How to always round up a XX.5 in numpy

Question

I read that numpy is unbiased in rounding and that it works the way its designed. That \"if you always round 0.5 up to the next largest number, then the average of a bunch

Answer 1

The answer is almost never np.vectorize. You can, and should, do this in a fully vectorized manner. Let's say that for x >= 0, you want r = floor(x + 0.5). If you want negative numbers to round towards zero, the same formula applies for x < 0. So let's say that you always want to round away from zero. In that case, you are looking for ceil(x - 0.5) for x < 0.

To implement that for an entire array without calling np.vectorize, you can use masking:

def round_half_up(x):
    mask = (x >= 0)
    out = np.empty_like(x)
    out[mask] = np.floor(x[mask] + 0.5)
    out[~mask] = np.ceil(x[~mask] - 0.5)
    return out

Notice that you don't need to use a mask if you round all in one direction:

def round_up(x):
    return np.floor(x + 0.5)

Now if you want to make this really efficient, you can get rid of all the temp arrays. This will use the full power of ufuncs:

def round_half_up(x):
    out = x.copy()
    mask = (out >= 0)
    np.add(out, 0.5, where=mask, out=out)
    np.floor(out, where=mask, out=out)
    np.invert(mask, out=mask)
    np.subtract(out, 0.5, where=mask, out=out)
    np.ceil(out, where=mask, out=out)
    return out

And:

def round_up(x):
    out = x + 0.5
    np.floor(out, out=out)
    return out

Answer 2

import numpy as np
A = [ [1.0, 1.5, 3.0], [2.5, 13.4, 4.1], [13.4, 41.3, 5.1]]
A = np.array(A)

print(A)

def rounder(x):
    if (x-int(x) >= 0.5):
        return np.ceil(x)
    else:
        return np.floor(x)

rounder_vec = np.vectorize(rounder)
whole = rounder_vec(A)
print(whole)

Alternatively, you can also look at numpy.ceil, numpy.floor, numpy.trunc for other rounding styles