How to properly round-up half float numbers?
PythonPython 3.xFloating PointRoundingPrecisionPython Problem Overview
I am facing a strange behavior of the round() function:
for i in range(1, 15, 2):
n = i / 2
print(n, "=>", round(n))
This code prints:
0.5 => 0
1.5 => 2
2.5 => 2
3.5 => 4
4.5 => 4
5.5 => 6
6.5 => 6
I expected the floating values to be always rounded up, but instead, it is rounded to the nearest even number.
Why such behavior, and what is the best way to get the correct result?
I tried to use the fractions but the result is the same.
Python Solutions
Solution 1 - Python
The Numeric Types section documents this behaviour explicitly:
> round(x[, n])
> x rounded to n digits, rounding half to even. If n is omitted, it defaults to 0.
Note the rounding half to even. This is also called bankers rounding; instead of always rounding up or down (compounding rounding errors), by rounding to the nearest even number you average out rounding errors.
If you need more control over the rounding behaviour, use the decimal module, which lets you specify exactly what rounding strategy should be used.
For example, to round up from half:
>>> from decimal import localcontext, Decimal, ROUND_HALF_UP
>>> with localcontext() as ctx:
... ctx.rounding = ROUND_HALF_UP
... for i in range(1, 15, 2):
... n = Decimal(i) / 2
... print(n, '=>', n.to_integral_value())
...
0.5 => 1
1.5 => 2
2.5 => 3
3.5 => 4
4.5 => 5
5.5 => 6
6.5 => 7
Solution 2 - Python
For example:
from decimal import Decimal, ROUND_HALF_UP
Decimal(1.5).quantize(0, ROUND_HALF_UP)
# This also works for rounding to the integer part:
Decimal(1.5).to_integral_value(rounding=ROUND_HALF_UP)
Solution 3 - Python
You can use this:
import math
def normal_round(n):
if n - math.floor(n) < 0.5:
return math.floor(n)
return math.ceil(n)
It will round number up or down properly.
Solution 4 - Python
round() will round either up or down, depending on if the number is even or odd. A simple way to only round up is:
int(num + 0.5)
If you want this to work properly for negative numbers use:
((num > 0) - (num < 0)) * int(abs(num) + 0.5)
Note, this can mess up for large numbers or really precise numbers like 5000000000000001.0 and 0.49999999999999994.
Solution 5 - Python
The behavior you are seeing is typical IEEE 754 rounding behavior. If it has to choose between two numbers that are equally different from the input, it always picks the even one. The advantage of this behavior is that the average rounding effect is zero - equally many numbers round up and down. If you round the half way numbers in a consistent direction the rounding will affect the expected value.
The behavior you are seeing is correct if the objective is fair rounding, but that is not always what is needed.
One trick to get the type of rounding you want is to add 0.5 and then take the floor. For example, adding 0.5 to 2.5 gives 3, with floor 3.
Solution 6 - Python
Love the fedor2612 answer. I expanded it with an optional "decimals" argument for those who want to use this function to round any number of decimals (say for example if you want to round a currency $26.455 to $26.46).
import math
def normal_round(n, decimals=0):
expoN = n * 10 ** decimals
if abs(expoN) - abs(math.floor(expoN)) < 0.5:
return math.floor(expoN) / 10 ** decimals
return math.ceil(expoN) / 10 ** decimals
oldRounding = round(26.455,2)
newRounding = normal_round(26.455,2)
print(oldRounding)
print(newRounding)
Output:
> 26.45 > > 26.46
Solution 7 - Python
Why make it so complicated?
def HalfRoundUp(value):
return int(value + 0.5)
You could of course make it into a lambda which would be:
HalfRoundUp = lambda value: int(value + 0.5)
Solution 8 - Python
Short version: use the decimal module. It can represent numbers like 2.675 precisely, unlike Python floats where 2.675 is really 2.67499999999999982236431605997495353221893310546875 (exactly). And you can specify the rounding you desire: ROUND_CEILING, ROUND_DOWN, ROUND_FLOOR, ROUND_HALF_DOWN, ROUND_HALF_EVEN, ROUND_HALF_UP, ROUND_UP, and ROUND_05UP are all options.
Solution 9 - Python
Rounding to the nearest even number has become common practice in numerical disciplines. "Rounding up" produces a slight bias towards larger results.
So, from the perspective of the scientific establishment, round has the correct behavior.
Solution 10 - Python
Here is another solution. It will work as normal rounding in excel.
from decimal import Decimal, getcontext, ROUND_HALF_UP
round_context = getcontext()
round_context.rounding = ROUND_HALF_UP
def c_round(x, digits, precision=5):
tmp = round(Decimal(x), precision)
return float(tmp.__round__(digits))
c_round(0.15, 1) -> 0.2, c_round(0.5, 0) -> 1
Solution 11 - Python
The following solution achieved "school fashion rounding" without using the decimal module (which turns out to be slow).
def school_round(a_in,n_in):
''' python uses "banking round; while this round 0.05 up" '''
if (a_in * 10 ** (n_in + 1)) % 10 == 5:
return round(a_in + 1 / 10 ** (n_in + 1), n_in)
else:
return round(a_in, n_in)
e.g.
print(round(0.005,2)) # 0
print(school_round(0.005,2)) #0.01
Solution 12 - Python
In the question this is basically an issue when dividing a positive integer by 2. The easisest way is int(n + 0.5) for individual numbers.
However we cannot apply this to series, therefore what we then can do for example for a pandas dataframe, and without going into loops, is:
import numpy as np
df['rounded_division'] = np.where(df['some_integer'] % 2 == 0, round(df['some_integer']/2,0), round((df['some_integer']+1)/2,0))
Solution 13 - Python
So just to make sure there is a crystal clear working example here, I wrote a small convenience function
def round_half_up(x: float, num_decimals: int) -> float:
"""Use explicit ROUND HALF UP. See references, for an explanation.
This is the proper way to round, as taught in school.
Args:
x:
num_decimals:
Returns:
https://stackoverflow.com/questions/33019698/how-to-properly-round-up-half-float-numbers-in-python
"""
if num_decimals < 0:
raise ValueError("Num decimals needs to be at least 0.")
target_precision = "1." + "0" * num_decimals
rounded_x = float(Decimal(x).quantize(Decimal(target_precision), ROUND_HALF_UP))
return rounded_x
And an appropriate set of test cases
def test_round_half_up():
x = 1.5
y = round_half_up(x, 0)
assert y == 2.0
y = round_half_up(x, 1)
assert y == 1.5
x = 1.25
y = round_half_up(x, 1)
assert y == 1.3
y = round_half_up(x, 2)
assert y == 1.25
Solution 14 - Python
A small addition as the rounding half up with some of the solutions might not work as expected in some cases.
Using the function from above for instance:
from decimal import Decimal, ROUND_HALF_UP
def round_half_up(x: float, num_decimals: int) -> float:
if num_decimals < 0:
raise ValueError("Num decimals needs to be at least 0.")
target_precision = "1." + "0" * num_decimals
rounded_x = float(Decimal(x).quantize(Decimal(target_precision), ROUND_HALF_UP))
return rounded_x
round_half_up(1.35, 1)
1.4
round_half_up(4.35, 1)
4.3
Where I was expecting 4.4. What did the trick for me was converting x into a string first.
from decimal import Decimal, ROUND_HALF_UP
def round_half_up(x: float, num_decimals: int) -> float:
if num_decimals < 0:
raise ValueError("Num decimals needs to be at least 0.")
target_precision = "1." + "0" * num_decimals
rounded_x = float(Decimal(str(x)).quantize(Decimal(target_precision), ROUND_HALF_UP))
return rounded_x
round_half_up(4.35, 1)
4.4
Solution 15 - Python
You can use:
from decimal import Decimal, ROUND_HALF_UP
for i in range(1, 15, 2):
n = i / 2
print(n, "=>", Decimal(str(n)).quantize(Decimal("1"), rounding=ROUND_HALF_UP))
Solution 16 - Python
A classical mathematical rounding without any libraries
def rd(x,y=0):
''' A classical mathematical rounding by Voznica '''
m = int('1'+'0'*y) # multiplier - how many positions to the right
q = x*m # shift to the right by multiplier
c = int(q) # new number
i = int( (q-c)*10 ) # indicator number on the right
if i >= 5:
c += 1
return c/m
Compare:
print( round(0.49), round(0.51), round(0.5), round(1.5), round(2.5), round(0.15,1)) # 0 1 0 2 2 0.1
print( rd(0.49), rd(0.51), rd(0.5), rd(1.5), rd(2.5), rd(0.15,1)) # 0 1 1 2 3 0.2
Solution 17 - Python
Knowing that round(9.99,0) rounds to int=10 and int(9.99) rounds to int=9 brings success:
Goal: Provide lower and higher round number depending on value
def get_half_round_numers(self, value):
"""
Returns dict with upper_half_rn and lower_half_rn
:param value:
:return:
"""
hrns = {}
if not isinstance(value, float):
print("Error>Input is not a float. None return.")
return None
value = round(value,2)
whole = int(value) # Rounds 9.99 to 9
remainder = (value - whole) * 100
if remainder >= 51:
hrns['upper_half_rn'] = round(round(value,0),2) # Rounds 9.99 to 10
hrns['lower_half_rn'] = round(round(value,0) - 0.5,2)
else:
hrns['lower_half_rn'] = round(int(value),2)
hrns['upper_half_rn'] = round(int(value) + 0.5,2)
return hrns
Some testing:
yw
Solution 18 - Python
import math
# round tossing n digits from the end
def my_round(n, toss=1):
def normal_round(n):
if isinstance(n, int):
return n
intn, dec = str(n).split(".")
if int(dec[-1]) >= 5:
if len(dec) == 1:
return math.ceil(n)
else:
return float(intn + "." + str(int(dec[:-1]) + 1))
else:
return float(intn + "." + dec[:-1])
while toss >= 1:
n = normal_round(n)
toss -= 1
return n
for n in [1.25, 7.3576, 30.56]:
print(my_round(n, 2))
1.0
7.36
31
Solution 19 - Python
You can try this
def round(num):
return round(num + 10**(-9))
it will work since num = x.5 will always will be x.5 + 0.00...01 in the process which its closer to x+1 hence the round function will work properly and it will round x.5 to x+1