Need help understanding rescale()

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Adrian Velasquez
Adrian Velasquez le 1 Août 2022
Commenté : Adrian Velasquez le 2 Août 2022
I have an array x that I'm running rescale on. However, I'm not understanding the output that I'm getting. Should it not be [1.000 0.9000 0.9500 0.9500; 0.7000, ect] I'm not sure if it's a problem with my code or my understanding of how rescale() works. Any tips help. Thank!
function X = rescale_scores()
x = [100 90 95 95; 70 50 60 60; 80 70 90 80]
[m,n] = size(x);
A = rescale(x)
end
% OUTPUT THAT I'M GETTING
x =
100 90 95 95
70 50 60 60
80 70 90 80
A =
1.0000 0.8000 0.9000 0.9000
0.4000 0 0.2000 0.2000
0.6000 0.4000 0.8000 0.6000

Réponse acceptée

Jeffrey Clark
Jeffrey Clark le 1 Août 2022
@Adrian Velasquez, according to rescale documentation each element should be rescaled to [0..1] using this method:
l + [(A-inmin)./(inmax-inmin)].*(u-l)
to scale the elements of an array A when the values of A are within the bounds of inmin and inmax.
  • If l and u are not specified, then rescale uses the default values 0 and 1, respectively.
  • If the 'InputMin' name-value pair is not specified, then rescale sets its value to the default min(A(:)).
  • If the 'InputMax' name-value pair is not specified, then rescale sets its value to the default max(A(:)).
In your case since l = 0, u = 1, InputMin = min(x(:)) and InputMax = max(x(:) it becomes:
0+[(x-50)./(100-50)].*(1-0) == (x-50)/50 ==> 100 becomes 1, 50 becomes 0 and so on.
So rescale works as advertised.
  1 commentaire
Adrian Velasquez
Adrian Velasquez le 2 Août 2022
Yes that makes sense! I didn't realize that the minimum defaulted to the minimum of the input if it is not explicitely specified. Thanks for the help!

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