Python 3.x Shifting Ranges

Question

suppose I have a range like this:

x = range(10)

which would have the following values as a list:

list(x)      # Prints [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

I would like to shift this range (possibly multiple times)
and iterate over the results e.g.

             #        [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]

Creating an equivalent list is not a problem. But I’d like
to know if it is possible to create something like this as
a range to save some space in memory and of course it would
be nice if the solution could be about as performant as:

for i in range(1000000)

Asked By: Nima Mousavi

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Source

Answer 1

You can wrap the range in a generator expression, applying the shift and modulo on the fly:

def shifted_range(rangeob, shift):
    size, shift = rangeob.stop, shift * rangeob.step
    return ((i + shift) % size for i in rangeob)

Demo:

>>> def shifted_range(rangeob, shift):
...     size, shift = rangeob.stop, shift * rangeobj.step
...     return ((i + shift) % size for i in rangeob)
... 
>>> range_10 = range(10)
>>> list(shifted_range(range_10, 3))
[3, 4, 5, 6, 7, 8, 9, 0, 1, 2]
>>> list(shifted_range(range_10, 7))
[7, 8, 9, 0, 1, 2, 3, 4, 5, 6]
>>> range_10_2 = range(0, 10, 2)
>>> list(shifted_range(range_10_2, 4))
[8, 0, 2, 4, 6]

You could make this a wrapper object as well:

class RangeShift:
    def __init__(self, rangeob, shift):
        self._range = rangeob
        self.shift = shift

    @property
    def start(self):
        r = self._range
        return (r.start + self.shift * r.step) % r.stop

    @property
    def stop(self):
        r = self._range
        return (r.stop + self.shift * r.step) % r.stop

    def index(self, value):
        idx = self._range.index(value)
        return (idx - self.shift) % len(self._range)

    def __getattr__(self, attr):
        return getattr(self._range, attr)

    def __getitem__(self, index):
        r = self._range
        return (r[index] + self.shift * r.step) % r.stop

    def __len__(self):
        return len(self._range)

    def __iter__(self):
        size, shift = self._range.stop, self.shift * self._range.step
        return ((i + shift) % size for i in self._range)

This will behave just like the original range, but applying a shift to all values produced. It even lets you alter the shift!

Demo:

>>> range_10 = range(10)
>>> shifted = RangeShift(range_10, 7)
>>> len(shifted)
10
>>> shifted.start
7
>>> shifted.stop
7
>>> shifted.step
1
>>> shifted[3]
0
>>> shifted[8]
5
>>> list(shifted)
[7, 8, 9, 0, 1, 2, 3, 4, 5, 6]
>>> shifted.shift = 3
>>> list(shifted)
[3, 4, 5, 6, 7, 8, 9, 0, 1, 2]
>>> range_10_2 = range(0, 10, 2)
>>> shifted_10_2 = RangeShift(range_10_2, 4)
>>> list(shifted_10_2)
[8, 0, 2, 4, 6]

Best trick this wrapper now supports: reversing the shifted range:

>>> list(reversed(shifted))
[2, 1, 0, 9, 8, 7, 6, 5, 4, 3]
>>> list(reversed(shifted_10_2))
[6, 4, 2, 0, 8]

Answered By: Martijn Pieters

Answer 2

I guess the simplest way is to chain two ranges:

from itertools import chain

shifted = chain(range(7, 10), range(7))
for x in shifted:
    print(x)

Answered By: georg

Answer 3

You can use itertools to chain two ranges. This code works even if the ranges have step > 1.

import itertools

def shift_range(r, s):
    return itertools.chain(range(r.start + s*r.step, r.stop, r.step), 
                           range(r.start, r.start + s*r.step, r.step))

Testing:

>>> list(shift_range(range(10), 5))
[5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
>>> list(shift_range(range(3, 30, 3), 5))
[18, 21, 24, 27, 3, 6, 9, 12, 15]

Answered By: parchment

Answer 4

It’s an old question, but I just stumbled across it.
My solution:

  offset = 7
  shifted_range = range(x.start + offset, x.stop + offset, x.step)

Answered By: Peter

Python 3.x Shifting Ranges

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