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How do I round to 2 decimals in python?

How do I round to 2 decimals?

In python, you have floats and decimals that can be rounded. If you care about the accuracy of rounding, use decimal type. If you use floats, you will have issues with accuracy.

All the examples use demical types, except for the original value, which is automatically casted as a float.

To set the context of what we are working with, let's start with an original value.

Original Value

print 16.0/7

Output: 2.2857142857142856

1: Round decimal using round()

from decimal import Decimal

# First we take a float and convert it to a decimal
x = Decimal(16.0/7)

# Then we round it to 2 places
output = round(x,2)
print output

Output: 2.29

2: Round decimal with super rounding powers

from decimal import Decimal, ROUND_HALF_UP
# Here are all your options for rounding:
# This one offers the most out of the box control

our_value = Decimal(16.0/7)
output = Decimal(our_value.quantize(Decimal('.01'), rounding=ROUND_HALF_UP))

print output

Output: 2.29

3: Round decimal by setting precision

# If you use deimcal, you need to import
from decimal import getcontext, Decimal

# Set the precision.
getcontext().prec = 3

# Execute 1/7, however cast both numbers as decimals
output = Decimal(16.0)/Decimal(7)

# Your output will return w/ 6 decimal places, which
# we set above.
print output

Output: 2.29

In example 3, if we set the prec to 2, then we would have 2.3. If we set to 6, then we would have 2.28571.


Which approach is best? They are all viable. I am a fan of the second option, because it offers the most control. If you have a very specific use case (i.e. 2010 WMATA practice rounding habits of up and down to the .05 depending on the fare), you may have to customize this part in your code.


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@gonvaled gonvaled commented Sep 16, 2015

See here for an alternative using __format__


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@Serpens66 Serpens66 commented Oct 11, 2015

I think you have to use the Decimal() with strings ?
Decimal(1.1)+Decimal(2.2) == 3.300000000000000266453525910
Decimal(repr(1.1))+Decimal(repr(2.2)) == 3.3

Or am I wrong?


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@The-Loeki The-Loeki commented Dec 7, 2015

Decimal(float) will give you the decimal representation of a floating point number. What you're seeing is the entire raison d'etre for Decimals; floats are binary approximations of (supposedly really large) numbers, converting them to Decimal (or trying to round them etc.) will expose that approximation, and if you're not aware of those internals you'll run into these kinds of suprises.

See for example


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@JamieMcNaught JamieMcNaught commented Jun 20, 2016

I don't think rounding by setting the precision is viable as it takes into account the digits before the decimal point as well as after (hence why set to 3 and not 2). In you example if you do:

from decimal import getcontext, Decimal
getcontext().prec = 3
output = Decimal(160.0)/Decimal(7)
print output

Which I don't think is what you want? You can of course set prec to 4, but that's not really a solution.


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@rpxz rpxz commented Jun 24, 2016

if i'll write
v= Decimal(x)
x is a float() type ,then can i use round(v,2)?


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@SeanTankGarvey SeanTankGarvey commented Jul 23, 2017

@rpxz if you have added the line above it that says:
from decimal import Decimal


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@LePingKYXK LePingKYXK commented Jul 30, 2017

Hello, thank you very much for sharing.
I found a strange result, maybe it is a bug in the decimal module. The code is shown below.

>>>from decimal import Decimal, ROUND_HALF_UP
>>>val = Decimal(1.45)
>>>Decimal(val.quantize(Decimal('0.1'), rounding=ROUND_HALF_UP))


The correct result should be Decimal('1.5').
By the way, I am using Python 3.5.


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@dualbus dualbus commented Aug 6, 2017

You forgot to pass 1.45 as a string:

Python 3.5.4rc1 (default, Jul 25 2017, 08:53:34) 
[GCC 6.4.0 20170704] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> from decimal import Decimal, ROUND_HALF_UP
>>> val = Decimal('1.45')
>>> Decimal(val.quantize(Decimal('0.1'), rounding=ROUND_HALF_UP))
>>> Decimal(1.45) - Decimal('1.45')

Which is mentioned above in


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@FirasBayazed FirasBayazed commented Oct 8, 2017

if anyone could understand this videos can you please write for me the steps and the code so I can type it in paython interpreter
I will be very thankfull


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@nvictor nvictor commented Feb 16, 2018

nice 👍


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@nagimov nagimov commented Mar 5, 2018

3a. Round decimals by setting precision without Decimal

to round float to 3 significant figures without dealing with Decimal module:

>>> '%s' % float('%.3g' % 3.141592654)

%s required to prevent python from printing float in e- scientific notation



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@Baahcy Baahcy commented Sep 5, 2018

for Python 3

Depending on the number you want to round
Kindly use, Example

The codes start below

a = 16 / 7
print ("%.1f" % a)


#The '1 explains where it should starts from
just after the decimal point.
#The 'a' represents the value you assigned it to
If you want it to round on the second, use 2
Like this ----below is the code

print ("%.2f" % a)


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@Ankit-Chauhan10 Ankit-Chauhan10 commented Oct 8, 2018

how to get 3.25, 3.27, 3.29 to be considered as 3.25 and 3.20, 3.22, 3.24 to be considered as 3.20 ? any advice.


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@Mec-iS Mec-iS commented Dec 18, 2018

Decimal has a quite high impact on performances:

In [12]: timeit.timeit("round(7.325, 2)")
Out[12]: 0.39438656000129413

In [9]: timeit.timeit("from decimal import Decimal, ROUND_DOWN;round(Decimal('7.325'), 2)")
Out[9]: 0.9643819720076863

In [21]: timeit.timeit("float('%.2f' % 7.325)")
Out[21]: 0.3827699579996988

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@rblack rblack commented Feb 19, 2019

@Mec-iS While Decimal is indeed slower, I believe your example is not totally correct. You measure imports together with initializing Decimal object. IMO timeit.timeit(stmt="round(d, 2)", setup="from decimal import Decimal, ROUND_DOWN;d=Decimal('7.325')") would be closer analog of rounding floats.


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@weaming weaming commented Oct 15, 2019

Slow but correct, I choose Decimal .


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@HeyHugo HeyHugo commented Nov 8, 2019

Money class method from py-moneyed:

def round(self, ndigits=0):
    Rounds the amount using the current ``Decimal`` rounding algorithm.
    if ndigits is None:
        ndigits = 0
    return self.__class__(
        amount=self.amount.quantize(Decimal('1e' + str(-ndigits))),

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@orrinstimpson orrinstimpson commented Mar 12, 2020

python has a built in round function that allows you to specify the number of decimals to truncate too. See

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