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How to use digit separators for Python integer literals?

Is there any way to group digits in a Python code to increase code legibility? I've tried ' and _ which are digit separators of some other languages, but no avail.

A weird operator which concatenates its left hand side with its right hand side could also work out.

Update a few years later: Python 3.6 now supports PEP515, and so you can use _ for float and integer literal readability improvement.

Python 3.6.1 (v3.6.1:69c0db5, Mar 21 2017, 18:41:36) [MSC v.1900 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> 1_1000

For historical reference, you can look at the lexical analysis for strict definitions python2.7, python3.5 ...

For python3.6.0a2 and earlier, you should get an error message similar to:

Python 3.6.0a2 (v3.6.0a2:378893423552, Jun 13 2016, 14:44:21) 
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> 1_000
  File "<stdin>", line 1
SyntaxError: invalid syntax
>>> amount = 10_000_000.0
  File "<stdin>", line 1
    amount = 10_000_000.0
SyntaxError: invalid syntax
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Currently there is no thousands separator in Python, but you can use locale module to convert string with such separators to an int:

import locale
locale.setlocale(locale.LC_ALL, '')
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The closest thing I've seen in python is 12 * 1000 * 1000, which is not ideal, but can be useful if 12000000 is needed. Be advised though, while in C, those are equivalent, because at compile time it converts both to the same thing, python may not share this optimization.

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    • If they are literals, python will fold the constants. It won't do any folding of symbols though.
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    • @ThoAppelsin -- Well, that depends on whether you want your value to be a float or an int...
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    • @TemporalWolf -- You can always check these things by disassembling the source: dis.dis(lambda : 100 * 200 * 300)
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    • @mgilson: As it happens, this is one of those things where order matters a lot. If you use x * 100 * 1000 * 1000, then it won't fold (because x * 100 might not return an int if x isn't an int). But 100 * 1000 * 1000 * x folds, because the left to right evaluation is working with known constants.

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