How to scale/normalize values in a matrix to be between -1 and 1

if you have an unequal spread of your data between positive and negative numbers, than the above solution may revert the sign of your array mean. To honour the original spread of positive and negative values (e.g if your smallest negative number is -20 and your largest positive number is +40) you can use the following function. Using this function the -20 will become -0.5 and the +40 will be +1. The solution above has the -20 equates to -1 and +40 to +1.

function norm_value = normalised_diff( data )
% Normalise values of an array to be between -1 and 1
% original sign of the array values is maintained.
if abs(min(data)) > max(data)
      max_range_value = abs(min(data));
      min_range_value = min(data);
  else
      max_range_value = max(data);
      min_range_value = -max(data);
  end
norm_value = 2 .* data ./ (max_range_value - min_range_value);
end

Udaya Rongala 2019-8-1

Thanks @Robert Hus

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Answer 2

Steven Lord 2019-8-1

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https://ww2.mathworks.cn/matlabcentral/answers/154075-how-to-scale-normalize-values-in-a-matrix-to-be-between-1-and-1#answer_385765

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If you're using release R2018a or later, use the normalize function. Specify 'range' as the method and the range to which you want the data normalized (in this case [-1, 1]) as the methodtype.

x = 5*rand(1, 10)
n = normalize(x, 'range', [-1 1])
[minValue, maxValue] = bounds(n) % Should return -1 and 1

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Answer 3

JAY R 2015-5-3

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/154075-how-to-scale-normalize-values-in-a-matrix-to-be-between-1-and-1#answer_177586

编辑：JAY R 2015-5-3

function data = normalize(d)

% the data is normalized so that max is 1, and min is 0

data = (d -repmat(min(d,[],1),size(d,1),1))*spdiags(1./(max(d,[],1)-min(d,[],1))’,0,size(d,2),size(d,2));

taken from here

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Syed 2023-6-1

i need denormalized command

Steven Lord 2023-6-1

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Suppose I told you that I had a normalized data set. Here it is.

x = [0 1];

What was the un-normalized data that was used to generate x? Any of these sets could have resulted in this normalized data.

y1 = [0 1];
y2 = [-1 1];
y3 = [42 1e6];
normalize(y1, 'range', [0 1])
ans = 1×2
     0     1
normalize(y2, 'range', [0 1])
ans = 1×2
     0     1
normalize(y3, 'range', [0 1])
ans = 1×2
     0     1