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A comparison of methods for rounding up a dvision to the nearest whole number
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""" | |
A comparison of methods for rounding upwards on dvision operations. Suprisingly, the fastest method was multiplying | |
the numerator in the division operation by negative one, using floor division, and then multiplying again by | |
negative one to arrive at the rounded up answer | |
Further testing is required on larger numbers (and maybe complex numbers?) to see if the results scale well. | |
""" | |
import math | |
import numpy | |
# Hardware: FX 8350 @ 4.5GHz, 8Gb 2133 | |
def negative_one(N, D): # 855.73 us for ten thousand loops - *fastest* | |
for i in range(10001): | |
a = -(-N//D) | |
return a | |
def math_ceiling(N, D): # 4.83 ms for ten thousand loops | |
for i in range(10001): | |
a = math.ceil(N/D) | |
return a | |
def numpy_ceiling(N, D): # 12.11 ms for ten thousand loops | |
for i in range(10001): | |
a = numpy.ceil(N/D) | |
return a | |
def modulo_if(N, D): # 1.50 ms for ten thousand loops | |
for i in range(10001): | |
if N % D == 0: | |
a = N//D | |
else: | |
a = (N//D)+1 | |
return a | |
# test numbers | |
numerator = 10 | |
denominator = 3 | |
# round up functions | |
print(negative_one(numerator, denominator)) | |
print(math_ceiling(numerator, denominator)) | |
print(numpy_ceiling(numerator, denominator)) | |
print(modulo_if(numerator, denominator)) |
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