Problems when creating matrix

3 visualizzazioni (ultimi 30 giorni)
Juan Manuel Romero
Juan Manuel Romero il 21 Ott 2019
Modificato: Stephen23 il 22 Ott 2019
Hello,
I am trying to create a matrix by several means, using linespace, using matrix notation and stating every element. For some reason the results obtained are not equal for some of the values in the matrix:
categ1=linspace(0.85,1.35, 51)
categ2=[0.85:0.01:1.35]
categ3=[0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.1 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.2 1.21 1.22 1.23 1.24 1.25 1.26 1.27 1.28 1.29 1.3 1.31 1.32 1.33 1.34 1.35]
categ1==categ2
categ2==categ3

Risposte (1)

Daniel M
Daniel M il 21 Ott 2019
The results are equal up to double precision. Look at this:
max(abs(categ1-categ2))
% ans =
% 2.22044604925031e-16
eps
% ans =
% 2.22044604925031e-16
So the values are equal up to the limit that the computer can differentiate between them.
It seems that linspace and colon have slightly different implementations. If you need something with higher precision, then look into John D'Errico's VPI toolbox.
  2 Commenti
Juan Manuel Romero
Juan Manuel Romero il 22 Ott 2019
The problem is that when I create the matrix whit linspace and colon and later try to find the colation of the value equal to 1.2 get an empty matrix. Any idea of how can I solve that issue?
>> categ1=linspace(0.85,1.35, 51);
categ2=[0.85:0.01:1.35];
find(categ1==1.2)
find(categ2==1.2)
ans =
1×0 empty double row vector
ans =
1×0 empty double row vector
Stephen23
Stephen23 il 22 Ott 2019
Modificato: Stephen23 il 22 Ott 2019

Accedi per commentare.

Categorie

Scopri di più su Operating on Diagonal Matrices in Help Center e File Exchange

Prodotti

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by