Numpy method of returning values based on two different arrays

Question

I have two NumPy arrays, say num and denom. I need to return specific values, based on if the respective elements are zero or not, in num and denom,

        r2 = []
        for denind, denel in enumerate(denom):
            numel = num[denind]
            if denel:  # Denominator is not zero
                r2.append(1 - numel/denom)
            elif numel:  # Denominator is zero, but numerator is not zero
                r2.append(0.0)
            else:  # Both denominator and numerator are zero.
                r2.append(1.0)
        return np.array(r2)

Is there a "NumPy" way of doing such an iteration.

Answer 1

Yes, you can use a boolean array as an index to fix up an array after your division. And you can use isinf(a) and isnan(a) to get the desired boolean arrays for indexing a.

with numpy.errstate(divide='ignore'):
    res = 1.0 - num/double(den);
res[isinf(res)] = 0.0;   # zero denominator will give +- infinity
res[isnan(res)] = 1.0;   # both zero will give nan

The magic here is that the boolean-array index returns a view into res which only selects those elements for which the boolean was True. If you read from that selection, you see a flat array, but if you write to it (as above) then it modifies the original array.

Also, I explicitly cast den to double because if both operands were integers there would by horror.

Update: It seems you should cast den, if you cast num it will still work, but you get a warning. Safest thing is to cast both.

Answer 2

How about doing the minimal amount of division work?

r2 = np.ones_like(numel, dtype=float)
num_mask = (numel != 0)
den_mask = (denel != 0)
mask = num_mask & den_mask
r2[mask] = 1 - numel[mask]/denel[mask]
mask = ~den_mask & num_mask
r2[mask] = 0

Answer 3

You could use a nested np.where statement

numpy.where(denom !=0, 1 - numel/denom, np.where(numel != 0, 0, 1))