Numpy conditional arithmetic operations on two arrays

Question

Still trying to earn my numpy stripes: I want to perform an arithmetic operation on two numpy arrays, which is simple enough:

return 0.5 * np.sum(((array1 - array2) ** 2) / (array1 + array2))

Problem is, I need to be able to specify the condition that, if both arrays are element-wise 0 at the same element i, don't perform the operation at all--would be great just to return 0 on this one--so as not to divide by 0.

However, I have no idea how to specify this condition without resorting to the dreaded nested for-loop. Thank you in advance for your assistance.

Edit: Would also be ideal not to have to resort to a pseudocount of +1.

Sven Marnach · Accepted Answer · 2011-07-20 09:41:28Z

3

Just replace np.sum() by np.nansum():

return 0.5 * np.nansum(((array1 - array2) ** 2) / (array1 + array2))

np.nansum() treats nans as zero.

edited Jul 20, 2011 at 9:41

answered Mar 15, 2011 at 23:44

Sven Marnach

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Benjamin · Accepted Answer · 2011-03-15 23:29:41Z

3

return numpy.select([array1 == array2, array1 != array2], [0.5 * np.sum(((array1 - array2) ** 2) / (array1 + array2)), 0])

should do the trick... numpy.where might also be used.

answered Mar 15, 2011 at 23:29

Benjamin

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2 Comments

Magsol Over a year ago

This is an excellent idea (didn't know I could do this), but shouldn't it be numpy.select([array1 == 0.0 and array2 == 0.0, array1 != array2], ...) for the first condition?

Benjamin Over a year ago

Yes, I didn't catch that the first time. You can do (array1 == 0) * (array2 == 0) for the first condition. However, Sven's solution seems even better.

JoshAdel · Accepted Answer · 2011-03-15 23:55:29Z

0

You could also try post-applying numpy.nan_to_num:

http://docs.scipy.org/doc/numpy/reference/generated/numpy.nan_to_num.html

although I found that when I have a divide by zero in your code, it gives a warning, but fills that element with zero (when doing integer math) and NaN when using floats.

If you want to skip the sum when you have a divide by zero, you could also just do the calculation and then test for the NaN before returning:

xx = np.sum(((array1 - array2) ** 2) / (array1 + array2))
if np.isnan(xx):
    return 0
else:
    return xx

Edit: To silence warnings you could try messing around with numpy.seterr:

http://docs.scipy.org/doc/numpy/reference/generated/numpy.seterr.html

edited Mar 15, 2011 at 23:55

answered Mar 15, 2011 at 23:35

JoshAdel

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4 Comments

Magsol Over a year ago

This is very good suggestion. My only complaint here is that I'd still have to silence the cascading warnings when a divide-by-zero occurs.

Sven Marnach Over a year ago

@Magsol: Why are there warnings? NumPy should actually divide by zero without complaining, and just yield NaN.

JoshAdel Over a year ago

@Sven: I believe the warning is new as of numpy 1.5

eat Over a year ago

No messing around is needed, just old_setting= np.seterr(divide= 'ignore') should be enough. Thanks

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