- 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(:)).
Need help understanding rescale()
20 次查看(过去 30 天)
显示 更早的评论
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
0 个评论
采纳的回答
Jeffrey Clark
2022-8-1
@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.
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.
更多回答(0 个)
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!