Scaling the natrix in specific range
古いコメントを表示
I have a 300x1 matrix
where the minmum value is
-2.1188e+003
maximum is
460.3171
i want to scale in region from -1 to 1
回答 (2 件)
Nasser M. Abbasi
2012 年 12 月 1 日
Since I just wrote the code, might as well post it. A quick function to scale it
function C = nma_rescale(A,new_min,new_max)
%Nasser M. Abbasi 011212
%NO ERROR CHECKING DONE ON INPUT. Rescale a matrix or a vector A
current_max = max(A(:));
current_min = min(A(:));
C =((A-current_min)*(new_max-new_min))/(current_max-current_min) + new_min;
end
to use:
EDU>> C=rand(3);
EDU>> B=nma_rescale(C,-1,1)
B =
-0.7625 0.8960 -0.2033
-0.3188 -0.8938 -1.0000
-0.8317 1.0000 -0.3316
6 件のコメント
kash
2012 年 12 月 1 日
Nasser M. Abbasi
2012 年 12 月 1 日
I am not sure I understand. To scale a matrix A to be between -1 and 1, just type
C = nma_rescale(A,-1,1)
I do not understand what B role is here. May be you can update your question and make it more clear what is needed
Walter Roberson
2012 年 12 月 1 日
The code in my Answer does that, at least when given vectors (the problems statement involved vectors.) But before you go further you should decide whether the negative range is to be scaled independently of the positive range (as in my first solution), or if the entire range is to be treated as a linear block to be re-centered at the average of the max and min (as in my second solution.)
Dario Dematties
2017 年 4 月 18 日
Hello, with respect to the solution proposed by Nasser M. Abbasi. What should I do when current_max=current_min? What is the conventional way to solve this problem?
Jan
2017 年 4 月 18 日
@Dario: There is no convention. It depends on your problem, if the result of a scaled scalar is -1, 0, +1 or Inf, perhaps NaN.
Dario Dematties
2017 年 4 月 18 日
Thanks Jan, Does anyone know which is the reaction of libsvm software on matlab when it is fed with vectors with missing data (NaN)?
Walter Roberson
2012 年 12 月 1 日
編集済み: Walter Roberson
2012 年 12 月 1 日
Does 0 need to remain 0, or should it become the midpoint of your [-2118.8 to 460.317] range?
0 remains 0:
matmin = min(mat);
matmax = max(mat);
newmat = mat;
newmat(mat < 0) = mat(mat < 0) ./ abs(minmat);
newmat(mat > 0) = mat(mat > 0) ./ maxmat;
Midpoint becomes 0:
matmin = min(mat);
matmax = max(mat);
newmat = 2 * (mat - matmin) ./ (matmax - matmin) - 1;
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