Wsrtjtyk

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 rounded numbers is likely to be slightly larger than the average of the unrounded numbers: this bias or drift can have very bad effects on some numerical algorithms and make them inaccurate."

Disregarding this information and assuming that I always want to round up, how can I do it in numpy? Assuming my array can be quite large.

For simplicity, lets assume i have the array:

import numpy as np



A = [ [10, 15, 30], [25, 134, 41], [134, 413, 51]]

A = np.array(A, dtype=np.int16)



decimal = A * .1

whole = np.round(decimal)

decimal looks like:

[[  1.    1.5   3. ]

 [  2.5  13.4   4.1]

 [ 13.4  41.3   5.1]]

whole looks like:

[[  1.   2.   3.]

 [  2.  13.   4.]

 [ 13.  41.   5.]]

As you can see, 1.5 rounded to 2 and 2.5 also rounded to 2. How can I force to always get a round up answer for a XX.5? I know I can loop through the array and use python round() but that would definitely be much slower. Was wondering if there is a way to do it using numpy functions

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

In [25]: def yourroundfunction(number):

    ...:     if(int(number)+0.5<=number):

    ...:         number=int(number)+1

    ...:     else:

    ...:         number=int(number)

    ...:     return number

    ...: 

    ...: 



In [26]:     

In [26]: customround=np.vectorize(yourroundfunction)



In [27]: customround(A)

Out[27]: 

array([[ 1,  2,  3],

       [ 3, 13,  4],

       [13, 41,  5]])

Well you could define a customround function and then pass it to np.vectorize and then get your expected output . I wrote this custom function because calling round in ipython doesnt work exactly the same as round in python