How can I check which values in 2 matrices are identical?

Dear all,

I have 2 matrices (attached) - as their name suggests, the 1st contains mean values while the 2nd contains trimmed means values. I am interested in finding out which means and trimmed means differ (upon eyeballing both arrays, lots of values seem identical, which makes sense to me since I'm trimming my data only ever so slightly). In order to do this, I've done this:

 difference= means_IB-trimmean_IB;
 distinction= find (d~=0);

However, upon eyeballing the difference matrix, I can't explain some values - for example,

means_IB(3,2)
ans =
-32.2740

 trimmean_IB(3,2)
ans =
-32.2740

 difference(3,2)
ans =
-1.4211e-14

I have a slight hunch the result of difference(3,2) has to do with decimal places not captured by the matrices but I don't think my intuition is right.

Would you please be so kind as to help me?

Thank you in advance, Bianca

[Jan]

What is your question? Do I assume correctly, that you want an explanation of the -1.42e-14 value?

[Bianca Elena Ivanof]

I am interested in finding out which means and trimmed means differ (which implies the explanation for the -1.42e-14 value).

[Jan]

You did not show yet, how this value is created. All we know, is that you got it from some calculations.

Answer 1

[Jan]

means_IB(3,2) - trimmean_IB(3,2)
% >> -1.4211e-14

This means, that this is the distance between the two values. An explanation of this effect would concern the calculation of trimmean_IB, but without seeing the corresponding code, it is hard to guess the details.

Most likely the routine for the trimming leads to different rounding "errors", which are not errors, but the effect of storing values in the double format with a limited precision. According to the "IEEE754 standard" doubles contain about 16 digits. In consequence arithmetic operations can differ by eps (see: doc eps) according to the order of the operations:

1 + 1e-17 - 1
1 - 1 + 1e-17

You can control (but not solve!) this by using a tolerance:

distinction= find(d > 10 * eps(means_IB));

The used limit depends on the problem.