Beginner question about for loops and indexing

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Callum Taylor
Callum Taylor 2018년 11월 20일
댓글: Callum Taylor 2018년 11월 21일
I want to perform a function on each number in a series 'a' depending on whether it's odd or even. Then output this series of numbers 'b' as an array.
I don't know what to do next. So far my script doesn't output anything.
a = 1:5:150; % array before function applied
b = zeros(1,30); % used to store numbers from a after functions
counter = 0;
for series = [1:30]
r = rem(a,2); % r = 0 if number is even, r = 1 if number is odd
counter = counter + 1;
if r == 0
b = a(counter)*4 % if number in b is even, multiply it by 4
elseif r == 1
b = a(counter)^3 % if number in a is odd, cube it
end
end
I am an absolute beginner, really have no idea what I'm doing, please help?

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Brian Hart
Brian Hart 2018년 11월 20일
Hi Callum,
You're really close.
First, you can move the "r = ... " line outside the loop to do the operation on the whole array at once.
Then inside the loop, you need to index the other variables the same way you did with "a(counter)".
So it looks like this:
a = 1:5:150; % array before function applied
b = zeros(1,30); % used to store numbers from a after functions
counter = 0;
r = rem(a,2); % r = 0 if number is even, r = 1 if number is odd
for series = [1:30]
counter = counter + 1;
if r(counter) == 0
b(counter) = a(counter)*4; % if number in b is even, multiply it by 4
elseif r(counter) == 1
b(counter) = a(counter)^3; % if number in a is odd, cube it
end
end
  댓글 수: 2
Brian Hart
Brian Hart 2018년 11월 21일
To encourage you to learn more about MATLAB, here's a different solution that does the same thing with few lines of code...
a = 1:5:150; % array before function applied
b = zeros(1,length(a)); % used to store result
b(rem(a,2)==0) = a(rem(a,2)==0) * 4; % Using logical indexing and vectorization
b(rem(a,2)==1)=a(rem(a,2)==1).^3;
Callum Taylor
Callum Taylor 2018년 11월 21일
Thank you so much! It works just fine now. I didn't know you had to index every variable, including 'r'.
I definitely need more practice, thanks for the help!

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