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How to round a percentage values column in python dataframe? [duplicate]

I was just re-reading What’s New In Python 3.0 and it states:
The round() function rounding strategy and return type have changed.
Exact halfway cases are now rounded to the nearest even result instead
of away from zero. (For example, round(2.5) now returns 2 rather than
3.)
and
the documentation for round:
For the built-in types supporting round(), values are rounded to the
closest multiple of 10 to the power minus n; if two multiples are
equally close, rounding is done toward the even choice
So, under v2.7.3:
In [85]: round(2.5)
Out[85]: 3.0
In [86]: round(3.5)
Out[86]: 4.0
as I'd have expected. However, now under v3.2.3:
In [32]: round(2.5)
Out[32]: 2
In [33]: round(3.5)
Out[33]: 4
This seems counter-intuitive and contrary to what I understand about
rounding (and bound to trip up people). English isn't my native language but
until I read this I thought I knew what rounding meant :-/ I am sure
at the time v3 was introduced there must have been some discussion of
this, but I was unable to find a good reason in my search.
Does anyone have insight into why this was changed to this?
Are there any other mainstream programming languages (e.g., C, C++, Java, Perl, ..) that do this sort of (to me inconsistent) rounding?
What am I missing here?
UPDATE: #Li-aungYip's comment re "Banker's rounding" gave me the right search term/keywords to search for and I found this SO question: Why does .NET use banker's rounding as default?, so I will be reading that carefully.

Python 3's way (called "round half to even" or "banker's rounding") is considered the standard rounding method these days, though some language implementations aren't on the bus yet.
The simple "always round 0.5 up" technique results in a slight bias toward the higher number. With large numbers of calculations, this can be significant. The Python 3.0 approach eliminates this issue.
There is more than one method of rounding in common use. IEEE 754, the international standard for floating-point math, defines five different rounding methods (the one used by Python 3.0 is the default). And there are others.
This behavior is not as widely known as it ought to be. AppleScript was, if I remember correctly, an early adopter of this rounding method. The round command in AppleScript offers several options, but round-toward-even is the default as it is in IEEE 754. Apparently the engineer who implemented the round command got so fed up with all the requests to "make it work like I learned in school" that he implemented just that: round 2.5 rounding as taught in school is a valid AppleScript command. :-)

You can control the rounding you get in Py3000 using the Decimal module:
>>> decimal.Decimal('3.5').quantize(decimal.Decimal('1'),
rounding=decimal.ROUND_HALF_UP)
>>> Decimal('4')
>>> decimal.Decimal('2.5').quantize(decimal.Decimal('1'),
rounding=decimal.ROUND_HALF_EVEN)
>>> Decimal('2')
>>> decimal.Decimal('3.5').quantize(decimal.Decimal('1'),
rounding=decimal.ROUND_HALF_DOWN)
>>> Decimal('3')

Just to add here an important note from documentation:
https://docs.python.org/dev/library/functions.html#round
Note
The behavior of round() for floats can be surprising: for example,
round(2.675, 2) gives 2.67 instead of the expected 2.68. This is not a
bug: it’s a result of the fact that most decimal fractions can’t be
represented exactly as a float. See Floating Point Arithmetic: Issues
and Limitations for more information.
So don't be surprised to get following results in Python 3.2:
>>> round(0.25,1), round(0.35,1), round(0.45,1), round(0.55,1)
(0.2, 0.3, 0.5, 0.6)
>>> round(0.025,2), round(0.035,2), round(0.045,2), round(0.055,2)
(0.03, 0.04, 0.04, 0.06)

Python 3.x rounds .5 values to a neighbour which is even
assert round(0.5) == 0
assert round(1.5) == 2
assert round(2.5) == 2
import decimal
assert decimal.Decimal('0.5').to_integral_value() == 0
assert decimal.Decimal('1.5').to_integral_value() == 2
assert decimal.Decimal('2.5').to_integral_value() == 2
however, one can change decimal rounding "back" to always round .5 up, if needed :
decimal.getcontext().rounding = decimal.ROUND_HALF_UP
assert decimal.Decimal('0.5').to_integral_value() == 1
assert decimal.Decimal('1.5').to_integral_value() == 2
assert decimal.Decimal('2.5').to_integral_value() == 3
i = int(decimal.Decimal('2.5').to_integral_value()) # to get an int
assert i == 3
assert type(i) is int

I recently had problems with this, too. Hence, I have developed a python 3 module that has 2 functions trueround() and trueround_precision() that address this and give the same rounding behaviour were are used to from primary school (not banker's rounding). Here is the module. Just save the code and copy it in or import it. Note: the trueround_precision module can change the rounding behaviour depending on needs according to the ROUND_CEILING, ROUND_DOWN, ROUND_FLOOR, ROUND_HALF_DOWN, ROUND_HALF_EVEN, ROUND_HALF_UP, ROUND_UP, and ROUND_05UP flags in the decimal module (see that modules documentation for more info). For the functions below, see the docstrings or use help(trueround) and help(trueround_precision) if copied into an interpreter for further documentation.
#! /usr/bin/env python3
# -*- coding: utf-8 -*-
def trueround(number, places=0):
'''
trueround(number, places)
example:
>>> trueround(2.55, 1) == 2.6
True
uses standard functions with no import to give "normal" behavior to
rounding so that trueround(2.5) == 3, trueround(3.5) == 4,
trueround(4.5) == 5, etc. Use with caution, however. This still has
the same problem with floating point math. The return object will
be type int if places=0 or a float if places=>1.
number is the floating point number needed rounding
places is the number of decimal places to round to with '0' as the
default which will actually return our interger. Otherwise, a
floating point will be returned to the given decimal place.
Note: Use trueround_precision() if true precision with
floats is needed
GPL 2.0
copywrite by Narnie Harshoe <signupnarnie#gmail.com>
'''
place = 10**(places)
rounded = (int(number*place + 0.5if number>=0 else -0.5))/place
if rounded == int(rounded):
rounded = int(rounded)
return rounded
def trueround_precision(number, places=0, rounding=None):
'''
trueround_precision(number, places, rounding=ROUND_HALF_UP)
Uses true precision for floating numbers using the 'decimal' module in
python and assumes the module has already been imported before calling
this function. The return object is of type Decimal.
All rounding options are available from the decimal module including
ROUND_CEILING, ROUND_DOWN, ROUND_FLOOR, ROUND_HALF_DOWN, ROUND_HALF_EVEN,
ROUND_HALF_UP, ROUND_UP, and ROUND_05UP.
examples:
>>> trueround(2.5, 0) == Decimal('3')
True
>>> trueround(2.5, 0, ROUND_DOWN) == Decimal('2')
True
number is a floating point number or a string type containing a number on
on which to be acted.
places is the number of decimal places to round to with '0' as the default.
Note: if type float is passed as the first argument to the function, it
will first be converted to a str type for correct rounding.
GPL 2.0
copywrite by Narnie Harshoe <signupnarnie#gmail.com>
'''
from decimal import Decimal as dec
from decimal import ROUND_HALF_UP
from decimal import ROUND_CEILING
from decimal import ROUND_DOWN
from decimal import ROUND_FLOOR
from decimal import ROUND_HALF_DOWN
from decimal import ROUND_HALF_EVEN
from decimal import ROUND_UP
from decimal import ROUND_05UP
if type(number) == type(float()):
number = str(number)
if rounding == None:
rounding = ROUND_HALF_UP
place = '1.'
for i in range(places):
place = ''.join([place, '0'])
return dec(number).quantize(dec(place), rounding=rounding)
Hope this helps,
Narnie

Python 2 rounding behaviour in python 3.
Adding 1 at the 15th decimal places.
Accuracy upto 15 digits.
round2=lambda x,y=None: round(x+1e-15,y)
Not right for 175.57. For that it should be added in the 13th decimal place as the number is grown. Switching to Decimal is better than reinventing the same wheel.
from decimal import Decimal, ROUND_HALF_UP
def round2(x, y=2):
prec = Decimal(10) ** -y
return float(Decimal(str(round(x,3))).quantize(prec, rounding=ROUND_HALF_UP))
Not used y

Some cases:
in: Decimal(75.29 / 2).quantize(Decimal('0.01'), rounding=ROUND_HALF_UP)
in: round(75.29 / 2, 2)
out: 37.65 GOOD
in: Decimal(85.55 / 2).quantize(Decimal('0.01'), rounding=ROUND_HALF_UP)
in: round(85.55 / 2, 2)
out: 42.77 BAD
For fix:
in: round(75.29 / 2 + 0.00001, 2)
out: 37.65 GOOD
in: round(85.55 / 2 + 0.00001, 2)
out: 42.78 GOOD
If you want more decimals, for example 4, you should add (+ 0.0000001).
Work for me.

Sample Reproduction:
['{} => {}'.format(x+0.5, round(x+0.5)) for x in range(10)]
['0.5 => 0', '1.5 => 2', '2.5 => 2', '3.5 => 4', '4.5 => 4', '5.5 => 6', '6.5 => 6', '7.5 => 8', '8.5 => 8', '9.5 => 10']
API: https://docs.python.org/3/library/functions.html#round
States:
Return number rounded to ndigits precision after the decimal point. If
ndigits is omitted or is None, it returns the nearest integer to its
input.
For the built-in types supporting round(), values are rounded to the
closest multiple of 10 to the power minus ndigits; if two multiples
are equally close, rounding is done toward the even choice (so, for
example, both round(0.5) and round(-0.5) are 0, and round(1.5) is 2).
Any integer value is valid for ndigits (positive, zero, or negative).
The return value is an integer if ndigits is omitted or None.
Otherwise the return value has the same type as number.
For a general Python object number, round delegates to
number.round.
Note The behavior of round() for floats can be surprising: for
example, round(2.675, 2) gives 2.67 instead of the expected 2.68. This
is not a bug: it’s a result of the fact that most decimal fractions
can’t be represented exactly as a float. See Floating Point
Arithmetic: Issues and Limitations for more information.
Given this insight you can use some math to resolve it
import math
def my_round(i):
f = math.floor(i)
return f if i - f < 0.5 else f+1
now you can run the same test with my_round instead of round.
['{} => {}'.format(x + 0.5, my_round(x+0.5)) for x in range(10)]
['0.5 => 1', '1.5 => 2', '2.5 => 3', '3.5 => 4', '4.5 => 5', '5.5 => 6', '6.5 => 7', '7.5 => 8', '8.5 => 9', '9.5 => 10']

I propose custom function which would work for a DataFrame:
def dfCustomRound(df, dec):
d = 1 / 10 ** dec
df = round(df, dec + 2)
return (((df % (1 * d)) == 0.5 * d).astype(int) * 0.1 * d * np.sign(df) + df).round(dec)

# round module within numpy when decimal is X.5 will give desired (X+1)
import numpy as np
example_of_some_variable = 3.5
rounded_result_of_variable = np.round(example_of_some_variable,0)
print (rounded_result_of_variable)

Try this code:
def roundup(input):
demo = input if str(input)[-1] != "5" else str(input).replace("5","6")
place = len(demo.split(".")[1])-1
return(round(float(demo),place))
The result will be:
>>> x = roundup(2.5)
>>> x
3.0
>>> x = roundup(2.05)
>>> x
2.1
>>> x = roundup(2.005)
>>> x
2.01
Ooutput you can check here:
https://i.stack.imgur.com/QQUkS.png

The easiest way to round in Python 3.x as taught in school is using an auxiliary variable:
n = 0.1
round(2.5 + n)
And these will be the results of the series 2.0 to 3.0 (in 0.1 steps):
>>> round(2 + n)
>>> 2
>>> round(2.1 + n)
>>> 2
>>> round(2.2 + n)
>>> 2
>>> round(2.3 + n)
>>> 2
>>> round(2.4 + n)
>>> 2
>>> round(2.5 + n)
>>> 3
>>> round(2.6 + n)
>>> 3
>>> round(2.7 + n)
>>> 3
>>> round(2.8 + n)
>>> 3
>>> round(2.9 + n)
>>> 3
>>> round(3 + n)
>>> 3

You can control the rounding you using the math.ceil module:
import math
print(math.ceil(2.5))
> 3

How to print as integer instead of float [duplicate]

In Python, how can one print a number that might be an integer or real type, when the latter case would require me to limit my printout to a certain amount of digits?
Long story short, say we have the following example:
print("{0:.3f}".format(num)) # I cannot do print("{}".format(num))
# because I don't want all the decimals
Is there a "Pythy" way to ensure e.g. in case num == 1 that I print 1 instead of 1.000 (I mean other than cluttering my code with if statements)

With Python 3*, you can just use round() because in addition to rounding floats, when applied to an integer it will always return an int:
>>> num = 1.2345
>>> round(num,3)
1.234
>>> num = 1
>>> round(num,3)
1
This behavior is documented in help(float.__round__):
Help on method_descriptor:
__round__(...)
Return the Integral closest to x, rounding half toward even.
When an argument is passed, work like built-in round(x, ndigits).
And help(int.__round__):
Help on method_descriptor:
__round__(...)
Rounding an Integral returns itself.
Rounding with an ndigits argument also returns an integer.
* With Python 2, round() always returns a float.

If you need to maintain a fixed-width for float values, you could use the printf-style formatting, like this:
>>> num = 1
>>> print('%0.*f' % (isinstance(num, float) * 3, num))
1
>>> num = 1.2345
>>> print('%0.*f' % (isinstance(num, float) * 3, num))
1.234
>>> num = 1.2
>>> print('%0.*f' % (isinstance(num, float) * 3, num))
1.200

If you use a fix number of floating point, you could just use a replace to remove the extra 0. For instance this would do the trick:
print("{:.3f}".format(1).replace(".000", ""))

For fix number of decimal point:
>>> num = 0.2
>>> print('%.04*f' % num)
0.2000
>>> num = 3.102
>>> print('%.02*f' % num)
3.10

getting Ceil() of Decimal in python?

Is there a way to get the ceil of a high precision Decimal in python?
>>> import decimal;
>>> decimal.Decimal(800000000000000000001)/100000000000000000000
Decimal('8.00000000000000000001')
>>> math.ceil(decimal.Decimal(800000000000000000001)/100000000000000000000)
8.0
math rounds the value and returns non precise value

The most direct way to take the ceiling of a Decimal instance x is to use x.to_integral_exact(rounding=ROUND_CEILING). There's no need to mess with the context here. Note that this sets the Inexact and Rounded flags where appropriate; if you don't want the flags touched, use x.to_integral_value(rounding=ROUND_CEILING) instead. Example:
>>> from decimal import Decimal, ROUND_CEILING
>>> x = Decimal('-123.456')
>>> x.to_integral_exact(rounding=ROUND_CEILING)
Decimal('-123')
Unlike most of the Decimal methods, the to_integral_exact and to_integral_value methods aren't affected by the precision of the current context, so you don't have to worry about changing precision:
>>> from decimal import getcontext
>>> getcontext().prec = 2
>>> x.to_integral_exact(rounding=ROUND_CEILING)
Decimal('-123')
By the way, in Python 3.x, math.ceil works exactly as you want it to, except that it returns an int rather than a Decimal instance. That works because math.ceil is overloadable for custom types in Python 3. In Python 2, math.ceil simply converts the Decimal instance to a float first, potentially losing information in the process, so you can end up with incorrect results.

x = decimal.Decimal('8.00000000000000000000001')
with decimal.localcontext() as ctx:
ctx.prec=100000000000000000
ctx.rounding=decimal.ROUND_CEILING
y = x.to_integral_exact()

You can do this using the precision and rounding mode option of the Context constructor.
ctx = decimal.Context(prec=1, rounding=decimal.ROUND_CEILING)
ctx.divide(decimal.Decimal(800000000000000000001), decimal.Decimal(100000000000000000000))
EDIT: You should consider changing the accepted answer.. Although the prec can be increased as needed, to_integral_exact is a simpler solution.

>>> decimal.Context(rounding=decimal.ROUND_CEILING).quantize(
... decimal.Decimal(800000000000000000001)/100000000000000000000, 0)
Decimal('9')

def decimal_ceil(x):
int_x = int(x)
if x - int_x == 0:
return int_x
return int_x + 1

Just use potency to make this.
import math
def lo_ceil(num, potency=0): # Use 0 for multiples of 1, 1 for multiples of 10, 2 for 100 ...
n = num / (10.0 ** potency)
c = math.ceil(n)
return c * (10.0 ** potency)
lo_ceil(8.0000001, 1) # return 10

How do you round UP a number?

How does one round a number UP in Python?
I tried round(number) but it rounds the number down. Example:
round(2.3) = 2.0
and not 3, as I would like.
The I tried int(number + .5) but it round the number down again! Example:
int(2.3 + .5) = 2

The math.ceil (ceiling) function returns the smallest integer higher or equal to x.
For Python 3:
import math
print(math.ceil(4.2))
For Python 2:
import math
print(int(math.ceil(4.2)))

I know this answer is for a question from a while back, but if you don't want to import math and you just want to round up, this works for me.
>>> int(21 / 5)
4
>>> int(21 / 5) + (21 % 5 > 0)
5
The first part becomes 4 and the second part evaluates to "True" if there is a remainder, which in addition True = 1; False = 0. So if there is no remainder, then it stays the same integer, but if there is a remainder it adds 1.

If working with integers, one way of rounding up is to take advantage of the fact that // rounds down: Just do the division on the negative number, then negate the answer. No import, floating point, or conditional needed.
rounded_up = -(-numerator // denominator)
For example:
>>> print(-(-101 // 5))
21

Interesting Python 2.x issue to keep in mind:
>>> import math
>>> math.ceil(4500/1000)
4.0
>>> math.ceil(4500/1000.0)
5.0
The problem is that dividing two ints in python produces another int and that's truncated before the ceiling call. You have to make one value a float (or cast) to get a correct result.
In javascript, the exact same code produces a different result:
console.log(Math.ceil(4500/1000));
5

You might also like numpy:
>>> import numpy as np
>>> np.ceil(2.3)
3.0
I'm not saying it's better than math, but if you were already using numpy for other purposes, you can keep your code consistent.
Anyway, just a detail I came across. I use numpy a lot and was surprised it didn't get mentioned, but of course the accepted answer works perfectly fine.

Use math.ceil to round up:
>>> import math
>>> math.ceil(5.4)
6.0
NOTE: The input should be float.
If you need an integer, call int to convert it:
>>> int(math.ceil(5.4))
6
BTW, use math.floor to round down and round to round to nearest integer.
>>> math.floor(4.4), math.floor(4.5), math.floor(5.4), math.floor(5.5)
(4.0, 4.0, 5.0, 5.0)
>>> round(4.4), round(4.5), round(5.4), round(5.5)
(4.0, 5.0, 5.0, 6.0)
>>> math.ceil(4.4), math.ceil(4.5), math.ceil(5.4), math.ceil(5.5)
(5.0, 5.0, 6.0, 6.0)

I am surprised nobody suggested
(numerator + denominator - 1) // denominator
for integer division with rounding up. Used to be the common way for C/C++/CUDA (cf. divup)

The syntax may not be as pythonic as one might like, but it is a powerful library.
https://docs.python.org/2/library/decimal.html
from decimal import *
print(int(Decimal(2.3).quantize(Decimal('1.'), rounding=ROUND_UP)))

For those who want to round up a / b and get integer:
Another variant using integer division is
def int_ceil(a, b):
return (a - 1) // b + 1
>>> int_ceil(19, 5)
4
>>> int_ceil(20, 5)
4
>>> int_ceil(21, 5)
5
Note: a and b must be non-negative integers

Here is a way using modulo and bool
n = 2.3
int(n) + bool(n%1)
Output:
3

Try this:
a = 211.0
print(int(a) + ((int(a) - a) != 0))

Be shure rounded value should be float
a = 8
b = 21
print math.ceil(a / b)
>>> 0
but
print math.ceil(float(a) / b)
>>> 1.0

The above answers are correct, however, importing the math module just for this one function usually feels like a bit of an overkill for me. Luckily, there is another way to do it:
g = 7/5
g = int(g) + (not g.is_integer())
True and False are interpreted as 1 and 0 in a statement involving numbers in python. g.is_interger() basically translates to g.has_no_decimal() or g == int(g). So the last statement in English reads round g down and add one if g has decimal.

In case anyone is looking to round up to a specific decimal place:
import math
def round_up(n, decimals=0):
multiplier = 10 ** decimals
return math.ceil(n * multiplier) / multiplier

Without importing math // using basic envionment:
a) method / class method
def ceil(fl):
return int(fl) + (1 if fl-int(fl) else 0)
def ceil(self, fl):
return int(fl) + (1 if fl-int(fl) else 0)
b) lambda:
ceil = lambda fl:int(fl)+(1 if fl-int(fl) else 0)

>>> def roundup(number):
... return round(number+.5)
>>> roundup(2.3)
3
>>> roundup(19.00000000001)
20
This function requires no modules.

x * -1 // 1 * -1
Confusing but it works: For x=7.1, you get 8.0. For x = -1.1, you get -1.0
No need to import a module.

For those who doesn't want to use import.
For a given list or any number:
x = [2, 2.1, 2.5, 3, 3.1, 3.5, 2.499,2.4999999999, 3.4999999,3.99999999999]
You must first evaluate if the number is equal to its integer, which always rounds down. If the result is True, you return the number, if is not, return the integer(number) + 1.
w = lambda x: x if x == int(x) else int(x)+1
[w(i) for i in z]
>>> [2, 3, 3, 3, 4, 4, 3, 3, 4, 4]
Math logic:
If the number has decimal part: round_up - round_down == 1, always.
If the number doens't have decimal part: round_up - round_down == 0.
So:
round_up == x + round_down
With:
x == 1 if number != round_down
x == 0 if number == round_down
You are cutting the number in 2 parts, the integer and decimal. If decimal isn't 0, you add 1.
PS:I explained this in details since some comments above asked for that and I'm still noob here, so I can't comment.

If you don't want to import anything, you can always write your own simple function as:
def RoundUP(num):
if num== int(num):
return num
return int(num + 1)

To do it without any import:
>>> round_up = lambda num: int(num + 1) if int(num) != num else int(num)
>>> round_up(2.0)
2
>>> round_up(2.1)
3

I know this is from quite a while back, but I found a quite interesting answer, so here goes:
-round(-x-0.5)
This fixes the edges cases and works for both positive and negative numbers, and doesn't require any function import
Cheers

I'm surprised I haven't seen this answer yet round(x + 0.4999), so I'm going to put it down. Note that this works with any Python version. Changes made to the Python rounding scheme has made things difficult. See this post.
Without importing, I use:
def roundUp(num):
return round(num + 0.49)
testCases = list(x*0.1 for x in range(0, 50))
print(testCases)
for test in testCases:
print("{:5.2f} -> {:5.2f}".format(test, roundUp(test)))
Why this works
From the docs
For the built-in types supporting round(), values are rounded to the closest multiple of 10 to the power minus n; if two multiples are equally close, rounding is done toward the even choice
Therefore 2.5 gets rounded to 2 and 3.5 gets rounded to 4. If this was not the case then rounding up could be done by adding 0.5, but we want to avoid getting to the halfway point. So, if you add 0.4999 you will get close, but with enough margin to be rounded to what you would normally expect. Of course, this will fail if the x + 0.4999 is equal to [n].5000, but that is unlikely.

You could use round like this:
cost_per_person = round(150 / 2, 2)

You can use floor devision and add 1 to it.
2.3 // 2 + 1

when you operate 4500/1000 in python, result will be 4, because for default python asume as integer the result, logically:
4500/1000 = 4.5 --> int(4.5) = 4
and ceil of 4 obviouslly is 4
using 4500/1000.0 the result will be 4.5 and ceil of 4.5 --> 5
Using javascript you will recieve 4.5 as result of 4500/1000, because javascript asume only the result as "numeric type" and return a result directly as float
Good Luck!!

I think you are confusing the working mechanisms between int() and round().
int() always truncates the decimal numbers if a floating number is given; whereas round(), in case of 2.5 where 2 and 3 are both within equal distance from 2.5, Python returns whichever that is more away from the 0 point.
round(2.5) = 3
int(2.5) = 2

My share
I have tested print(-(-101 // 5)) = 21 given example above.
Now for rounding up:
101 * 19% = 19.19
I can not use ** so I spread the multiply to division:
(-(-101 //(1/0.19))) = 20

I'm basically a beginner at Python, but if you're just trying to round up instead of down why not do:
round(integer) + 1

round() doesn't seem to be rounding properly

The documentation for the round() function states that you pass it a number, and the positions past the decimal to round. Thus it should do this:
n = 5.59
round(n, 1) # 5.6
But, in actuality, good old floating point weirdness creeps in and you get:
5.5999999999999996
For the purposes of UI, I need to display 5.6. I poked around the Internet and found some documentation that this is dependent on my implementation of Python. Unfortunately, this occurs on both my Windows dev machine and each Linux server I've tried. See here also.
Short of creating my own round library, is there any way around this?

I can't help the way it's stored, but at least formatting works correctly:
'%.1f' % round(n, 1) # Gives you '5.6'

Formatting works correctly even without having to round:
"%.1f" % n

If you use the Decimal module you can approximate without the use of the 'round' function. Here is what I've been using for rounding especially when writing monetary applications:
from decimal import Decimal, ROUND_UP
Decimal(str(16.2)).quantize(Decimal('.01'), rounding=ROUND_UP)
This will return a Decimal Number which is 16.20.

round(5.59, 1) is working fine. The problem is that 5.6 cannot be represented exactly in binary floating point.
>>> 5.6
5.5999999999999996
>>>
As Vinko says, you can use string formatting to do rounding for display.
Python has a module for decimal arithmetic if you need that.

You get '5.6' if you do str(round(n, 1)) instead of just round(n, 1).

You can switch the data type to an integer:
>>> n = 5.59
>>> int(n * 10) / 10.0
5.5
>>> int(n * 10 + 0.5)
56
And then display the number by inserting the locale's decimal separator.
However, Jimmy's answer is better.

Take a look at the Decimal module
Decimal “is based on a floating-point
model which was designed with people
in mind, and necessarily has a
paramount guiding principle –
computers must provide an arithmetic
that works in the same way as the
arithmetic that people learn at
school.” – excerpt from the decimal
arithmetic specification.
and
Decimal numbers can be represented
exactly. In contrast, numbers like 1.1
and 2.2 do not have an exact
representations in binary floating
point. End users typically would not
expect 1.1 + 2.2 to display as
3.3000000000000003 as it does with binary floating point.
Decimal provides the kind of operations that make it easy to write apps that require floating point operations and also need to present those results in a human readable format, e.g., accounting.

Floating point math is vulnerable to slight, but annoying, precision inaccuracies. If you can work with integer or fixed point, you will be guaranteed precision.

It's a big problem indeed. Try out this code:
print "%.2f" % (round((2*4.4+3*5.6+3*4.4)/8,2),)
It displays 4.85. Then you do:
print "Media = %.1f" % (round((2*4.4+3*5.6+3*4.4)/8,1),)
and it shows 4.8. Do you calculations by hand the exact answer is 4.85, but if you try:
print "Media = %.20f" % (round((2*4.4+3*5.6+3*4.4)/8,20),)
you can see the truth: the float point is stored as the nearest finite sum of fractions whose denominators are powers of two.

printf the sucker.
print '%.1f' % 5.59 # returns 5.6

I would avoid relying on round() at all in this case. Consider
print(round(61.295, 2))
print(round(1.295, 2))
will output
61.3
1.29
which is not a desired output if you need solid rounding to the nearest integer. To bypass this behavior go with math.ceil() (or math.floor() if you want to round down):
from math import ceil
decimal_count = 2
print(ceil(61.295 * 10 ** decimal_count) / 10 ** decimal_count)
print(ceil(1.295 * 10 ** decimal_count) / 10 ** decimal_count)
outputs
61.3
1.3
Hope that helps.

You can use the string format operator %, similar to sprintf.
mystring = "%.2f" % 5.5999

I am doing:
int(round( x , 0))
In this case, we first round properly at the unit level, then we convert to integer to avoid printing a float.
so
>>> int(round(5.59,0))
6
I think this answer works better than formating the string, and it also makes more sens to me to use the round function.

Works Perfect
format(5.59, '.1f') # to display
float(format(5.59, '.1f')) #to round

Another potential option is:
def hard_round(number, decimal_places=0):
"""
Function:
- Rounds a float value to a specified number of decimal places
- Fixes issues with floating point binary approximation rounding in python
Requires:
- `number`:
- Type: int|float
- What: The number to round
Optional:
- `decimal_places`:
- Type: int
- What: The number of decimal places to round to
- Default: 0
Example:
```
hard_round(5.6,1)
```
"""
return int(number*(10**decimal_places)+0.5)/(10**decimal_places)

Code:
x1 = 5.63
x2 = 5.65
print(float('%.2f' % round(x1,1))) # gives you '5.6'
print(float('%.2f' % round(x2,1))) # gives you '5.7'
Output:
5.6
5.7

The problem is only when last digit is 5. Eg. 0.045 is internally stored as 0.044999999999999... You could simply increment last digit to 6 and round off. This will give you the desired results.
import re
def custom_round(num, precision=0):
# Get the type of given number
type_num = type(num)
# If the given type is not a valid number type, raise TypeError
if type_num not in [int, float, Decimal]:
raise TypeError("type {} doesn't define __round__ method".format(type_num.__name__))
# If passed number is int, there is no rounding off.
if type_num == int:
return num
# Convert number to string.
str_num = str(num).lower()
# We will remove negative context from the number and add it back in the end
negative_number = False
if num < 0:
negative_number = True
str_num = str_num[1:]
# If number is in format 1e-12 or 2e+13, we have to convert it to
# to a string in standard decimal notation.
if 'e-' in str_num:
# For 1.23e-7, e_power = 7
e_power = int(re.findall('e-[0-9]+', str_num)[0][2:])
# For 1.23e-7, number = 123
number = ''.join(str_num.split('e-')[0].split('.'))
zeros = ''
# Number of zeros = e_power - 1 = 6
for i in range(e_power - 1):
zeros = zeros + '0'
# Scientific notation 1.23e-7 in regular decimal = 0.000000123
str_num = '0.' + zeros + number
if 'e+' in str_num:
# For 1.23e+7, e_power = 7
e_power = int(re.findall('e\+[0-9]+', str_num)[0][2:])
# For 1.23e+7, number_characteristic = 1
# characteristic is number left of decimal point.
number_characteristic = str_num.split('e+')[0].split('.')[0]
# For 1.23e+7, number_mantissa = 23
# mantissa is number right of decimal point.
number_mantissa = str_num.split('e+')[0].split('.')[1]
# For 1.23e+7, number = 123
number = number_characteristic + number_mantissa
zeros = ''
# Eg: for this condition = 1.23e+7
if e_power >= len(number_mantissa):
# Number of zeros = e_power - mantissa length = 5
for i in range(e_power - len(number_mantissa)):
zeros = zeros + '0'
# Scientific notation 1.23e+7 in regular decimal = 12300000.0
str_num = number + zeros + '.0'
# Eg: for this condition = 1.23e+1
if e_power < len(number_mantissa):
# In this case, we only need to shift the decimal e_power digits to the right
# So we just copy the digits from mantissa to characteristic and then remove
# them from mantissa.
for i in range(e_power):
number_characteristic = number_characteristic + number_mantissa[i]
number_mantissa = number_mantissa[i:]
# Scientific notation 1.23e+1 in regular decimal = 12.3
str_num = number_characteristic + '.' + number_mantissa
# characteristic is number left of decimal point.
characteristic_part = str_num.split('.')[0]
# mantissa is number right of decimal point.
mantissa_part = str_num.split('.')[1]
# If number is supposed to be rounded to whole number,
# check first decimal digit. If more than 5, return
# characteristic + 1 else return characteristic
if precision == 0:
if mantissa_part and int(mantissa_part[0]) >= 5:
return type_num(int(characteristic_part) + 1)
return type_num(characteristic_part)
# Get the precision of the given number.
num_precision = len(mantissa_part)
# Rounding off is done only if number precision is
# greater than requested precision
if num_precision <= precision:
return num
# Replace the last '5' with 6 so that rounding off returns desired results
if str_num[-1] == '5':
str_num = re.sub('5$', '6', str_num)
result = round(type_num(str_num), precision)
# If the number was negative, add negative context back
if negative_number:
result = result * -1
return result

Here's where I see round failing. What if you wanted to round these 2 numbers to one decimal place?
23.45
23.55
My education was that from rounding these you should get:
23.4
23.6
the "rule" being that you should round up if the preceding number was odd, not round up if the preceding number were even.
The round function in python simply truncates the 5.

Here is an easy way to round a float number to any number of decimal places, and it still works in 2021!
float_number = 12.234325335563
rounded = round(float_number, 3) # 3 is the number of decimal places to be returned.You can pass any number in place of 3 depending on how many decimal places you want to return.
print(rounded)
And this will print;
12.234

What about:
round(n,1)+epsilon