Why I get 1×0 empty double row vector?

Hi, I have problem with my code. I don't know why fidx1 is 1x0 empty double row vector?

min(oh1) and max(oh1) results are:

>>min(oh1)

ans = 102.8890

>> max(oh1)

ans = 106.2470

d1=[0 min(oh1) min(oh1)];
e1=[max(oh1) max(oh1) max([oh1 oh2 oh3 oh4 oh5 oh7 oh11 oh13 oh19 oh23 oh25 oh29 oh35 oh37])];
X1=[xq1];
Y1=[100*r1];
lidx1 = find(X1 == max(oh1));
Ylidx1 = Y1(lidx1);
fidx1 = find(X1 == min(oh1));
Yfidx1 = Y1(fidx1);

[Walter Roberson]

Why do you assume that there is any relationship between xq1 and oh1? Why do you assume that there is bit-for-bit exact matches?

[Aleksandra Pawlak]

Relationship between xq1 and oh1:

oh1L1=[104.154	103.962	104.171	104.119	104.093	104.006,......]
oh1L2=[103.735	103.517	103.727	103.718	103.753	103.674,......];
oh1L3=[103.666	103.491	103.674	103.657	103.666	103.596,......];
oh1=[oh1L1 oh1L2 oh1L3];
xq1 = (0:(max(oh1))/200:max(oh1));

[Aleksandra Pawlak]

My code works only for max value.

There's something wrong with part:

fidx1 = find(X1 == min(oh1));
Yfidx1 = Y1(fidx1);

I can't understand this

[Image Analyst]

A thorough discussion and solution is in the FAQ:

https://matlab.fandom.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

Answer 1

[Ive J]

xq1 (and therefore X1) simply may don't contain min(oh1) since you are discretizing [0, max(oh1)]. What you can do is to find the idx corresponding to the closest value to min_oh1

max_oh1 = 104;
xq1 = 0:104/200:104;
[~, fidx1] = min(abs(xq1 - min(oh1))); 

[Aleksandra Pawlak]

104 result from rounding values in oh1 vector?

[Ive J]

no it's just a simple example.

oh1 = [104.154	103.962	104.171	104.119	104.093	104.006 103.666	...
    103.491	103.674	103.657	103.666	103.596];
xq1 = 0:max(oh1)/200:max(oh1);
find(xq1 == min(oh1)) % nothing's here
ans =

  1×0 empty double row vector
  
[~, fidx1] = min(abs(xq1 - min(oh1)));
[xq1(fidx1), min(oh1)] % to compare
ans =

  103.6501  103.4910

[Aleksandra Pawlak]

It works, thank you!

But I have small question by the way: When I should use something like [~, fidx1]= instead of just fidx1= ?

[Ive J]

The first output argument is the minimum value itself, and the second one is it's index.

[Aleksandra Pawlak]

Ok, get it, thanks!