Indexing with min and numel

vector1 = 1:5
vector1 = 1×5
     1     2     3     4     5
vector2 = 2:7
vector2 = 1×6
     2     3     4     5     6     7
N = max(numel(vector1), numel(vector2))
N = 6

I'm not sure what you mean by indexing. But the above seems perfectly logical. And there is no indexing involved. N is simply the larger of the two vector lengths. It must be the last line you show that is in question.

[n,I] = min([numel(vector1),numel(vector2)])
n = 5
I = 1

To understand this, you need to break it down. What did you do here?

[numel(vector1),numel(vector2)]
ans = 1×2
     5     6

You created a VECTOR of length 2.

Now which element is the smaller one? The first one. That is what I tells you. Since that vector only has two elements in it, then I will only ever be 1 or 2.

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Answer 3

Les Beckham 2023-11-17

1
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2048657-indexing-with-min-and-numel#answer_1355067

在 MATLAB Online 中打开

vector1 = 1:5
vector1 = 1×5
     1     2     3     4     5
vector2 = 2:7
vector2 = 1×6
     2     3     4     5     6     7
N = max(numel(vector1), numel(vector2))
N = 6
[n,I] = min([numel(vector1),numel(vector2)])
n = 5
I = 1

I'm not sure what you are asking. In the last line of code min returns 5 for n because numel(vector1) is 5 and returns 1 for I because numel(vector1) is in the first position of the vector constructed by concatenation: [numel(vector1),numel(vector2)].

Does that answer your question?