I'm doing my own thing just for fun, and I decided to try to make a program find the fraction of a reoccurring decimal. I have tried this program quite a bit and got it to always work until we reach numbers that are greater than 1
whole = number // 1
number -= whole
repeating = ''
final_repeating = ''
copy_of_number = str(number)
stop = False
number = str(number)[2:]
for digit in number:
repeating += digit
number = number[1:]
for checker in range(len(repeating)):
try:
if repeating[checker] == number[0]:
final_repeating = repeating[checker:]
times_it_appears = number.count(final_repeating)
length = len(number)
length_of_final = len(final_repeating)
if times_it_appears == length // length_of_final and (length % length_of_final == 0 or number[(length % length_of_final * -1):] == final_repeating[:(length % length_of_final)]):
stop = True
break
else:
final_repeating = ''
except IndexError:
print('this isn\'t a reoccuring decimal')
quit()
if stop == True:
break
If number = 1.166666 or any other amount of 6's that is >= 2, then it final_repeating should equal '6' and the code should move on
The actual output is that it isn't a reoccurring decimal, and if I check number. after the program finishes, there are a lot of 0's and some random single number at the end a user doesn't type in
When you subtract whole from number, Python approximates the result. You can round the number when subtracting whole from number to avoid this.
Replace the first two lines of your code with this:
copy_of_number=str(number)
whole = int(number // 1)
copy_of_whole=str(whole)
number -= whole
number = round(number, len(copy_of_number)-len(copy_of_whole)-1)
I have made a program that converts "any" given rational number into a fraction using a function that receives a string representing a rational number such as "2.5(37)", "-7.8" or "4" and returns a string with the correspondent fraction.
Eg.: from_string_to_fraction("-2.5(37)") returns "-1256/495". It needs the functions gcd() and number_of_decimal_places_of().
This only works with rational numbers because rational numbers are all the numbers and the only numbers that can be written as the divison of two integers, that's the definition of ratinal numbers. Note that recurring decimals (dízimas infinitas periódicas) are rational numbers, and that 0,(9)=1.
"""
Author: #t3m2.
Date of creation: 09/07/2019, , dd/mm/yyyy, (july).
Version: 14/12/2019, dd/mm/yyyy, (december).
Language: Python.
This program defines three functions which can be used to
converts rational numbers into fractions:
from_string_to_fraction(),
gcd() and number_of_decimal_places_of().
I have built the function from_string_to_fraction()
that receives a string representing a rational number
in decimal notation (including repeating decimals),
such as "2.5(37)", "-7.8" or "4"
and returns a string with the correspondent fraction
in form "n/m" where n and m are integers.
Eg.: from_string_to_fraction("-2.5(37)") returns "-1256/495".
If the number is a repeating decimal,
its peridod should appear between round brackets.
It needs the functions gcd() and number_of_decimal_places_of()!
Read the functions' docstrings for more information.
"""
__author__ = "t3m2"
__date__ = "09/07/2019, , dd/mm/yyyy, (july)"
__version__ = "14/12/2019, dd/mm/yyyy, (december)"
def gcd(a=1, b=1):
"""Returns the greatest common divisor
between two positive integers.
(Recursive solution)
Make sure a and b are positive int's
because this function assumes that."""
if b == 0:
return a
else:
return gcd(b, a%b)
def number_of_decimal_places_of(x=0):
"""Returns the number of decimal places of a float or int.
Make sure x is a float or int
because this function assumes that."""
if x == int(x):
return 0
return len(str(x)[str(x).index('.')+1:])
def from_string_to_fraction(x='0'):
"""Receives a string representing a rational number in decimal notation,
(including repeating decimals) such as "-2.5(37)", "-7.8" or "4", and
returns a string with the correspondent fraction in form "n/m",
where both n and m are integers.
Eg.: from_string_to_fraction("-2.5(37)") returns "-1256/495".
It needs the functions gcd() and number_of_decimal_places_of()!
This only works with rational numbers because rational numbers
are all the numbers and the only numbers that can be written as
the divison of two integers, that's the definition of rational numbers.
Note that recurring decimals are rational numbers, and that 0,(9)=1.
Make sure that: (input restrictions)
- the argument is a valid string representing
a rational number in decimal notation;
- the decimal separator (if there) is a '.' or a ',';
- if the input is a recurring decimal, the period comes between round brackets.
(12.431111111111111... is represented as "12.43(1)".)
Because this function assumes that!"""
# The input string can have a ',' or a '.' separating the int and the decimal part:
x = x.replace(',', '.', 1)
sign = 1
if x[0] == '-':
# iff x represent a negative number, this turns x into
# a string representing positive number,
# because if it easier to work with positives numbers.
# And, in the end, we turn the result into negative again
# by making something like: "final_result*=sign".
sign = -1
x = x[1:]
### Getting the finit part (f) and the period (p): ###
# I will explain this with an example:
# If x == "2.5(37)"; then I set f, the finit part, to 2.5 and p, the period, to 37.
# If the number is non-recurring, f = x, since it has no period.
# Eg: if x == "-3.4"; then f = -3.4 and p = 0.
# Note that x, our argument, is still a 'string'.
try: # This will be executed iff x is a non-recurring decimal:
f = eval(x) # eval(x) "turns" the string x into a float or int.
p = 0
except TypeError: # This will be executed iff x is a recurring decimal:
f = float(x[:x.index('(')]) # The finit part of the dizim is all the way until '('.
p = int(x[x.index('(')+1:x.index(')')]) # The period of the dizim is
# the part of the number between '(' and ')').
### Getting the numerator and denominator: ###
# With f and p defined, I have to discover the numerator and the denominator:
# Here is a method that can be used in order to discover the fraction:
# If y=2,5(37): (mathematical notation)
# 1000y - 10y = 2537,(37) - 25,(37) <=>
# 1000y - 10y = 2537 - 25 <=>
# (1000-10) * y = 2537 - 25 <=>
# <=> y = (2537-25) / (1000-10) <=>
# <=> y = 2512 / 990 =>
# => y = numerator / denominator => # Then we need to simplify the fraction,
# => y = 1256 / 495 # and this will be the final result.
# Note that both numerator and denominator are integers.
# I implemented this with f and p:
numerator = f*10**(number_of_decimal_places_of(f)+len(str(p)))+p \
- f*10**number_of_decimal_places_of(f)
denominator = 10**(number_of_decimal_places_of(f)+len(str(p))) \
- 10**number_of_decimal_places_of(f)
### Simplifying the fraction: ###
# Here I am slimplifying the fraction, if it's possible:
factor = gcd(numerator, denominator)
# "sign*" is used to get the correct sign of the final answer,
# ie, the same sign of x.
numerator = sign*(numerator/factor)
denominator = denominator/factor
return "%d/%d" % (numerator, denominator)
### TESTING ###
print("This program turns \"any\" rational number in decimal notation \
into a fraction, for example: -2.5(37) = -2.537373737373737... = -1256/495\n\n")
while 1:
try:
x = input("Enter a rational number in decimal notation (exit: 'b'): ")
if x == 'b':
break
print("%s = %s" % (x, from_string_to_fraction(x)))
except:
print("Error: probably, invalid input.")
print()
If it doesn't help, I am sorry, but it will probably help others in the future.
This answer took me lots of time, I hope that someone in the future will use this to convert rational numbers in tithe notation to fractions.
The program seems to be a little bit big but this is due to the comments.
Repository of this program in GitHub.
Working on Repl.it.
I want to remove digits from a float to have a fixed number of digits after the dot, like:
1.923328437452 → 1.923
I need to output as a string to another function, not print.
Also I want to ignore the lost digits, not round them.
round(1.923328437452, 3)
See Python's documentation on the standard types. You'll need to scroll down a bit to get to the round function. Essentially the second number says how many decimal places to round it to.
First, the function, for those who just want some copy-and-paste code:
def truncate(f, n):
'''Truncates/pads a float f to n decimal places without rounding'''
s = '{}'.format(f)
if 'e' in s or 'E' in s:
return '{0:.{1}f}'.format(f, n)
i, p, d = s.partition('.')
return '.'.join([i, (d+'0'*n)[:n]])
This is valid in Python 2.7 and 3.1+. For older versions, it's not possible to get the same "intelligent rounding" effect (at least, not without a lot of complicated code), but rounding to 12 decimal places before truncation will work much of the time:
def truncate(f, n):
'''Truncates/pads a float f to n decimal places without rounding'''
s = '%.12f' % f
i, p, d = s.partition('.')
return '.'.join([i, (d+'0'*n)[:n]])
Explanation
The core of the underlying method is to convert the value to a string at full precision and then just chop off everything beyond the desired number of characters. The latter step is easy; it can be done either with string manipulation
i, p, d = s.partition('.')
'.'.join([i, (d+'0'*n)[:n]])
or the decimal module
str(Decimal(s).quantize(Decimal((0, (1,), -n)), rounding=ROUND_DOWN))
The first step, converting to a string, is quite difficult because there are some pairs of floating point literals (i.e. what you write in the source code) which both produce the same binary representation and yet should be truncated differently. For example, consider 0.3 and 0.29999999999999998. If you write 0.3 in a Python program, the compiler encodes it using the IEEE floating-point format into the sequence of bits (assuming a 64-bit float)
0011111111010011001100110011001100110011001100110011001100110011
This is the closest value to 0.3 that can accurately be represented as an IEEE float. But if you write 0.29999999999999998 in a Python program, the compiler translates it into exactly the same value. In one case, you meant it to be truncated (to one digit) as 0.3, whereas in the other case you meant it to be truncated as 0.2, but Python can only give one answer. This is a fundamental limitation of Python, or indeed any programming language without lazy evaluation. The truncation function only has access to the binary value stored in the computer's memory, not the string you actually typed into the source code.1
If you decode the sequence of bits back into a decimal number, again using the IEEE 64-bit floating-point format, you get
0.2999999999999999888977697537484345957637...
so a naive implementation would come up with 0.2 even though that's probably not what you want. For more on floating-point representation error, see the Python tutorial.
It's very rare to be working with a floating-point value that is so close to a round number and yet is intentionally not equal to that round number. So when truncating, it probably makes sense to choose the "nicest" decimal representation out of all that could correspond to the value in memory. Python 2.7 and up (but not 3.0) includes a sophisticated algorithm to do just that, which we can access through the default string formatting operation.
'{}'.format(f)
The only caveat is that this acts like a g format specification, in the sense that it uses exponential notation (1.23e+4) if the number is large or small enough. So the method has to catch this case and handle it differently. There are a few cases where using an f format specification instead causes a problem, such as trying to truncate 3e-10 to 28 digits of precision (it produces 0.0000000002999999999999999980), and I'm not yet sure how best to handle those.
If you actually are working with floats that are very close to round numbers but intentionally not equal to them (like 0.29999999999999998 or 99.959999999999994), this will produce some false positives, i.e. it'll round numbers that you didn't want rounded. In that case the solution is to specify a fixed precision.
'{0:.{1}f}'.format(f, sys.float_info.dig + n + 2)
The number of digits of precision to use here doesn't really matter, it only needs to be large enough to ensure that any rounding performed in the string conversion doesn't "bump up" the value to its nice decimal representation. I think sys.float_info.dig + n + 2 may be enough in all cases, but if not that 2 might have to be increased, and it doesn't hurt to do so.
In earlier versions of Python (up to 2.6, or 3.0), the floating point number formatting was a lot more crude, and would regularly produce things like
>>> 1.1
1.1000000000000001
If this is your situation, if you do want to use "nice" decimal representations for truncation, all you can do (as far as I know) is pick some number of digits, less than the full precision representable by a float, and round the number to that many digits before truncating it. A typical choice is 12,
'%.12f' % f
but you can adjust this to suit the numbers you're using.
1Well... I lied. Technically, you can instruct Python to re-parse its own source code and extract the part corresponding to the first argument you pass to the truncation function. If that argument is a floating-point literal, you can just cut it off a certain number of places after the decimal point and return that. However this strategy doesn't work if the argument is a variable, which makes it fairly useless. The following is presented for entertainment value only:
def trunc_introspect(f, n):
'''Truncates/pads the float f to n decimal places by looking at the caller's source code'''
current_frame = None
caller_frame = None
s = inspect.stack()
try:
current_frame = s[0]
caller_frame = s[1]
gen = tokenize.tokenize(io.BytesIO(caller_frame[4][caller_frame[5]].encode('utf-8')).readline)
for token_type, token_string, _, _, _ in gen:
if token_type == tokenize.NAME and token_string == current_frame[3]:
next(gen) # left parenthesis
token_type, token_string, _, _, _ = next(gen) # float literal
if token_type == tokenize.NUMBER:
try:
cut_point = token_string.index('.') + n + 1
except ValueError: # no decimal in string
return token_string + '.' + '0' * n
else:
if len(token_string) < cut_point:
token_string += '0' * (cut_point - len(token_string))
return token_string[:cut_point]
else:
raise ValueError('Unable to find floating-point literal (this probably means you called {} with a variable)'.format(current_frame[3]))
break
finally:
del s, current_frame, caller_frame
Generalizing this to handle the case where you pass in a variable seems like a lost cause, since you'd have to trace backwards through the program's execution until you find the floating-point literal which gave the variable its value. If there even is one. Most variables will be initialized from user input or mathematical expressions, in which case the binary representation is all there is.
The result of round is a float, so watch out (example is from Python 2.6):
>>> round(1.923328437452, 3)
1.923
>>> round(1.23456, 3)
1.2350000000000001
You will be better off when using a formatted string:
>>> "%.3f" % 1.923328437452
'1.923'
>>> "%.3f" % 1.23456
'1.235'
n = 1.923328437452
str(n)[:4]
At my Python 2.7 prompt:
>>> int(1.923328437452 * 1000)/1000.0
1.923
The truely pythonic way of doing it is
from decimal import *
with localcontext() as ctx:
ctx.rounding = ROUND_DOWN
print Decimal('1.923328437452').quantize(Decimal('0.001'))
or shorter:
from decimal import Decimal as D, ROUND_DOWN
D('1.923328437452').quantize(D('0.001'), rounding=ROUND_DOWN)
Update
Usually the problem is not in truncating floats itself, but in the improper usage of float numbers before rounding.
For example: int(0.7*3*100)/100 == 2.09.
If you are forced to use floats (say, you're accelerating your code with numba), it's better to use cents as "internal representation" of prices: (70*3 == 210) and multiply/divide the inputs/outputs.
Simple python script -
n = 1.923328437452
n = float(int(n * 1000))
n /=1000
def trunc(num, digits):
sp = str(num).split('.')
return '.'.join([sp[0], sp[1][:digits]])
This should work. It should give you the truncation you are looking for.
So many of the answers given for this question are just completely wrong. They either round up floats (rather than truncate) or do not work for all cases.
This is the top Google result when I search for 'Python truncate float', a concept which is really straightforward, and which deserves better answers. I agree with Hatchkins that using the decimal module is the pythonic way of doing this, so I give here a function which I think answers the question correctly, and which works as expected for all cases.
As a side-note, fractional values, in general, cannot be represented exactly by binary floating point variables (see here for a discussion of this), which is why my function returns a string.
from decimal import Decimal, localcontext, ROUND_DOWN
def truncate(number, places):
if not isinstance(places, int):
raise ValueError("Decimal places must be an integer.")
if places < 1:
raise ValueError("Decimal places must be at least 1.")
# If you want to truncate to 0 decimal places, just do int(number).
with localcontext() as context:
context.rounding = ROUND_DOWN
exponent = Decimal(str(10 ** - places))
return Decimal(str(number)).quantize(exponent).to_eng_string()
>>> from math import floor
>>> floor((1.23658945) * 10**4) / 10**4
1.2365
# divide and multiply by 10**number of desired digits
If you fancy some mathemagic, this works for +ve numbers:
>>> v = 1.923328437452
>>> v - v % 1e-3
1.923
I did something like this:
from math import trunc
def truncate(number, decimals=0):
if decimals < 0:
raise ValueError('truncate received an invalid value of decimals ({})'.format(decimals))
elif decimals == 0:
return trunc(number)
else:
factor = float(10**decimals)
return trunc(number*factor)/factor
You can do:
def truncate(f, n):
return math.floor(f * 10 ** n) / 10 ** n
testing:
>>> f=1.923328437452
>>> [truncate(f, n) for n in range(5)]
[1.0, 1.9, 1.92, 1.923, 1.9233]
Just wanted to mention that the old "make round() with floor()" trick of
round(f) = floor(f+0.5)
can be turned around to make floor() from round()
floor(f) = round(f-0.5)
Although both these rules break around negative numbers, so using it is less than ideal:
def trunc(f, n):
if f > 0:
return "%.*f" % (n, (f - 0.5*10**-n))
elif f == 0:
return "%.*f" % (n, f)
elif f < 0:
return "%.*f" % (n, (f + 0.5*10**-n))
def precision(value, precision):
"""
param: value: takes a float
param: precision: int, number of decimal places
returns a float
"""
x = 10.0**precision
num = int(value * x)/ x
return num
precision(1.923328437452, 3)
1.923
Short and easy variant
def truncate_float(value, digits_after_point=2):
pow_10 = 10 ** digits_after_point
return (float(int(value * pow_10))) / pow_10
>>> truncate_float(1.14333, 2)
>>> 1.14
>>> truncate_float(1.14777, 2)
>>> 1.14
>>> truncate_float(1.14777, 4)
>>> 1.1477
When using a pandas df this worked for me
import math
def truncate(number, digits) -> float:
stepper = 10.0 ** digits
return math.trunc(stepper * number) / stepper
df['trunc'] = df['float_val'].apply(lambda x: truncate(x,1))
df['trunc']=df['trunc'].map('{:.1f}'.format)
int(16.5);
this will give an integer value of 16, i.e. trunc, won't be able to specify decimals, but guess you can do that by
import math;
def trunc(invalue, digits):
return int(invalue*math.pow(10,digits))/math.pow(10,digits);
Here is an easy way:
def truncate(num, res=3):
return (floor(num*pow(10, res)+0.5))/pow(10, res)
for num = 1.923328437452, this outputs 1.923
def trunc(f,n):
return ('%.16f' % f)[:(n-16)]
A general and simple function to use:
def truncate_float(number, length):
"""Truncate float numbers, up to the number specified
in length that must be an integer"""
number = number * pow(10, length)
number = int(number)
number = float(number)
number /= pow(10, length)
return number
There is an easy workaround in python 3. Where to cut I defined with an help variable decPlace to make it easy to adapt.
f = 1.12345
decPlace= 4
f_cut = int(f * 10**decPlace) /10**decPlace
Output:
f = 1.1234
Hope it helps.
Most answers are way too complicated in my opinion, how about this?
digits = 2 # Specify how many digits you want
fnum = '122.485221'
truncated_float = float(fnum[:fnum.find('.') + digits + 1])
>>> 122.48
Simply scanning for the index of '.' and truncate as desired (no rounding).
Convert string to float as final step.
Or in your case if you get a float as input and want a string as output:
fnum = str(122.485221) # convert float to string first
truncated_float = fnum[:fnum.find('.') + digits + 1] # string output
I think a better version would be just to find the index of decimal point . and then to take the string slice accordingly:
def truncate(number, n_digits:int=1)->float:
'''
:param number: real number ℝ
:param n_digits: Maximum number of digits after the decimal point after truncation
:return: truncated floating point number with at least one digit after decimal point
'''
decimalIndex = str(number).find('.')
if decimalIndex == -1:
return float(number)
else:
return float(str(number)[:decimalIndex+n_digits+1])
int(1.923328437452 * 1000) / 1000
>>> 1.923
int(1.9239 * 1000) / 1000
>>> 1.923
By multiplying the number by 1000 (10 ^ 3 for 3 digits) we shift the decimal point 3 places to the right and get 1923.3284374520001. When we convert that to an int the fractional part 3284374520001 will be discarded. Then we undo the shifting of the decimal point again by dividing by 1000 which returns 1.923.
use numpy.round
import numpy as np
precision = 3
floats = [1.123123123, 2.321321321321]
new_float = np.round(floats, precision)
Something simple enough to fit in a list-comprehension, with no libraries or other external dependencies. For Python >=3.6, it's very simple to write with f-strings.
The idea is to let the string-conversion do the rounding to one more place than you need and then chop off the last digit.
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nout+1}f}'[:-1] for x in [2/3, 4/5, 8/9, 9/8, 5/4, 3/2]]
['0.666', '0.800', '0.888', '1.125', '1.250', '1.500']
Of course, there is rounding happening here (namely for the fourth digit), but rounding at some point is unvoidable. In case the transition between truncation and rounding is relevant, here's a slightly better example:
>>> nacc = 6 # desired accuracy (maximum 15!)
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nacc}f}'[:-(nacc-nout)] for x in [2.9999, 2.99999, 2.999999, 2.9999999]]
>>> ['2.999', '2.999', '2.999', '3.000']
Bonus: removing zeros on the right
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nout+1}f}'[:-1].rstrip('0') for x in [2/3, 4/5, 8/9, 9/8, 5/4, 3/2]]
['0.666', '0.8', '0.888', '1.125', '1.25', '1.5']
The core idea given here seems to me to be the best approach for this problem.
Unfortunately, it has received less votes while the later answer that has more votes is not complete (as observed in the comments). Hopefully, the implementation below provides a short and complete solution for truncation.
def trunc(num, digits):
l = str(float(num)).split('.')
digits = min(len(l[1]), digits)
return l[0] + '.' + l[1][:digits]
which should take care of all corner cases found here and here.
Am also a python newbie and after making use of some bits and pieces here, I offer my two cents
print str(int(time.time()))+str(datetime.now().microsecond)[:3]
str(int(time.time())) will take the time epoch as int and convert it to string and join with...
str(datetime.now().microsecond)[:3] which returns the microseconds only, convert to string and truncate to first 3 chars
# value value to be truncated
# n number of values after decimal
value = 0.999782
n = 3
float(int(value*1en))*1e-n
When you convert a float to Decimal, the Decimal will contain as accurate a representation of the binary number that it can. It's nice to be accurate, but it isn't always what you want. Since many decimal numbers can't be represented exactly in binary, the resulting Decimal will be a little off - sometimes a little high, sometimes a little low.
>>> from decimal import Decimal
>>> for f in (0.1, 0.3, 1e25, 1e28, 1.0000000000001):
print Decimal(f)
0.1000000000000000055511151231257827021181583404541015625
0.299999999999999988897769753748434595763683319091796875
10000000000000000905969664
9999999999999999583119736832
1.000000000000099920072216264088638126850128173828125
Ideally we'd like the Decimal to be rounded to the most likely decimal equivalent.
I tried converting to str since a Decimal created from a string will be exact. Unfortunately str rounds a little too much.
>>> for f in (0.1, 0.3, 1e25, 1e28, 1.0000000000001):
print Decimal(str(f))
0.1
0.3
1E+25
1E+28
1.0
Is there a way of getting a nicely rounded Decimal from a float?
It turns out that repr does a better job of converting a float to a string than str does. It's the quick-and-easy way to do the conversion.
>>> for f in (0.1, 0.3, 1e25, 1e28, 1.0000000000001):
print Decimal(repr(f))
0.1
0.3
1E+25
1E+28
1.0000000000001
Before I discovered that, I came up with a brute-force way of doing the rounding. It has the advantage of recognizing that large numbers are accurate to 15 digits - the repr method above only recognizes one significant digit for the 1e25 and 1e28 examples.
from decimal import Decimal,DecimalTuple
def _increment(digits, exponent):
new_digits = [0] + list(digits)
new_digits[-1] += 1
for i in range(len(new_digits)-1, 0, -1):
if new_digits[i] > 9:
new_digits[i] -= 10
new_digits[i-1] += 1
if new_digits[0]:
return tuple(new_digits[:-1]), exponent + 1
return tuple(new_digits[1:]), exponent
def nearest_decimal(f):
sign, digits, exponent = Decimal(f).as_tuple()
if len(digits) > 15:
round_up = digits[15] >= 5
exponent += len(digits) - 15
digits = digits[:15]
if round_up:
digits, exponent = _increment(digits, exponent)
while digits and digits[-1] == 0 and exponent < 0:
digits = digits[:-1]
exponent += 1
return Decimal(DecimalTuple(sign, digits, exponent))
>>> for f in (0.1, 0.3, 1e25, 1e28, 1.0000000000001):
print nearest_decimal(f)
0.1
0.3
1.00000000000000E+25
1.00000000000000E+28
1.0000000000001
Edit: I discovered one more reason to use the brute-force rounding. repr tries to return a string that uniquely identifies the underlying float bit representation, but it doesn't necessarily ensure the accuracy of the last digit. By using one less digit, my rounding function will more often be the number you would expect.
>>> print Decimal(repr(2.0/3.0))
0.6666666666666666
>>> print dec.nearest_decimal(2.0/3.0)
0.666666666666667
The decimal created with repr is actually more accurate, but it implies a level of precision that doesn't exist. The nearest_decimal function delivers a better match between precision and accuracy.
I have implemented this in Pharo Smalltalk, in a Float method named asMinimalDecimalFraction.
It is exactly the same problem as printing the shortest decimal fraction that would be re-interpreted as the same float/double, assuming correct rounding (to nearest).
See my answer at Count number of digits after `.` in floating point numbers? for more references