Two ways of iterating a vector

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Dawid Najda
Dawid Najda 2020-1-6
评论： Stephen23 2020-1-6
I have a problem with iterating a vector.
Here is part of my code (values of Best, Choices and Follower are not hardcoded ofc, I just put them not to include the entire file)
Best = 3;
Choices = [1;2];
Follower = [0;1];
for iter = 1:length(Choices)
fprintf('i=%i\n', Choices(iter))
fprintf('outcome=(%f, %f)\n', Best, Follower(Choices(iter)))
end
for item = Choices
fprintf('i=%i\n', item)
fprintf('outcome=(%f, %f)\n', Best, Follower(item))
end
When iterating by index, it works properly (result is the same as analytical solution of the problem):
i=1
outcome=(3.000000, 0.000000)
i=2
outcome=(3.000000, 1.000000)
But when I try to iterate over values, the result is:
i=1
i=2
outcome=(3.000000, 0.000000)
outcome=(1.000000, >>
So the lines are mixed, value of Best changes? or is missing? and the last fprintf doesn't finish...
Please explain if I am doing something wrong, misunderstood sth or is it a bug?
I just noticed that the second version works properly for horizontal vectors ( eg. [1 2] instead of [1;2])...
  1 个评论
Stephen23
Stephen23 2020-1-6
Note that this is simple and efficient without a loop:
>> M = [1,2;3,3;0,1]
M =
1 2
3 3
0 1
>> fprintf('i=%i\noutcome=(%f,%f)\n',M)
i=1
outcome=(3.000000,0.000000)
i=2
outcome=(3.000000,1.000000)

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采纳的回答

Mustafa Abu-Mallouh
In MATLAB, the for loop iterates over columns. When you define item as Choices, you're defining it as a single column. Change your loop to begin with
for item = Choices'
fprintf('i=%i\n', item)
fprintf('outcome=(%f, %f)\n', Best, Follower(item))
end
And it will work properly.
You could also define Choices as a row vector from the start.
See here for more information.
  1 个评论
Dawid Najda
Dawid Najda 2020-1-6
Oh right, I changed the definition to
Follower = zeros(1, hght);
and then the part that fills it accepted it, and iterating worked too.
Thanks for the explaination.

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