I want a to be rounded to 13.95. I tried using round, but I get:
>>> a
13.949999999999999
>>> round(a, 2)
13.949999999999999
For the analogous issue with the standard library Decimal class, see How can I format a decimal to always show 2 decimal places?.
You are running into the old problem with floating point numbers that not all numbers can be represented exactly. The command line is just showing you the full floating point form from memory.
With floating point representation, your rounded version is the same number. Since computers are binary, they store floating point numbers as an integer and then divide it by a power of two so 13.95 will be represented in a similar fashion to 125650429603636838/(2**53).
Double precision numbers have 53 bits (16 digits) of precision and regular floats have 24 bits (8 digits) of precision. The floating point type in Python uses double precision to store the values.
For example,
>>> 125650429603636838/(2**53)
13.949999999999999
>>> 234042163/(2**24)
13.949999988079071
>>> a = 13.946
>>> print(a)
13.946
>>> print("%.2f" % a)
13.95
>>> round(a,2)
13.949999999999999
>>> print("%.2f" % round(a, 2))
13.95
>>> print("{:.2f}".format(a))
13.95
>>> print("{:.2f}".format(round(a, 2)))
13.95
>>> print("{:.15f}".format(round(a, 2)))
13.949999999999999
If you are after only two decimal places (to display a currency value, for example), then you have a couple of better choices:
Use integers and store values in cents, not dollars and then divide by 100 to convert to dollars.
Or use a fixed point number like decimal.
There are new format specifications, String Format Specification Mini-Language:
You can do the same as:
"{:.2f}".format(13.949999999999999)
Note 1: the above returns a string. In order to get as float, simply wrap with float(...):
float("{:.2f}".format(13.949999999999999))
Note 2: wrapping with float() doesn't change anything:
>>> x = 13.949999999999999999
>>> x
13.95
>>> g = float("{:.2f}".format(x))
>>> g
13.95
>>> x == g
True
>>> h = round(x, 2)
>>> h
13.95
>>> x == h
True
The built-in round() works just fine in Python 2.7 or later.
Example:
>>> round(14.22222223, 2)
14.22
Check out the documentation.
Let me give an example in Python 3.6's f-string/template-string format, which I think is beautifully neat:
>>> f'{a:.2f}'
It works well with longer examples too, with operators and not needing parentheses:
>>> print(f'Completed in {time.time() - start:.2f}s')
I feel that the simplest approach is to use the format() function.
For example:
a = 13.949999999999999
format(a, '.2f')
13.95
This produces a float number as a string rounded to two decimal points.
Most numbers cannot be exactly represented in floats. If you want to round the number because that's what your mathematical formula or algorithm requires, then you want to use round. If you just want to restrict the display to a certain precision, then don't even use round and just format it as that string. (If you want to display it with some alternate rounding method, and there are tons, then you need to mix the two approaches.)
>>> "%.2f" % 3.14159
'3.14'
>>> "%.2f" % 13.9499999
'13.95'
And lastly, though perhaps most importantly, if you want exact math then you don't want floats at all. The usual example is dealing with money and to store 'cents' as an integer.
Use
print"{:.2f}".format(a)
instead of
print"{0:.2f}".format(a)
Because the latter may lead to output errors when trying to output multiple variables (see comments).
Try the code below:
>>> a = 0.99334
>>> a = int((a * 100) + 0.5) / 100.0 # Adding 0.5 rounds it up
>>> print a
0.99
TLDR ;)
The rounding problem of input and output has been solved definitively by Python 3.1 and the fix is backported also to Python 2.7.0.
Rounded numbers can be reversibly converted between float and string back and forth:
str -> float() -> repr() -> float() ... or Decimal -> float -> str -> Decimal
>>> 0.3
0.3
>>> float(repr(0.3)) == 0.3
True
A Decimal type is not necessary for storage anymore.
Results of arithmetic operations must be rounded again because rounding errors could accumulate more inaccuracy than that is possible after parsing one number. That is not fixed by the improved repr() algorithm (Python >= 3.1, >= 2.7.0):
>>> 0.1 + 0.2
0.30000000000000004
>>> 0.1, 0.2, 0.3
(0.1, 0.2, 0.3)
The output string function str(float(...)) was rounded to 12 valid digits in Python < 2.7x and < 3.1, to prevent excessive invalid digits similar to unfixed repr() output. That was still insufficientl after subtraction of very similar numbers and it was too much rounded after other operations. Python 2.7 and 3.1 use the same length of str() although the repr() is fixed. Some old versions of Numpy had also excessive invalid digits, even with fixed Python. The current Numpy is fixed. Python versions >= 3.2 have the same results of str() and repr() function and also output of similar functions in Numpy.
Test
import random
from decimal import Decimal
for _ in range(1000000):
x = random.random()
assert x == float(repr(x)) == float(Decimal(repr(x))) # Reversible repr()
assert str(x) == repr(x)
assert len(repr(round(x, 12))) <= 14 # no excessive decimal places.
Documentation
See the Release notes Python 2.7 - Other Language Changes the fourth paragraph:
Conversions between floating-point numbers and strings are now correctly rounded on most platforms. These conversions occur in many different places: str() on floats and complex numbers; the float and complex constructors; numeric formatting; serializing and de-serializing floats and complex numbers using the marshal, pickle and json modules; parsing of float and imaginary literals in Python code; and Decimal-to-float conversion.
Related to this, the repr() of a floating-point number x now returns a result based on the shortest decimal string that’s guaranteed to round back to x under correct rounding (with round-half-to-even rounding mode). Previously it gave a string based on rounding x to 17 decimal digits.
The related issue
More information: The formatting of float before Python 2.7 was similar to the current numpy.float64. Both types use the same 64 bit IEEE 754 double precision with 52 bit mantissa. A big difference is that np.float64.__repr__ is formatted frequently with an excessive decimal number so that no bit can be lost, but no valid IEEE 754 number exists between 13.949999999999999 and 13.950000000000001. The result is not nice and the conversion repr(float(number_as_string)) is not reversible with numpy. On the other hand: float.__repr__ is formatted so that every digit is important; the sequence is without gaps and the conversion is reversible. Simply: If you perhaps have a numpy.float64 number, convert it to normal float in order to be formatted for humans, not for numeric processors, otherwise nothing more is necessary with Python 2.7+.
Use:
float_number = 12.234325335563
round(float_number, 2)
This will return;
12.23
Explanation:
The round function takes two arguments;
The number to be rounded and the number of decimal places to be returned. Here I returned two decimal places.
You can modify the output format:
>>> a = 13.95
>>> a
13.949999999999999
>>> print "%.2f" % a
13.95
With Python < 3 (e.g. 2.6 or 2.7), there are two ways to do so.
# Option one
older_method_string = "%.9f" % numvar
# Option two (note ':' before the '.9f')
newer_method_string = "{:.9f}".format(numvar)
But note that for Python versions above 3 (e.g. 3.2 or 3.3), option two is preferred.
For more information on option two, I suggest this link on string formatting from the Python documentation.
And for more information on option one, this link will suffice and has information on the various flags.
Reference: Convert floating point number to a certain precision, and then copy to string
You can use format operator for rounding the value up to two decimal places in Python:
print(format(14.4499923, '.2f')) // The output is 14.45
As Matt pointed out, Python 3.6 provides f-strings, and they can also use nested parameters:
value = 2.34558
precision = 2
width = 4
print(f'result: {value:{width}.{precision}f}')
which will display result: 2.35
In Python 2.7:
a = 13.949999999999999
output = float("%0.2f"%a)
print output
We multiple options to do that:
Option 1:
x = 1.090675765757
g = float("{:.2f}".format(x))
print(g)
Option 2:
The built-in round() supports Python 2.7 or later.
x = 1.090675765757
g = round(x, 2)
print(g)
The Python tutorial has an appendix called Floating Point Arithmetic: Issues and Limitations. Read it. It explains what is happening and why Python is doing its best. It has even an example that matches yours. Let me quote a bit:
>>> 0.1
0.10000000000000001
you may be tempted to use the round()
function to chop it back to the single
digit you expect. But that makes no
difference:
>>> round(0.1, 1)
0.10000000000000001
The problem is that the binary
floating-point value stored for “0.1”
was already the best possible binary
approximation to 1/10, so trying to
round it again can’t make it better:
it was already as good as it gets.
Another consequence is that since 0.1
is not exactly 1/10, summing ten
values of 0.1 may not yield exactly
1.0, either:
>>> sum = 0.0
>>> for i in range(10):
... sum += 0.1
...
>>> sum
0.99999999999999989
One alternative and solution to your problems would be using the decimal module.
Use combination of Decimal object and round() method.
Python 3.7.3
>>> from decimal import Decimal
>>> d1 = Decimal (13.949999999999999) # define a Decimal
>>> d1
Decimal('13.949999999999999289457264239899814128875732421875')
>>> d2 = round(d1, 2) # round to 2 decimals
>>> d2
Decimal('13.95')
It's doing exactly what you told it to do and is working correctly. Read more about floating point confusion and maybe try decimal objects instead.
from decimal import Decimal
def round_float(v, ndigits=2, rt_str=False):
d = Decimal(v)
v_str = ("{0:.%sf}" % ndigits).format(round(d, ndigits))
if rt_str:
return v_str
return Decimal(v_str)
Results:
Python 3.6.1 (default, Dec 11 2018, 17:41:10)
>>> round_float(3.1415926)
Decimal('3.14')
>>> round_float(3.1445926)
Decimal('3.14')
>>> round_float(3.1455926)
Decimal('3.15')
>>> round_float(3.1455926, rt_str=True)
'3.15'
>>> str(round_float(3.1455926))
'3.15'
The simple solution is here
value = 5.34343
rounded_value = round(value, 2) # 5.34
Use a lambda function like this:
arred = lambda x,n : x*(10**n)//1/(10**n)
This way you could just do:
arred(3.141591657, 2)
and get
3.14
orig_float = 232569 / 16000.0
14.5355625
short_float = float("{:.2f}".format(orig_float))
14.54
For fixing the floating point in type-dynamic languages such as Python and JavaScript, I use this technique
# For example:
a = 70000
b = 0.14
c = a * b
print c # Prints 980.0000000002
# Try to fix
c = int(c * 10000)/100000
print c # Prints 980
You can also use Decimal as following:
from decimal import *
getcontext().prec = 6
Decimal(1) / Decimal(7)
# Results in 6 precision -> Decimal('0.142857')
getcontext().prec = 28
Decimal(1) / Decimal(7)
# Results in 28 precision -> Decimal('0.1428571428571428571428571429')
It's simple like:
use decimal module for fast correctly-rounded decimal floating point arithmetic:
d = Decimal(10000000.0000009)
to achieve rounding:
d.quantize(Decimal('0.01'))
will result with Decimal('10000000.00')
make the above DRY:
def round_decimal(number, exponent='0.01'):
decimal_value = Decimal(number)
return decimal_value.quantize(Decimal(exponent))
or
def round_decimal(number, decimal_places=2):
decimal_value = Decimal(number)
return decimal_value.quantize(Decimal(10) ** -decimal_places)
PS: critique of others: formatting is not rounding.
Here is the simple solution using the format function.
float(format(num, '.2f'))
Note: We are converting numbers to float, because the format method is returning a string.
If you want to handle money, use the Python decimal module:
from decimal import Decimal, ROUND_HALF_UP
# 'amount' can be integer, string, tuple, float, or another Decimal object
def to_money(amount) -> Decimal:
money = Decimal(amount).quantize(Decimal('.00'), rounding=ROUND_HALF_UP)
return money
lambda x, n:int(x*10^n + 0.5)/10^n
has worked for me for many years in many languages.
To round a number to a resolution, the best way is the following one, which can work with any resolution (0.01 for two decimals or even other steps):
>>> import numpy as np
>>> value = 13.949999999999999
>>> resolution = 0.01
>>> newValue = int(np.round(value/resolution))*resolution
>>> print newValue
13.95
>>> resolution = 0.5
>>> newValue = int(np.round(value/resolution))*resolution
>>> print newValue
14.0
The answers I saw didn't work with the float(52.15) case. After some tests, there is the solution that I'm using:
import decimal
def value_to_decimal(value, decimal_places):
decimal.getcontext().rounding = decimal.ROUND_HALF_UP # define rounding method
return decimal.Decimal(str(float(value))).quantize(decimal.Decimal('1e-{}'.format(decimal_places)))
(The conversion of the 'value' to float and then string is very important, that way, 'value' can be of the type float, decimal, integer or string!)
Hope this helps anyone.
This question already has answers here:
Why 0.2 is not equal to 0.2 when using the decimal method?
(2 answers)
Closed 1 year ago.
I'm reading up on the Python Decimal module. I have a need to make a large number of precise calculations, often with lots of decimal places, where being off by a small amount adds up over time. Enter the Decimal library.
Step 1: Read the intro to Decimal library (added bold):
Decimal numbers can be represented exactly. In contrast, numbers like 1.1 and 2.2 do not have 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.
Step 2: Plug a decimal in to Python. This seems to be imprecise - off by a very similar margin as the float calculation.
>>> from decimal import *
>>> 1.1 + 2.2
3.3000000000000003
>>> Decimal(3.3)
Decimal('3.29999999999999982236431605997495353221893310546875')
What's going on?
Per the documentation:
Construction from an integer or a float performs an exact conversion of the value of that integer or float.
The exact value of the float literal 3.3 is not 3.3 = 33/10, but the binary approximation 3715469692580659 / 250, whose exact value is what you see in your screenshot. If this is not what you want, then pass a str instead of a float to the constructor.
>>> from decimal import *
>>> Decimal(3.3)
Decimal('3.29999999999999982236431605997495353221893310546875')
>>> Decimal('3.3')
Decimal('3.3')
Also remember while that Decimal is exact at representing base-ten fractions like 1/10, 1/100, or 1/1000, other fractions are approximated (albeit to more precision than float).
>>> Decimal(1) / Decimal(3)
Decimal('0.3333333333333333333333333333')
>>> _ * 3
Decimal('0.9999999999999999999999999999')
If this is an issue for you, then use the Fraction class instead of Decimal.
>>> from fractions import *
>>> Fraction(1) / Fraction(3)
Fraction(1, 3)
>>> _ * 3
Fraction(1, 1)
This question already has answers here:
Limiting floats to two decimal points
(35 answers)
Closed 2 years ago.
I want to round the number below to two decimal places. The result must be 33.39 but Python gives 33.38 because the 5 rounds to even and hence to 8.
round(33.385, 2)
This actually has nothing to do with round-to-nearest/even. 33.385 is a decimal number, but is represented in your hardware as an approximation in binary floating point. The decimal module can show you the exact decimal value of that binary approximation:
>>> import decimal
>>> decimal.Decimal(33.385)
Decimal('33.38499999999999801048033987171947956085205078125')
That's why it rounds to 33.38: the exact value stored is slightly closer to 33.38 than to 33.39.
If you need exact decimal results, don't use your hardware's binary floating point. For example, you could use the decimal module for this with the ROUND_HALF_UP rounding mode.
For example,
>>> from decimal import Decimal as D
>>> x = D("33.385")
>>> x
Decimal('33.385')
>>> twodigits = D("0.01")
>>> x.quantize(twodigits) # nearest/even is the default
Decimal('33.38')
>>> x.quantize(twodigits, rounding=decimal.ROUND_HALF_EVEN) # same thing
Decimal('33.38')
>>> x.quantize(twodigits, rounding=decimal.ROUND_HALF_UP) # what you want
Decimal('33.39')
>>> float(_) # back to binary float
33.39
>>> D(_) # but what the binary 33.39 _really_ is
Decimal('33.3900000000000005684341886080801486968994140625')
Actually round() rounds to the closest decimal point you specified. There are some incosistencies with floating point numbers in programming languages, because the computer can't create specific enough decimals with the given bits, so the programming language already shows you what you need to see, but not the real number. You can see the real number by using the decimal library. That way you can see why sometimes numbers that end in 5 round either up or down.
If you want to round down then you can use the math module to do it easily.
import math
math.floor(33.385 * 100) / 100
And if you want to round up then you can do the same with math.ceil
import math
math.ceil(33.385 * 100) / 100
Or if you wanted you could still use the round() function, but change it a bit
Round up:
decimal_point = 2
change = 0.3 / 10**decimal_point
round(33.385 + change, decimal_point)
Round down:
decimal_point = 2
change = 0.3 / 10**decimal_point
round(33.385 - change, decimal_point)
Just a result of floating-point approximation by computers.
Computers understand binary and not all floating point values have accurate binary rep.
A famous example:
print(0.1 + 0.2 == 0.3)
False
Wonder why?
The result is 0.30000000000000004 and it's because 0.2 in binary goes as in 0.00110011001100...
Use #Rfroes87's suggestion to convert your numbers into an integer-precision scale and then round off to nearest integer)
You can do the following:
import math
math.ceil(33.385 * 100.0) / 100.0
Reference: Round up to Second Decimal Place in Python
I'm trying to round to two decimal places with error.
How could I round, eg:
7.1450, should be: 7.15
7.144, should be 7.14
>>> round(7.1450,2)
7.14
I will appreciate your help, I have been looking for a long time, without finding the solution.
I just found the answer of this problem in the Python docs.
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.
If you need more accurate method, you can use decimal module.
>>> import decimal
>>> decimal.getcontext().rounding = decimal.ROUND_HALF_UP # change round setting of decimal.
>>> round(decimal.Decimal('7.1450'), 2)
Decimal('7.15')
The decimal module lets you control what rounding mode is used. You're looking for ROUND_HALF_UP.
I'm not too familiar with decimal myself, but this is the simplest way I found to do it:
import decimal
with decimal.localcontext() as ctx:
ctx.rounding = decimal.ROUND_HALF_UP
a = decimal.Decimal('7.145')
print(round(a, 2)) # -> 7.15
You can add a very small value to make it a bit more than .5.
Here I use sys.float_info.min, which is the minimum representable positive normalized float:
import sys
round(4.115 + sys.float_info.min, 2)
How about this
def round2(number, digit=0):
return int(number*10**digit+0.5)/10**digit
Python stores a decimal in base 2 format. For example,
0.5 is stored as (1/2).
0.75 is stored as (1/2) + (1/4).
You observed? All decimals cannot be stored in this way. So, It maintains a large precision. For example, (0.1) cannot be stored exactly in this way. It rounds to nearest possible decimal.
For example, 7.1450 is stored as 7.144999999999999573674358543939888477325439453125. So, Naturally, when round() is applied on this It gives 7.14.
So, How to resolve this?
Add a very low value (Say 0.000000001) to your decimals before rounding it.
It does the job.
try this
import numpy as np
x = 2.78
print(np.round(x,1))
It will give desire output such as
>>> print(np.round(x,1))
2.8
And also try this.
import math
def round_nearest(num, a):
return round(round(num / a) * a, -int(math.floor(math.log10(a))))
round_nearest(1.145, 0.05)
>>> round_nearest(1.145, 0.05)
1.15
This question already has answers here:
Python rounding error with float numbers [duplicate]
(2 answers)
Python weird addition bug [duplicate]
(4 answers)
Closed 9 years ago.
I'm trying to add decimal numbers a decimal number and it works correctly but when I do 1.1 + 0.1 I get 1.2000000000000002 but all I want it to equal to is 1.2. When I do 1.0 + 0.1 I get 1.1 which is perfect but i don't get that for 1.1 + 0.1. So is there a way that I can get rid of the 000000000000002 from 1.2000000000000002?
Thanks.
As has been stated countless times, 0.1 cannot be represented exactly in IEEE 754 floating point. You can read all about why in What Every Computer Scientist Should Know About Floating-Point Arithmetic or The Floating Point Guide
You can trucate or round the value:
>>> round(1.1+.1,2)
1.2
>>> "%.*f" % (1, 1.1+.1 )
'1.2'
>>> s=str(1.1+.1)
>>> s[0:s.find('.')+2]
'1.2'
If you want exact representation of those values, consider using the Decimal module:
>>> import decimal
>>> decimal.Decimal('1.1')+decimal.Decimal('.1')
Decimal('1.2')
Note that you need to start with the string representation of your float, '0.1' since 0.1 is not exactly representable in binary in IEEE floating point:
>>> decimal.Decimal(.1)
Decimal('0.1000000000000000055511151231257827021181583404541015625')
To then get a string representation back after you calculate, you can use str:
>>> str(sum(map(decimal.Decimal,['.1','.1','.5','.5'])))
'1.2'
Another alternative is to use a rational number library such as Fractions:
>>> from fractions import Fraction as Fr
>>> Fr(11,10)+Fr(1,10)
Fraction(6, 5)
With that result, you will still need to round, truncate, or use an arbitrary precision arithmetic package to get an exact number (depending on the inputs...)
You can try string formatting, documentation here.
>>> "%0.2f" % float(1.1 + 0.1)
'1.20'
Or Even:
>>> "%0.1f" % float(1.1 + 0.1)
'1.2'
As to why, it is explicitly described on PEP 327 here.
This is the literal answer to your question:
float(str(1.1 + 0.1)[0:3])
If you're interested in the "why" of the problem then refer to the links provided in the question comments.