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How to properly round-up half float numbers?
(21 answers)
Closed 7 months ago.
In Python 3, I'm trying to round the value 4800.5, so I was expecting it to 4801
but it's giving me 4800. I'm not able to track why this is happening.
Any help will be appreciated.
That's by design.
If you have a look at round function documentation (https://docs.python.org/3/library/functions.html#round) you will find that:
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).
In simple words, 0.5 is a special case which is always rounded toward an even number.
But there're more interesting things. Please have a look at that example:
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.
What you might want to do is to use Decimal for more conventional rounding logic: https://docs.python.org/3/library/decimal.html
For example:
>>> Decimal('7.325').quantize(Decimal('.01'), rounding=ROUND_DOWN)
Decimal('7.32')
>>> Decimal('7.325').quantize(Decimal('.01'), rounding=ROUND_UP)
Decimal('7.33')
There are lot of ways to round a number. round() behaves according to a particular rounding strategy which may or may not be the one you need for a given situation (see the first comment to your question).
If you want to round your number to the upper, you can try this:
import math
n = 4800.5
print(math.ceil(n))
you can do something like this:
from decimal import Decimal, ROUND_HALF_UP
def round_half_up(decimal_number, places=0):
if places == 0:
exp = Decimal('1')
else:
exp_str = '0' * places
exp_str = exp_str[:-1] + '1'
exp = Decimal('.{}'.format(exp_str))
return Decimal(decimal_number).quantize(exp, rounding=ROUND_HALF_UP)
print(round_half_up(4800.5)) -> 4801
print(round_half_up(4800.555, 2)) -> 4800.56
Round() function will round up to next value, if decimal is >.5
upto .5 it would round up to just the integer part.
Related
When working with the built in decimal module in python I can round decimals as follows.
Decimal(50.212345).quantize(Decimal('0.01'))
> Decimal('50.21')
But I can also round the same number with the built in round function
round(Decimal(50.212345), 2)
> Decimal('50.21')
Why would I use one instead of the other when rounding Decimals? In previous answers about rounding decimals, users suggested to use quantize because the built in round function would return a value of type float. Based on my testing, these both return a Decimal. Other than syntax, is there a reason to choose one over the other?
The return types aren't always the same. round() used with a single argument actually returns an int:
>>> round(5.3)
5
>>> round(decimal.Decimal("5.3"))
5
Other than that, suit yourself. quantize() is especially handy if you want a deoimal rounded to "the same" precision as another decimal you already have.
>>> x = decimal.Decimal("123.456")
>>> x*x
Decimal('15241.383936')
>>> (x*x).quantize(x)
Decimal('15241.384')
See? The code doing this doesn't have to know that x originally had 3 digits after the decimal point. Just passing x to quantize() forces the function to round back to the same precision as the original x, regardless of what that may be.
quantize() is also necessary if you want to use a rounding mode other than the default nearest/even.
>>> (x*x).quantize(x, decimal.ROUND_FLOOR)
Decimal('15241.383')
I know that questions about rounding in python have been asked multiple times already, but the answers did not help me. I'm looking for a method that is rounding a float number half up and returns a float number. The method should also accept a parameter that defines the decimal place to round to. I wrote a method that implements this kind of rounding. However, I think it does not look elegant at all.
def round_half_up(number, dec_places):
s = str(number)
d = decimal.Decimal(s).quantize(
decimal.Decimal(10) ** -dec_places,
rounding=decimal.ROUND_HALF_UP)
return float(d)
I don't like it, that I have to convert float to a string (to avoid floating point inaccuracy) and then work with the decimal module.
Do you have any better solutions?
Edit: As pointed out in the answers below, the solution to my problem is not that obvious as correct rounding requires correct representation of numbers in the first place and this is not the case with float. So I would expect that the following code
def round_half_up(number, dec_places):
d = decimal.Decimal(number).quantize(
decimal.Decimal(10) ** -dec_places,
rounding=decimal.ROUND_HALF_UP)
return float(d)
(that differs from the code above just by the fact that the float number is directly converted into a decimal number and not to a string first) to return 2.18 when used like this: round_half_up(2.175, 2) But it doesn't because Decimal(2.175) will return Decimal('2.17499999999999982236431605997495353221893310546875'), the way the float number is represented by the computer.
Suprisingly, the first code returns 2.18 because the float number is converted to string first. It seems that the str() function conducts an implicit rounding to the number that was initially meant to be rounded. So there are two roundings taking place. Even though this is the result that I would expect, it is technically wrong.
Rounding is surprisingly hard to do right, because you have to handle floating-point calculations very carefully. If you are looking for an elegant solution (short, easy to understand), what you have like like a good starting point. To be correct, you should replace decimal.Decimal(str(number)) with creating the decimal from the number itself, which will give you a decimal version of its exact representation:
d = Decimal(number).quantize(...)...
Decimal(str(number)) effectively rounds twice, as formatting the float into the string representation performs its own rounding. This is because str(float value) won't try to print the full decimal representation of the float, it will only print enough digits to ensure that you get the same float back if you pass those exact digits to the float constructor.
If you want to retain correct rounding, but avoid depending on the big and complex decimal module, you can certainly do it, but you'll still need some way to implement the exact arithmetics needed for correct rounding. For example, you can use fractions:
import fractions, math
def round_half_up(number, dec_places=0):
sign = math.copysign(1, number)
number_exact = abs(fractions.Fraction(number))
shifted = number_exact * 10**dec_places
shifted_trunc = int(shifted)
if shifted - shifted_trunc >= fractions.Fraction(1, 2):
result = (shifted_trunc + 1) / 10**dec_places
else:
result = shifted_trunc / 10**dec_places
return sign * float(result)
assert round_half_up(1.49) == 1
assert round_half_up(1.5) == 2
assert round_half_up(1.51) == 2
assert round_half_up(2.49) == 2
assert round_half_up(2.5) == 3
assert round_half_up(2.51) == 3
Note that the only tricky part in the above code is the precise conversion of a floating-point to a fraction, and that can be off-loaded to the as_integer_ratio() float method, which is what both decimals and fractions do internally. So if you really want to remove the dependency on fractions, you can reduce the fractional arithmetic to pure integer arithmetic; you stay within the same line count at the expense of some legibility:
def round_half_up(number, dec_places=0):
sign = math.copysign(1, number)
exact = abs(number).as_integer_ratio()
shifted = (exact[0] * 10**dec_places), exact[1]
shifted_trunc = shifted[0] // shifted[1]
difference = (shifted[0] - shifted_trunc * shifted[1]), shifted[1]
if difference[0] * 2 >= difference[1]: # difference >= 1/2
shifted_trunc += 1
return sign * (shifted_trunc / 10**dec_places)
Note that testing these functions brings to spotlight the approximations performed when creating floating-point numbers. For example, print(round_half_up(2.175, 2)) prints 2.17 because the decimal number 2.175 cannot be represented exactly in binary, so it is replaced by an approximation that happens to be slightly smaller than the 2.175 decimal. The function receives that value, finds it smaller than the actual fraction corresponding to the 2.175 decimal, and decides to round it down. This is not a quirk of the implementation; the behavior derives from properties of floating-point numbers and is also present in the round built-in of Python 3 and 2.
I don't like it, that I have to convert float to a string (to avoid
floating point inaccuracy) and then work with the decimal module. Do
you have any better solutions?
Yes; use Decimal to represent your numbers throughout your whole program, if you need to represent numbers such as 2.675 exactly and have them round to 2.68 instead of 2.67.
There is no other way. The floating point number which is shown on your screen as 2.675 is not the real number 2.675; in fact, it is very slightly less than 2.675, which is why it gets rounded down to 2.67:
>>> 2.675 - 2
0.6749999999999998
It only shows in string form as '2.675' because that happens to be the shortest string such that float(s) == 2.6749999999999998. Note that this longer representation (with lots of 9s) isn't exact either.
However you write your rounding function, it is not possible for my_round(2.675, 2) to round up to 2.68 and also for my_round(2 + 0.6749999999999998, 2) to round down to 2.67; because the inputs are actually the same floating point number.
So if your number 2.675 ever gets converted to a float and back again, you have already lost the information about whether it should round up or down. The solution is not to make it float in the first place.
After trying for a very long time to produce an elegant one-line function, I ended up getting something that is comparable to a dictionary in size.
I would say the simplest way to do this is just to
def round_half_up(inp,dec_places):
return round(inp+0.0000001,dec_places)
i would acknowledge that this is not accurate in every cases, but should work if you just want a simple quick workaround.
I have to translate euro's (in a string) to euro cents (int):
Examples:
'12,1' => 1210
'14,51' => 1451
I use this python function:
int(round(float(amount.replace(',', '.')), 2) * 100)
But with this amount '1229,84' the result is : 122983
Update
I use the solution from Wim, bacause I use integers in both Python / Jinja and javascript for currency artitmetic. See also the answer from Chepner.
int(round(100 * float(amout.replace(',', '.')), 2))
My questions was anwered by Mr. Me, who explained the above result.
What the Docs Say, and a simple explanation
I tried it out, and was surprised that this was happening. So I turned to the documentation, and there is a little note in there that says.
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.
Now what does that mean, most decimal fractions can't be represented as a float. Well the documentations follows up with a great link at explains this, but since you probably didn't come here to read a nerdy technical document, let me summarize what is going on.
Python uses the IEEE-754 floating point standard to represent floats. This standard compromises accuracy for speed. Some numbers cannot be accurately represented. For example .1 is actually represented as 0.1000000000000000055511151231257827021181583404541015625. Interestingly, .1 in binary is actually an infinitely repeating number, just like 1/3 is an infinitely repeating .333333.
An Under the Hood Case Study
Now on to your particular case. This was pretty fun to look into, and this is what I discovered.
first lets simplify what you where trying to do
>>> amount = '1229,84'
>>> int(round(float(amount.replace(',', '.')), 2) * 100)
>>> 122983
to
>>>int(1229.84 * 100)
>>> 122983
Sometimes Python1 is unable to 100% accurately display binary floating point numbers, for the same reason we are unable to display the fraction 1/3 as a decimal. When this happens Python hides any extra digits. .1 is actually stored as -0.100000000000000092, but Python will display it as .1 if you type it into the console. We can see those extra digits by doing int(1.1) - 1.13. we can apply this int(myNum) - myNum formula to most floating point numbers to see the extra hidden digits behind them.4. In your case we would do the following.
>>> int(1229.84) - 1229.84
-0.8399999999999181
1229.84 is actually 1229.8399999999999181. Continuing on.5
>>> 1229.84, 2) * 100
122983.99999999999 #there's all of our hidden digits showing up.
Now on to the last step. This is the part we are concerned about. Changing it back to an integer.
>>> int(122983.99999999999)
122983
It rounds downwards instead of upwards, however, if we never had multiplied it by 100, we would still have 2 more 9s at the end, and Python would round up.
>>> int(122983.9999999999999)
122984
??? Now what is going on. Why is Python rounding 122983.99999999999 down, but it rounds 122983.9999999999999 up? Well whenever Python turns a float into a integer it rounds down. However, you have to remember that to Python 122983.9999999999999 with the extra two 99s at the end is the same thing as 122984.0 For example.
>>> 122983.9999999999999
122984.0
>>> a = 122983.9999999999999
>>> int(a) - a
0.0
and without the two extra 99s on the end.
>>> 122983.99999999999
122983.99999999999
>>> a=122983.99999999999
>>> int(a) - a
-0.9999999999854481
Python is definitely treating 122983.9999999999999 as 122984.0 but not 122983.99999999999. Now back to casting 122983.99999999999 to an integer. Because we have created ourselves a decimal portion that is less than 122984 that Python sees as being a seperate number from 122984, and because casting to an integer always causes Python to round down, we get 122983 as a result.
Whew. That was a lot to go through, but I sure learned a lot writing this out, and I hope you did to. The solution to all of this is to use decimal numbers instead of floats which compromises speed for accuracy.
What about rounding? The original problem had some rounding in it as well -- it's useless. See appendix item 6.
The Solution
a) The easiest solution is to use the decimal module instead of floating point numbers. This is the preferred way of doing things in any finance or accounting program.
The documentation also mentioned the following solutions which I've summarized.
b) The exact value can be expressed and retrieved in a hexadecimal form via myFloat.hex() and float.fromhex(myHex)
c) The exact value can also be retrieved as a fraction through myFloat.as_integer_ratio()
d) The documentation briefly mentions using SciPy for floating point arithmitic, however this SO question mentions that SciPy's NumPy floats are nothing more than aliases to the built-in float type. The decimal module would be a better solution.
Appendix
1 - Even though I will often refer to Python's behavior, the things I talk about are part of the IEEE-754 floating point standard which is what the major programming languages use for their floating point numbers.
2 - int(1.1) - 1.1 gives me -0.10000000000000009, but according to the documentation .1 is really 0.1000000000000000055511151231257827021181583404541015625
3 - We used int(1.1) - 1.1 instead of int(.1) - .1 because int(.1) - .1 does not give us the hidden digits, but according to the documentation they should still be there for .1, hence I say int(someNum) -someNum works most of the time, but not all of the time.
4 - When we use the formula int(myNum) - myNum what is happening is that casting the number to an integer will round the number down so int(3.9) becomes 3, and when we minus 3 from 3.9 we are left with -.9. However, for some reason that I do not know, when we get rid of all the whole numbers, and we're just left with the decimal portion, Python decides to show us everything -- the whole mantissa.
5 - this does not really affect the outcome of our analysis, but when multiplying by 100, instead of the hidden digits being shifted over by 2 decimal places, they changed a little as well.
>>> a = 1229.84
>>> int(a) - a
-0.8399999999999181
>>> a = round(1229.84, 2) * 100
>>> int(a) - a
-0.9999999999854481 #I expected -0.9999999999918100?
6 - It may seem like we can get rid of all those extra digits by rounding to two decimal places.
>>> round(1229.84, 2) # which is really round(1229.8399999999999181, 2)
1229.84
But when we use our int(someNum) - someNum formula to see the hidden digits, they are still there.
>>> a = round(1229.84, 2)
>>> int(a) - a
-0.8399999999999181
This is because Python cannot store 1229.84 as a binary floating point number. It can't be done. So... rounding 1229.84 does absolutely nothing.
Don't use floating-point arithmetic for currency; rounding error for values that cannot be represented exactly will cause the type of loss you are seeing. Instead, convert the string representation to an integer number of cents, which you can convert to euros-and-cents for display as needed.
euros, cents = '12,1'.split(',') # '12,1' -> ('12', '1')
cents = 100*int(euros) + int(cents * 10 if len(cents) == 1 else 1) # ('12', '1') -> 1210
(Notice you'll need a check to handle cents without a trailing 0.)
display_str = '%d,%d' % divMod(cents, 100) # 1210 -> (12, 10) -> '12.10'
You can also use the Decimal class from the decimal module, which essentially encapsulates all the logic for using integers to represent fractional values.
As #wim mentions in a comment, use the Decimal type from the stdlib decimal module instead of the built in float type. Decimal objects do not have the binary rounding behavior that floats have and also have a precision that can be user defined.
Decimal should be used anywhere you are doing financial calculations or anywhere you need floating point calculations that behave like the decimal math people learn in school (as opposed to the binary floating point behavior of the built in float type).
How to check if a float value is within a range (0.50,150.00) and has 2 decimal digits?
For example, 15.22366 should be false (too many decimal digits). But 15.22 should be true.
I tried something like:
data= input()
if data in range(0.50,150.00):
return True
Is that you are looking for?
def check(value):
if 0.50 <= value <= 150 and round(value,2)==value:
return True
return False
Given your comment:
i input 15.22366 it is going to return true; that is why i specified the range; it should accept 15.22
Simply said, floating point values are imprecise. Many values don't have a precise representation. Say for example 1.40. It might be displayed "as it":
>>> f = 1.40
>>> print f
1.4
But this is an illusion. Python has rounded that value in order to nicely display it. The real value as referenced by the variable f is quite different:
>>> from decimal import Decimal
>>> Decimal(f)
Decimal('1.399999999999999911182158029987476766109466552734375')
According to your rule of having only 2 decimals, should f reference a valid value or not?
The easiest way to fix that issue is probably to use round(...,2) as I suggested in the code above. But this in only an heuristic -- only able to reject "largely wrong" values. See my point here:
>>> for v in [ 1.40,
... 1.405,
... 1.399999999999999911182158029987476766109466552734375,
... 1.39999999999999991118,
... 1.3999999999999991118]:
... print check(v), v
...
True 1.4
False 1.405
True 1.4
True 1.4
False 1.4
Notice how the last few results might seems surprising at first. I hope my above explanations put some light on this.
As a final advice, for your needs as I guess them from your question, you should definitively consider using "decimal arithmetic". Python provides the decimal module for that purpose.
float is the wrong data type to use for your case, Use Decimal instead.
Check python docs for issues and limitations. To quote from there (I've generalised the text in Italics)
Floating-point numbers are represented in computer hardware as base 2 (binary) fractions.
no matter how many base 2 digits you’re willing to use, some decimal value (like 0.1) cannot be represented exactly as a base 2 fraction.
Stop at any finite number of bits, and you get an approximation
On a typical machine running Python, there are 53 bits of precision available for a Python float, so the value stored internally when you enter a decimal number is the binary fraction which is close to, but not exactly equal to it.
The documentation for the built-in round() function says that it rounds to the nearest value, rounding ties away from zero.
And finally, it recommends
If you’re in a situation where you care which way your decimal halfway-cases are rounded, you should consider using the decimal module.
And this will hold for your case as well, as you are looking for a precision of 2 digits after decimal points, which float just can't guarantee.
EDIT Note: The answer below corresponds to original question related to random float generation
Seeing that you need 2 digits of sure shot precision, I would suggest generating integer random numbers in range [50, 15000] and dividing them by 100 to convert them to float yourself.
import random
random.randint(50, 15000)/100.0
Why don't you just use round?
round(random.uniform(0.5, 150.0), 2)
Probably what you want to do is not to change the value itself. As said by Cyber in the comment, even if your round a floating point number, it will always store the same precision. If you need to change the way it is printed:
n = random.uniform(0.5, 150)
print '%.2f' % n # 58.03
The easiest way is to first convert the decimal to string and split with '.' and check if the length of the character. If it is >2 then pass on. i.e. Convert use input number to check if it is in a given range.
a=15.22366
if len(str(a).split('.')[1])>2:
if 0.50 <= value <= 150:
<do your stuff>>
Sorry, but I really don't know what's the meaning of the defination of round in python 3.3.2 doc:
round(number[, ndigits])
Return the floating point value number rounded to ndigits digits after the decimal point. If ndigits is omitted, it defaults to zero. Delegates to number.__round__(ndigits).
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). The return value is an integer if called with one argument, otherwise of the same type as number.
I don't know how come the multiple of 10 and pow.
After reading the following examples, I think round(number,n) works like:
if let number be 123.456, let n be 2
round will get two number:123.45 and 123.46
round compares abs(number-123.45) (0.006) and abs(number-123.46) (0.004),and chooses the smaller one.
so, 123.46 is the result.
and if let number be 123.455, let n be 2:
round will get two number:123.45 and 123.46
round compares abs(number-123.45) (0.005) and abs(number-123.46) (0.005). They are equal. So round checks the last digit of 123.45 and 123.46. The even one is the result.
so, the result is 123.46
Am I right?
If not, could you offer a understandable version of values are rounded to the closest multiple of 10 to the power minus ndigits?
ndigits = 0 => pow(10, -ndigits) = 10^(-ndigits) = 1
ndigits = 1 => pow(10, -ndigits) = 10^(-ndigits) = 0.1
etc.
>>> for ndigits in range(6):
... print round(123.456789, ndigits) / pow(10, -ndigits)
123.0
1235.0
12346.0
123457.0
1234568.0
12345679.0
Basically, the number you get is always an integer multiple of 10^(-ndigits). For ndigits=0, that means the number you get is itself an integer, for ndigits=1 it means it won't have more than one non-zero value after the decimal point.
It helps to know that anything to the power of 0 equals 1. As ndigits increases, the function:
f(ndigits) = 10-ndigits gets smaller as you increase ndigits. Specifically as you increase ndigits by 1, you simply shift the decimal place of precision one left. e.g. 10^-0 = 1, 10^-1 = .1 and 10^-2 = .01. The place where the 1 is in the answer is the last point of precision for round.
For the part where it says
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).
This has unexpected behavior in Python 3 and it will not work for all floats. Consider the example you gave, round(123.455, 2) yields the value 123.45. This is not expected behavior because the closest even multiple of 10^-2 is 123.46, not 123.45!
To understand this, you have to pay special attention to the note below this:
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.
And that is why certain floats will round to the "wrong value" and there is really no easy workaround as far as I am aware. (sadface) You could use fractions (i.e. two variables representing the numerator and the denominator) to represent floats in a custom round function if you want to get different behavior than the unpredictable behavior for floats.