I have two numpy arrays and I am trying to divide one with the other and at the same time, I want to make sure that the entries where the divisor is 0, should just be replaced with 0.
So, I do something like:
log_norm_images = np.where(b_0 > 0, np.divide(diff_images, b_0), 0)
This gives me a run time warning of:
RuntimeWarning: invalid value encountered in true_divide
Now, I wanted to see what was going on and I did the following:
xx = np.isfinite(diff_images)
print (xx[xx == False])
xx = np.isfinite(b_0)
print (xx[xx == False])
However, both of these return empty arrays meaning that all the values in the arrays are finite. So, I am not sure where the invalid value is coming from. I am assuming checking b_0 > 0 in the np.where function takes care of the divide by 0.
The shape of the two arrays are (96, 96, 55, 64) and (96, 96, 55, 1)
4 Answers
You may have a NAN, INF, or NINF floating around somewhere. Try this:
np.isfinite(diff_images).all()
np.isfinite(b_0).all()
If one or both of those returns False, that's likely the cause of the runtime error.
The reason you get the runtime warning when running this:
log_norm_images = np.where(b_0 > 0, np.divide(diff_images, b_0), 0)
is that the inner expression
np.divide(diff_images, b_0)
gets evaluated first, and is run on all elements of diff_images and b_0 (even though you end up ignoring the elements that involve division-by-zero). In other words, the warning happens before the code that ignores those elements. That is why it's a warning and not an error: there are legitimate cases like this one where the division-by-zero is not a problem because it's being handled in a later operation.
Another useful Numpy command is nan_to_num(diff_images) By default it replaces in a Numpy array; NaN to zero, -INF to -(large number) and +INF to +(large number)
You can change the defaults, see
num = np.array([1,2,3,4,5])
den = np.array([1,1,0,1,1])
res = np.array([None]*5)
ix = (den!=0)
res[ix] = np.divide( num[ix], den[ix] )
print(res)
[1.0 2.0 None 4.0 5.0]
