Given a matrix A = [2 5 7; 4 1 8], write a for loop that will replace all the even numbers with 0.

I'm looking for something using a for loop (probably nested) and possibly mod(A,2) to determine the even numbers
What I tried: It took me a while to figure out how to format mod, but once I did I started with: for b=mod(A,2) I think tried to add a logical function (such as b==0) but it kept giving me errors. The closest thing that I could write that made sense in my head was:
for b=mod(A,2);
if b==0
A(b)=0
end
end
I would then get the error: Subscript indices must either be real positive integers or logicals.
I also couldn't figure out how to size a matrix to be the same size as a different matrix.

3 Kommentare

What have you done so far? Do you know how to write a loop? Do you know how to write if-else-end conditional code? What specific problems are you having with your current code?
Have you made an attempt that you could post here?
It took me a while to figure out how to format mod, but once I did I started with: for b=mod(A,2) I think tried to add a logical function (such as b==0) but it kept giving me errors. The closest thing that I could write that made sense in my head was:
% code
for b=mod(A,2);
if b==0
A(b)=0
end
end
I would then get the error: Subscript indices must either be real positive integers or logicals.
I also couldn't figure out how to size a matrix to be the same size as a different matrix.

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 Akzeptierte Antwort

This can be a solution using for loops:
[m,n] = size(A);
for i = 1:m
for j = 1:n
if mod(A(i,j),2) == 0
A(i,j) = 0;
end
end
end

2 Kommentare

This is obviously a homework post. It is discouraged to do homework problems for others on this forum. Helping OP with concepts, correcting their code, etc is all within the spirit of this forum. But doing the entire problem outright for them is not.
This problem is an exercise in a reading I had to do to understand a homework problem.

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Weitere Antworten (1)

A(rem(A,2)==0)=0;
or with loop
for ii = 1:numel(A)
if rem(A(ii),2) == 0
A(ii) == 0;
end
end

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