Let's say I have a numpy array [[0,1],[3,4],[5,6]] and want to divide it elementwise by [1,2,0].
The desired result will be [[0,1],[1.5,2],[0,0]]. So if the division is by zero, then the result is zero. I only found a way to do it in pandas dataframe with div command, but couldn't find it for numpy arrays and conversion to dataframe does not seem like a good solution.
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2 Answers
You could wrap your operation with np.where to assign the invalid values to 0:
>>> np.where(d[:,None], x/d[:,None], 0)
array([[0. , 1. ],
[1.5, 2. ],
[0. , 0. ]])
This will still raise a warning though because we're not avoiding the division by zero:
/usr/local/lib/python3.7/dist-packages/ipykernel_launcher.py:1:
RuntimeWarning: divide by zero encountered in `true_divide`
"""Entry point for launching an IPython kernel.
A better way is to provide a mask to np.divide with the where argument:
>>> np.divide(x, d[:,None], where=d[:,None] != 0)
array([[0. , 1. ],
[1.5, 2. ],
[0. , 0. ]])
Comments
I have worked out this solution:
[list(x/y) if y != 0 else len(x)*[0,] for x, y in zip(a1, a2)]
