Python round list to 2 decimal places

I have a list which consists of float values but they're too detailed to proceed. I know we can shorten them by using the ("%.f" % variable) operator, like:

result = [359.70000000000005]
result = "%.2f" % result
result = [359.70]

My question is how can I turn a list of values into their rounded equivalents without using an iterator. I've tried something, but it throws a TypeError:

list = [0.30000000000000004, 0.5, 0.20000000000000001]
list = "%.2f" % list
TypeError: not all arguments converted during string formatting

How can I provide a clean list like:

list = [0.30, 0.5, 0.20]

gandreadis

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asked Mar 16, 2011 at 13:37

1

"%.2f" does not return a clean float. It returns a string representing this float with two decimals.

my_list = [0.30000000000000004, 0.5, 0.20000000000000001]
my_formatted_list = [ '%.2f' % elem for elem in my_list ]

returns:

['0.30', '0.50', '0.20']

Also, don't call your variable list. This is a reserved word for list creation. Use some other name, for example my_list.

If you want to obtain [0.30, 0.5, 0.20] (or at least the floats that are the closest possible), you can try this:

my_rounded_list = [ round(elem, 2) for elem in my_list ]

returns:

[0.29999999999999999, 0.5, 0.20000000000000001]

gandreadis

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answered Mar 16, 2011 at 13:39

eumiroeumiro

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If you really want an iterator-free solution, you can use numpy and its array round function.

import numpy as np
myList = list(np.around(np.array(myList),2))

answered Mar 25, 2018 at 4:37

john ktejikjohn ktejik

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You might want to look at Python's decimal module, which can make using floating point numbers and doing arithmetic with them a lot more intuitive. Here's a trivial example of one way of using it to "clean up" your list values:

>>> from decimal import *
>>> mylist = [0.30000000000000004, 0.5, 0.20000000000000001]
>>> getcontext().prec = 2
>>> ["%.2f" % e for e in mylist]
['0.30', '0.50', '0.20']
>>> [Decimal("%.2f" % e) for e in mylist]
[Decimal('0.30'), Decimal('0.50'), Decimal('0.20')]
>>> data = [float(Decimal("%.2f" % e)) for e in mylist]
>>> data
[0.3, 0.5, 0.2]

answered Mar 16, 2011 at 14:08

martineaumartineau

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mylist = [0.30000000000000004, 0.5, 0.20000000000000001]
myRoundedList =  [round(x,2) for x in mylist] 
# [0.3, 0.5, 0.2]

pzaenger

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answered Oct 21, 2018 at 17:42

Silvia42Silvia42

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You can use the built-in map along with a lambda expression:

my_list = [0.2111111111, 0.5, 0.3777777777]
my_list_rounded = list(map(lambda x: round(x, ndigits=2), my_list))
my_list_rounded                                                                                                                                                                                                                 
Out[3]: [0.21, 0.5, 0.38]

Alternatively you could also create a named function for the rounding up to a specific digit using partial from the functools module for working with higher order functions:

from functools import partial

my_list = [0.2111111111, 0.5, 0.3777777777]
round_2digits = partial(round, ndigits=2)
my_list_rounded = list(map(round_2digits, my_list))
my_list_rounded                                                                                                                                                                                                                 
Out[6]: [0.21, 0.5, 0.38]

answered Sep 8, 2020 at 18:22

jojojojo

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Another option which doesn't require numpy is:

precision = 2  
myRoundedList = [int(elem*(10**precision)+delta)/(10.0**precision) for elem in myList]

# delta=0 for floor
# delta = 0.5 for round
# delta = 1 for ceil

answered May 16, 2018 at 15:55

Easiest way :

import numpy as np

np.round(list,2)

answered Jul 4 at 14:42

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