how to set conditional reference to range on Python 3, and xrange on Python 2

Question

I have a python-3 code that I would like to be compatible for both python-2 and python-3 keeping the code as is as much as possible. I would like to use the iteration behavior of range (to iterate over many items).

In order to get an iterator:

python-2 uses xrange(N)
python-3 uses range(N)

what is the best way to make it an iterator for python2 with minimal changes as possible?

looking at this link, it suggests a few ways for range and xrange but couldn't make it work

Detect the python version and use range = xrange if you find the user running python 2. Basically, deliberately redefine range — roganjosh
– roganjosh, Commented Dec 26, 2018 at 15:27
This might be semantics, but neither xrange in python 2 nor range in python 3 return iterators, they return an xrange or range object and you can't use them with next(). — cdarke
– cdarke, Commented Dec 26, 2018 at 15:35

iBug · Accepted Answer · 2018-12-26 15:31:40Z

3

Put this near the start of the program.

import sys
import builtins

if sys.version_info[0] == 2:
    range = xrange

a = range(100)  # Keep your Python 3 code unchanged

answered Dec 26, 2018 at 15:31

iBug

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ycx · Accepted Answer · 2018-12-26 15:30:06Z

0

try:
    n_range = xrange
except NameError:
    n_range = range

We can set n_range as xrange and if it fails, set it as range.

answered Dec 26, 2018 at 15:30

ycx

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1 Comment

Aaron_ab Over a year ago

I think it's much less efficient to catch exception rather than simple if

Nullman · Accepted Answer · 2018-12-26 15:32:19Z

0

you can detect the version of python at runtime:

from sys import version_info
if version_info[0] < 3:
    range = xrange

this overrides the regular range behavior in python 2 so make sure your code doesn't use the python 2 range

answered Dec 26, 2018 at 15:32

Nullman

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