How to calculate mean average percentage error in Matlab? (single array division)

I have 2 variables A(1200x1 double),B(1200x1 double), and I want to get the mean value of their division result :

C=A./B; M=mean(C)

Is it correct to calculate this way?

Thank you and wish you will have a nice Christmas!

[Image Analyst]

A division is simply a division - a ratio. So what is the error? What are you considering to the "the error" in that division operation? Is something the "true" value? If so, what? What are the actual values? The C values? So are we supposed to get the difference between C and some "true" (ground truth) values? How can we compute the error(s) if we don't have both the true values and the estimated/noisy/guessed values?

Answer 1

[Star Strider]

It would be nice to know how ‘A’ and ‘B’ relate to all this.

Assuming that ‘A’ is ‘experience_value’ and ‘B’ is ‘theoretical_value’, your equation becomes:

experience_value = A;
theoritical_value = B;
percentage error = (experience_value-theoritical_value)./theoritical_value*100;

Make appropriate changes to reflect the actual significance of your vectors.

___________________

EDIT — Answering with respect to the information suppied in the edited Question:

‘Is it correct to calculate this way?’

Yes.

Answer 2

[John BG]

'a priori' i see 4 possibilities, i hope you don't mind me using horizontal A and B instead of vertical:

A=[1:10];B=1:2:20;m1=mean(A\B)
sz1=size(A/B)
A=[1:10];B=1:2:20;m2=mean(A.\B)
sz2=size(A./B)
A=[1:10];B=1:2:20;m3=mean(A/B)
sz3=size(A\B)
A=[1:10];B=1:2:20;m4=mean(A./B)
sz4=size(A.\B)
m1 =
  Columns 1 through 3
   0.010000000000000   0.030000000000000   0.050000000000000
  Columns 4 through 6
   0.070000000000000   0.090000000000000   0.110000000000000
  Columns 7 through 9
   0.130000000000000   0.150000000000000   0.170000000000000
  Column 10
   0.190000000000000
sz1 =
     1     1
m2 =
   1.707103174603175
sz2 =
     1    10
m3 =
   0.537593984962406
sz3 =
    10    10
m4 =
   0.606662776507978
sz4 =
     1    10

or the actual value that you may need is

m5=nean(A)/mean(B)
m5 =
   0.550000000000000

   m6=mean(A)\mean(B)
  m6 =
     1.818181818181818

i hope you don't mind asking the following questions:

1. what is it the objective of your calculations?

2. what are A and B?

3. what is your choice, my guess woudl be, from your question, m1

but since in your question you are already guessing, a guess of a guess doesn't reduce the uncertainty, does it?

 awaiting answer to complete my answer

John BG

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