Round float to x decimals?

Question

Is there a way to round a python float to x decimals? For example:

>>> x = roundfloat(66.66666666666, 4)
66.6667
>>> x = roundfloat(1.29578293, 6)
1.295783

I've found ways to trim/truncate them (66.666666666 --> 66.6666), but not round (66.666666666 --> 66.6667).

Solution 1

Use the built-in function round():

In [23]: round(66.66666666666,4)
Out[23]: 66.6667

In [24]: round(1.29578293,6)
Out[24]: 1.295783

help on round():

round(number[, ndigits]) -> floating point number

Round a number to a given precision in decimal digits (default 0 digits). This always returns a floating point number. Precision may be negative.

Solution 2

I feel compelled to provide a counterpoint to Ashwini Chaudhary's answer. Despite appearances, the two-argument form of the round function does not round a Python float to a given number of decimal places, and it's often not the solution you want, even when you think it is. Let me explain...

The ability to round a (Python) float to some number of decimal places is something that's frequently requested, but turns out to be rarely what's actually needed. The beguilingly simple answer round(x, number_of_places) is something of an attractive nuisance: it looks as though it does what you want, but thanks to the fact that Python floats are stored internally in binary, it's doing something rather subtler. Consider the following example:

>>> round(52.15, 1)
52.1

With a naive understanding of what round does, this looks wrong: surely it should be rounding up to 52.2 rather than down to 52.1? To understand why such behaviours can't be relied upon, you need to appreciate that while this looks like a simple decimal-to-decimal operation, it's far from simple.

So here's what's really happening in the example above. (deep breath) We're displaying a decimal representation of the nearest binary floating-point number to the nearest n-digits-after-the-point decimal number to a binary floating-point approximation of a numeric literal written in decimal. So to get from the original numeric literal to the displayed output, the underlying machinery has made four separate conversions between binary and decimal formats, two in each direction. Breaking it down (and with the usual disclaimers about assuming IEEE 754 binary64 format, round-ties-to-even rounding, and IEEE 754 rules):

1. First the numeric literal 52.15 gets parsed and converted to a Python float. The actual number stored is 7339460017730355 * 2**-47, or 52.14999999999999857891452847979962825775146484375.

2. Internally as the first step of the round operation, Python computes the closest 1-digit-after-the-point decimal string to the stored number. Since that stored number is a touch under the original value of 52.15, we end up rounding down and getting a string 52.1. This explains why we're getting 52.1 as the final output instead of 52.2.

3. Then in the second step of the round operation, Python turns that string back into a float, getting the closest binary floating-point number to 52.1, which is now 7332423143312589 * 2**-47, or 52.10000000000000142108547152020037174224853515625.

4. Finally, as part of Python's read-eval-print loop (REPL), the floating-point value is displayed (in decimal). That involves converting the binary value back to a decimal string, getting 52.1 as the final output.

In Python 2.7 and later, we have the pleasant situation that the two conversions in step 3 and 4 cancel each other out. That's due to Python's choice of repr implementation, which produces the shortest decimal value guaranteed to round correctly to the actual float. One consequence of that choice is that if you start with any (not too large, not too small) decimal literal with 15 or fewer significant digits then the corresponding float will be displayed showing those exact same digits:

>>> x = 15.34509809234
>>> x
15.34509809234

Unfortunately, this furthers the illusion that Python is storing values in decimal. Not so in Python 2.6, though! Here's the original example executed in Python 2.6:

>>> round(52.15, 1)
52.200000000000003

Not only do we round in the opposite direction, getting 52.2 instead of 52.1, but the displayed value doesn't even print as 52.2! This behaviour has caused numerous reports to the Python bug tracker along the lines of "round is broken!". But it's not round that's broken, it's user expectations. (Okay, okay, round is a little bit broken in Python 2.6, in that it doesn't use correct rounding.)

Short version: if you're using two-argument round, and you're expecting predictable behaviour from a binary approximation to a decimal round of a binary approximation to a decimal halfway case, you're asking for trouble.

So enough with the "two-argument round is bad" argument. What should you be using instead? There are a few possibilities, depending on what you're trying to do.

• If you're rounding for display purposes, then you don't want a float result at all; you want a string. In that case the answer is to use string formatting:

  >>> format(66.66666666666, '.4f')
  '66.6667'
  >>> format(1.29578293, '.6f')
  '1.295783'
  

  Even then, one has to be aware of the internal binary representation in order not to be surprised by the behaviour of apparent decimal halfway cases.

  >>> format(52.15, '.1f')
  '52.1'
  

• If you're operating in a context where it matters which direction decimal halfway cases are rounded (for example, in some financial contexts), you might want to represent your numbers using the Decimal type. Doing a decimal round on the Decimal type makes a lot more sense than on a binary type (equally, rounding to a fixed number of binary places makes perfect sense on a binary type). Moreover, the decimal module gives you better control of the rounding mode. In Python 3, round does the job directly. In Python 2, you need the quantize method.

  >>> Decimal('66.66666666666').quantize(Decimal('1e-4'))
  Decimal('66.6667')
  >>> Decimal('1.29578293').quantize(Decimal('1e-6'))
  Decimal('1.295783')
  

• In rare cases, the two-argument version of round really is what you want: perhaps you're binning floats into bins of size 0.01, and you don't particularly care which way border cases go. However, these cases are rare, and it's difficult to justify the existence of the two-argument version of the round builtin based on those cases alone.

Comments

[gseattle]

Another route. Ignoring float(), what is this notation called? float('%.2f' % 1.29578293). That produces 1.3, it drops the zero. Or float('%.3f' % 1.29578293) is 1.296

[Mark Dickinson]

@gseattle: As of Python 3, using fixed-precision formatting and then converting to float should be entirely equivalent to using round. The formatting operation is again a conversion from binary floating-point to a decimal string, and it's using exactly the same machinery under the hood that round is using. (For Python 2, it's a bit different, because the half-way behaviour is different.)

[Serge Stroobandt]

As for MicroPython, this topic is discussed here.

[Maggyero]

Hi Mark. Since you seem to be very knowledgeable in floating-point arithmetic, would you like to share your expertise on this post?

[Canute S]

Using the above function with 2.5,0 gives the output as 2 which is the same as default behaviour. Is it what is expected?

[NMC]

Thank you Canute S, I used this method mostly for 2 decimal places (currencies) and seems that is not working properly for 0 decimal places

[NMC]

@CanuteS If you find a fully working solution please share with us!

[Canute S]

This code can't be formatted properly so posting as answer below

[NMC]

@CanuteS thanks! the "decimal.getcontext().rounding = decimal.ROUND_HALF_UP" does the trick! I will update my answer

[CodingMatters]

I hate the idea I like this simple answer, while knowing the explanation provided by @mark-dickinson is valid.