Python error in basic subtraction? [duplicate]

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Closed 12 years ago.

Possible Duplicate:
Python rounding error with float numbers
python maths is wrong

I can't get Python to correctly do the subtraction 1 - 0.8 and assign it. It keeps coming up with the incorrect answer, 0.19999999999999996.

I explored a bit:

sq = {}
sub = {}
for i in range(1000):
 sq[str(i/1000.)+'**2']=((i/1000.)**2)
 sub['1-'+str(i/1000.)]=(1.0-(i/1000.))

and discovered that this error happens with a somewhat random group of the floats between 0 and 1 to the third decimal place. A similar error also occurs when you square those floats, but to a different subset.

I'm hoping for an explanation of this and how to make Python do the arithmetic right. Using round(x,3) is the work-around I'm using for now, but it's not elegant.

Thanks!

This is a session in my Python 2.7.3 shell:

*** Python 2.7.3 (default, Apr 10 2012, 23:24:47) [MSC v.1500 64 bit (AMD64)] on win32. ***
*** Remote Python engine is active ***
>>> 1-0.8
0.19999999999999996
>>> print 1-0.8
0.2
>>> a = 1-0.8
>>> a
0.19999999999999996
>>> print a
0.2
>>> a = 0.2
>>> print a
0.2
>>> a
0.2
>>> 

Here's the code I put into a couple online interpreters:

def doit():
 d = {'a':1-0.8}
 return d
print doit()

and the output:

{'a': 0.19999999999999996}

• avasal: check yoda.arachsys.com/csharp/floatingpoint.html
• Paul Hankin: effbot.org/pyfaq/…

Answer 1

Use Decimal its designed just for this:

>>> from decimal import Decimal, getcontext
>>> Decimal(1) - Decimal(0.8)
Decimal('0.1999999999999999555910790150')
>>> getcontext().prec = 3
>>> Decimal(1) - Decimal(0.8)
Decimal('0.200')
>>> float(Decimal(1) - Decimal(0.8))
0.2

• wjandrea: Why are you inputting floats? The first subtraction has the same problem as OP because of that. Use strings instead: Decimal(1) - Decimal('0.8') -> Decimal('0.2')
• Wardy: Thank you! I didn't notice this in the doco. Makes perfect sense and solved my problem :)
• alper: This does not work for: float(Decimal(0.061157) - Decimal(0.060782)), which generates 0.00037500000000000033

Answer 2

Floating numbers don't work as you're expecting them to.

For starters, read the floating point guide. Long story short: computers represent floating point numbers as binary, and it turns out that storing a precise decimal fraction as binary is not possible (try it for yourself on paper to see why). For practical purposes, 0.19999999999999996 is "close enough" to 0.2. If you wanted to print it as 0.2, then you could do something like:

print "%0.1f" % floating_point_value

So what you're seeing isn't an error. It's expected behavior.

• maclian: Do you have any suggestions for how to use this work around when using jq? I am using jq for a simple math function and it's giving me these similar outputs for easy problems.

Answer 3

Python stores floats with 'bits', and some floats you just can't represent accurately, no matter how many bits of precision you have. This is the problem you have here. It's sorta like trying to write 1/3 in decimal with a limited amount of decimals places perfectly accurately.