Indexing with min and numel

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?

Answer 1

[Dyuman Joshi]

[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.

Answer 2

[John D'Errico]

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.

Answer 3

[Les Beckham]

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?