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

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Avi
Avi 2014-9-8
评论： Steven Lord 2023-6-1
I found the script that scale/normalize values in a matrix to be between 0 and 1
I = [ 1 2 3; 4 5 6]; % Some n x m matrix I that contains unscaled values.
scaledI = (I-min(I(:))) ./ (max(I(:)-min(I(:))));
min(scaledI(:)) % the min is 0
max(scaledI(:)) % the max 1
Was wondering if anyone could help me normalize values in matrix between -1 and +1 Thanks

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采纳的回答

José-Luis
José-Luis 2014-9-8
编辑：José-Luis 2014-9-8
Once you have your result:
scaledl = scaledl.*2 - 1;
Or directly:
result = -1 + 2.*(data - min(data))./(max(data) - min(data));
  2 个评论
Robert Hus
Robert Hus 2017-4-13
编辑：Robert Hus 2017-4-13
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

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更多回答（2 个）

Steven Lord
Steven Lord 2019-8-1
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
JAY R
JAY R 2015-5-3
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
  2 个评论
Steven Lord
Steven Lord 2023-6-1
Ran in:
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
So what additional information do you have that would let you "denormalize" x to generate a specific one of y1, y2, or y3?

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