Min-Max normalization for uniform vectors

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JohnDylon
JohnDylon el 9 de Oct. de 2016
Editada: JohnDylon el 19 de Nov. de 2016
Hi,
Can anyone have any point on how to normalize a single number, say 1000, into a range of [-1,1]? Or even a uniform vector of say [1000 1000 1000 1000] into the same range as suggested above?
Normalizing a non-uniform data is trivial. Say the data is v=[1 3 5 7] and we normalize input 5, then
(v(3)-(min(v)))*(1-(-1))/(max(v)-min(v))+(-1)
is the formula I should follow. How about, v is uniform, thus generates a zero denominator in the formula?
JD
  2 comentarios
Image Analyst
Image Analyst el 9 de Oct. de 2016
If it's uniform (all the same value), do you want the output value to be -1, 0, 1, or something else? It can be only one number, unless of course you just want to ignore it and replace the whole thing by random numbers or some other scheme.
JohnDylon
JohnDylon el 9 de Oct. de 2016
I thought to assign 0 for the uniform vector, however this made me think of what the difference should be between [1000 1000 1000] and [2 2 2] if I scale down them to same number between [-1,1]? If I do that, I loose information.

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Respuestas (2)

Jan
Jan el 9 de Oct. de 2016
A normalization requires a range of data. Either this range is predefined or it is determined by the values. A single number or a vector of equal numbers does not have a range. Therefore a normalization requires a predefined knowledge of the possible range of values. Without knowing "min(v)" and "max(v)", a normalization is not possible.
  1 comentario
JohnDylon
JohnDylon el 9 de Oct. de 2016
You are right, however from scaling up or down point of view, there should be a way to represent points that are not in a given interval at another one. I suspect a probability distribution does the job.

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KSSV
KSSV el 9 de Oct. de 2016
Other ways of normalizing are:
Divide each element by the norm of the array. doc norm for this.
Divide each element by maximum of the array.
  1 comentario
JohnDylon
JohnDylon el 9 de Oct. de 2016
Editada: JohnDylon el 19 de Nov. de 2016
I thought such as z-score normalization as well, however it also may gives zero variance, which would make normalization broken at some point.

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