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())
*= 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
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)