m)
T=[ones(n,m) ; (2*ones(n,m)) ; (3*ones(n,m))];
end
Rik
2020년 8월 17일
편집: Rik
2020년 8월 18일
I'm going to delete duplicate answers. I will consider the list below as the list of solutions. From the current answer section I will only leave the top one for each of these:
- all answer with general advice about how to solve the question
- answers with more than 1 vote
- allocating the full size array with ones or zeros and indexing into it, writing the correct values
- allocating the full size array with a non-standard function (like randi) and indexing into it, writing the correct values
- 3 and 4, but for the three parts separately, requiring concatenation
- kron (posted in a comment)
- repmat combined with implicit expansion
If new valid solutions are posted I will of course leave those as well, although I think the non-esoteric solutions may be exhausted.
채택된 답변
James Tursa
2019년 5월 1일
편집: James Tursa
2019년 5월 1일
Your row indexing is wrong.
The first n rows are 1:n which you have correct.
The second n rows indexing is n more that the first set, so simply add n: n + (1:n)
The third n rows indexing is n more than the second set, which I will let you figure out (it's pretty simple)
댓글 수: 3
Ammar
2023년 9월 27일
function T = trio (n, m)
T =randi (10, (3 * n) , m)
T ( 1:n , :) = 1
T ( (n+1):2*n , :) = 2
T ( (2*n+1):end, :) = 3
end
추가 답변 (6개)
AYUSH GURTU
2019년 5월 28일
function T = trio (n, m)
T = randi (10, (3 * n) , m);
T (1:n,:) = 1;
T ((n+(1:n)),:) = 2;
T (n+(n+(1:n)):end,:) = 3;
end
댓글 수: 2
sona rai
2020년 8월 9일
% sir this is right code instead of your code.
function T=trio(n,m)
T=randi(10,(3*n),m);
T(1:n,:)=1;
T((n+(1:n)),:)=2;
T((n+(n+(1:n))),:)=3;
end
t
PRAKASH ANAND
2019년 11월 8일
% That's my trio code.
%From India.
function T=trio(n,m)
x=ones(n,m);
y=2*x;
z=3*x;
T=[x;y;z];
end
댓글 수: 7
OUSSAMA El GABBARI
2022년 1월 19일
that function you made I see it's very restricted.. I wonder if it'd work for matrices of larger dimensions !
Juvraj Singh Sandhu
2020년 10월 4일
편집: Juvraj Singh Sandhu
2020년 10월 4일
this will work for all inputs
function T= trio(n,m);
T1= ones(n,m);
T2= 2*ones(n,m);
T3= 3*ones(n,m);
T= [T1;T2;T3];
mayank ghugretkar
2019년 6월 5일
function T=trio(n,m)
T(3*n,m)=3; % or you can use random no. generation...but since we are assigning alues anyway , this vl work fine !
T(1:n,:)=1;
T((n+1):2*n,:)=2;
T((2*n+1):3*n,:)=3;
end
hope this'll help, welcome !
댓글 수: 1
Doga Savas
2019년 8월 22일
function d = trio(n,m)
a = randi(1,n,m);
b = 2 + rand(n,m)*0;
c = 3 + rand(n,m)*0;
d = [a;b;c];
end
댓글 수: 3
Rik
2023년 9월 8일
trio_even_less_baloney = @(n,m) repelem((1:3).',n,m);
n=2;m=4;
trio_even_less_baloney(n,m)
Although your point is absolutely valid.
DGM
2023년 9월 8일
Yeah, maybe I should've called it trio_reduced_baloney()
I figured repelem() was already officially exhausted. Reshaping/permuting might not be minimal, but it was a reasonable basic option that was still fair game.
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