Index with min command

Hi

I am working with the minimums calculated from two arrays in the following way -

NE = 100;

t = linspace(0,2,2*NE);

T = linspace(0,1,2*NE);

for ii = 1:NE

    [val,index] = min(abs(t-T(ii)))
end

now i understand the value of "val", but i cannot figure out how the values for "index" are coming out. For my work the value of index is very important. I would really appreciate if someone help me figure it out.

Thanks

Hossain

[Matt J]

What doesn't make sense about the behavior you're seeing?

[SAZZAD HOSSAIN]

for each loop there is a min value. but i dont understad how the index assigned to the value is related to "t" or "T".

Answer 1

[Azzi Abdelmalek]

NE = 100;
t = linspace(0,2,2*NE);
T = linspace(0,1,2*NE);
val=[];
index=[];
for ii = 1:NE
  [val1,index1] = min(abs(t-T(ii)))
  val=[val val1];
  index=[index index1];
end

index is the position of the min value in the vector t

[Jan]

No, index is not the position of the minimal value of t. This would be [val, index] = min(t).

[Azzi Abdelmalek]

It's clear min value don't belong to t, it's the position of the value of t making t-T(ii) minimal. And I did'nt say it's a minimal value of t

[Jan]

Ok, then I confuse "index is the position of the min value in the vector t" with "...of the vector t". My English is not firm.

The problem of the OP is exactly, that the meaning of the index is not clear. So this point should be very clear, although it is almost trivial.

Answer 2

[Jan]

At first you create a temporary vector:

v = abs(t-T(ii));

Then you find its minimum value and the corresponding index:

 [val, index] = min(v)

such that v(index) has the value val.