# How to normalize a NumPy array to within a certain range?

StackOverflow

After doing some processing on an audio or image array, it needs to be normalized within a range before it can be written back to a file. This can be done like so:

``````# Normalize audio channels to between -1.0 and +1.0
audio[:,0] = audio[:,0]/abs(audio[:,0]).max()
audio[:,1] = audio[:,1]/abs(audio[:,1]).max()

# Normalize image to between 0 and 255
image = image/(image.max()/255.0)
``````

Is there a less verbose, convenience function way to do this? `matplotlib.colors.Normalize()` doesn"t seem to be related.

``````audio /= np.max(np.abs(audio),axis=0)
image *= (255.0/image.max())
``````

Using `/=` and `*=` allows you to eliminate an intermediate temporary array, thus saving some memory. Multiplication is less expensive than division, so

``````image *= 255.0/image.max()    # Uses 1 division and image.size multiplications
``````

is marginally faster than

``````image /= image.max()/255.0    # Uses 1+image.size divisions
``````

Since we are using basic numpy methods here, I think this is about as efficient a solution in numpy as can be.

In-place operations do not change the dtype of the container array. Since the desired normalized values are floats, the `audio` and `image` arrays need to have floating-point point dtype before the in-place operations are performed. If they are not already of floating-point dtype, you"ll need to convert them using `astype`. For example,

``````image = image.astype("float64")
``````

If the array contains both positive and negative data, I"d go with:

``````import numpy as np

a = np.random.rand(3,2)

# Normalised [0,1]
b = (a - np.min(a))/np.ptp(a)

# Normalised [0,255] as integer: don"t forget the parenthesis before astype(int)
c = (255*(a - np.min(a))/np.ptp(a)).astype(int)

# Normalised [-1,1]
d = 2.*(a - np.min(a))/np.ptp(a)-1
``````

If the array contains `nan`, one solution could be to just remove them as:

``````def nan_ptp(a):
return np.ptp(a[np.isfinite(a)])

b = (a - np.nanmin(a))/nan_ptp(a)
``````

However, depending on the context you might want to treat `nan` differently. E.g. interpolate the value, replacing in with e.g. 0, or raise an error.

Finally, worth mentioning even if it"s not OP"s question, standardization:

``````e = (a - np.mean(a)) / np.std(a)
``````