# Indexing with min and numel

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dat 2023 年 11 月 17 日

vector1 = 1:5
vector2 = 2:7
N = max(numel(vector1), numel(vector2))
[n,I] = min([numel(vector1),numel(vector2)]) *can someone kindly explain to me how indexing in this works? How is it either going to be 1 or 2?

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### 採用された回答

Dyuman Joshi 2023 年 11 月 17 日

[n,I] = min([numel(vector1),numel(vector2)])
There are 2 elements in the input array, as numel() returns a scalar value.
Then min() will output the minimum value of those, and the first index of where the minimum value occurs.
"How is it either going to be 1 or 2?"
Because there are only 2 elements in the array.

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### その他の回答 (2 件)

John D'Errico 2023 年 11 月 17 日

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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Les Beckham 2023 年 11 月 17 日
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)].

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