Normalize / Translate ndarray - Numpy / Python

Question

There is a simple way to normalize a ndarray (every values between 0.0, 1.0)?

For example, I have a matrix like:

a = [[1., 2., 3.],
     [4., 5., 6.],
     [7., 8., 9.]]

Until now I'm getting the max value with

max(max(p[1:]) for p in a)
a / p

Besides I think numpy may have a method to this in one line, this doesn't work if my data is something like this:

b = [[-1., -2., -3.],
     [-4., -5., -6.],
     [-7., -8., 0.]]

Which gives an error caused by zero division.

What I'm trying to do is that the max value became 1. So, I would like to do a translation such that 9 becomes 1 (in positive case just dividing the values by it max value), and 0 (when it is the max value) becomes 1 (with translation method, e.g), which I know hot to do, but I guess numpy may have a solution for do this thing in its package.

How can I perform this nicely with numpy?

Thank you in advance.

well how would you normalize the array if the maximum value is a 0? The error isn't because of the way you're doing it, it is legitimately a domain error in the way you've defined the problem. — brianmearns
– brianmearns, Commented Mar 17, 2014 at 16:08
What I'm trying to do is that the max value became 1 and the minimum 0. So, I would like to do a translation such that 0 becomes 1, which I know hot to do, but I guess numpy may have a solution inside its package. — pceccon
– pceccon, Commented Mar 17, 2014 at 16:15
Ok, so you want to normalize by the largest magnitude (i.e., absolute value). — brianmearns
– brianmearns, Commented Mar 17, 2014 at 16:16
@sh1ftst0rm -- Not exactly. Normalizing by the largest magnitude would put the values in the range from -1, 1 in the general case. There needs to be a shift and normalization by the peak to peak value. — mgilson
– mgilson, Commented Mar 17, 2014 at 16:20

mgilson · Accepted Answer · 2014-03-17 16:19:07Z

7

You could use np.ptp¹ (peak to peak) in conjunction with np.min to do this in the general case:

new_arr = (a - a.min())/np.ptp(a)

example:

>>> a = np.array([[-1., 0, 1], [0, 2, 1]])
>>> np.ptp(a)
3.0
>>> a
array([[-1.,  0.,  1.],
       [ 0.,  2.,  1.]])
>>> (a - a.min())/np.ptp(a)
array([[ 0.        ,  0.33333333,  0.66666667],
       [ 0.33333333,  1.        ,  0.66666667]])

Of course, this still would give an error if a consists of entirely zeros -- But the problem isn't well posed in that case.

^{¹IIRC, np.ptp calls np.max and np.min. If performance is really critical, you might what to create your own ptp and save np.min to a temporary variable so you don't calculate it twice.}

edited Mar 17, 2014 at 16:19

answered Mar 17, 2014 at 16:07

mgilson

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2 Comments

Cody Piersall Over a year ago

Note that the OP has a problem when the max value is 0; I don't think this helps in that case.

mgilson Over a year ago

@CodyPiersall -- Yeah, I realized that. Addressed.

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