Reversing an arbitrary slice in Python

Question

I'm looking for a general method on how to reverse a slice in Python. I read this comprehensive post, which has several nice explanations about how slicing works: Understanding Python's slice notation

Yet I cannot figure out a generalized rule on how to calculate a reversed slice which addresses exactly the same elements in reverse order. I was actually surprised not to find a builtin method doing this.

What I'm looking for is a method reversed_slice that works like this with arbitrary start, stop and step values including negative values:

>>> import numpy as np
>>> a = np.arange(30)
>>> s = np.s_[10:20:2]
>>> a[s]
array([10, 12, 14, 16, 18])
>>> a[reversed_slice(s,len(a))]
array([18, 16, 14, 12, 10])

What I've tried but doesn't work is this:

def reversed_slice(slice_, len_):
    """
    Reverses a slice (selection in array of length len_), 
    addressing the same elements in reverse order.
    """
    assert isinstance(slice_, slice)
    instart, instop, instep = slice_.indices(len_)
    if instep > 0:
        start, stop, step = instop - 1, instart - 1, -instep
    else:
        start, stop, step = instop + 1, instart + 1, -instep
    return slice(start, stop, step)

This works fine for step of 1 and when the last addressed element coincides with stop-1. For other cases it does not:

>>> import numpy as np
>>> a = np.arange(30)
>>> s = np.s_[10:20:2]
>>> a[s]
array([10, 12, 14, 16, 18])
>>> a[reversed_slice(s,len(a))]
array([19, 17, 15, 13, 11])

So it seems like I'm missing some relation like (stop - start) % step. Any help on how to write a general method is greatly appreciated.

Notes:

• I do know that there a other possibilities to get a sequence with the same elements reversed, like calling reversed(a[s]). This is not an option here, as I need to reverse the slice itself. The reason is that I work on h5py datasets which do not allow negative step values in slices.

• An easy but not very elegant way would be the use of coordinate lists, i.e. a[list(reversed(range(*s.indices(len(a)))))]. This is also not an option due to the h5py requirement that indices in the list must be given in increasing order.

Answer 1

You can specify negative values for step.

>>> s = np.s_[20-2:10-2:-2]
>>> a[s]
array([18, 16, 14, 12, 10])

So you can build the reversed_slice function as follows

>>> def reversed_slice(s):
...     """
...     Reverses a slice 
...     """
...     m = (s.stop-s.start) % s.step or s.step
...     return slice(s.stop-m, s.start-m, -s.step)
... 
>>> a = np.arange(30)
>>> s = np.s_[10:20:2]
>>> a[reversed_slice(s)]
array([18, 16, 14, 12, 10])
>>> 
>>> a[reversed_slice(reversed_slice(s))]
array([10, 12, 14, 16, 18])
>>> 

Answer 2

I just wanted to use the answer to this question but while testing found that there are still some cases that would silently give the wrong result -

The following definition of the reversed_slice function developed from the other answers seems to cover those cases correctly -

def reversed_slice(s, len_):
    """
    Reverses a slice selection on a sequence of length len_, 
    addressing the same elements in reverse order.
    """
    assert isinstance(s, slice)
    instart, instop, instep = s.indices(len_)

    if (instop < instart and instep > 0) or (instop > instart and instep < 0) \
      or (instop == 0 and instart == 0) :
        return slice(0,0,None)

    overstep = abs(instop-instart) % abs(instep)

    if overstep == 0 :
        overstep = abs(instep)

    if instep > 0:
        start = instop - overstep
        stop = instart - 1
    else :
        start = instop + overstep
        stop = instart + 1

    if stop < 0 :
        stop = None

    return slice(start, stop, -instep)

Answer 3

You made some mistakes with start/stop math:

overstep = abs(instop-instart) % abs(instep)
if overstep == 0 :
    overstep = abs(instep)

if instep > 0:
    start = instop - overstep
    stop = instart - 1
else :
    start = instop + overstep
    stop = instart + 1

step = -instep

Once you put this in your code, everything should work just fine.

Answer 4

So far I also didn't find any builtin, but what worked even for negative steps:

def invert_slice(start, stop, step=1):
    distance = stop - start
    step_distance = distance // step
    expected_distance = step_distance * step

    if expected_distance != distance:
        expected_distance += step

    new_start = start + expected_distance - step
    new_stop = start - step

    return slice(new_start, new_stop, -step)

This gives you

>>> import numpy as np
>>> a = np.arange(30)
>>> s = np.s_[24:10:-1]
>>> expected = list(reversed(a[s]))

[18, 16, 14, 12, 10]

>>> # resulting slice
>>> result = invert_slice(s.start, s.stop, s.step)

slice(18, 8, -2)

>>> assert np.allclose(expected, a[result]), "Invalid Slice %s" % result
>>> a[result]

[18 16 14 12 10] They are equal ;-)

Answer 5

I found a working solution based on Sunitha's answer (EDIT: Also implemented Warwick's answer):

def reversed_slice(s, len_):
    """
    Reverses a slice selection on a sequence of length len_, 
    addressing the same elements in reverse order.
    """
    assert isinstance(s, slice)
    instart, instop, instep = s.indices(len_)

    if (instop < instart and instep > 0) or (instop > instart and instep < 0) \
            or (instop == 0 and instart == 0):
        return slice(0, 0, None)

    m = (instop - instart) % instep or instep

    if instep > 0 and instart - m < 0:
        outstop = None
    else:
        outstop = instart - m
    if instep < 0 and instop - m > len_:
        outstart = None
    else:
        outstart = instop - m

    return slice(outstart, outstop, -instep)

It uses the slice.indices(len) method to expand the functionality so it can also be used with None entries in the slice, e.g. with [::-1]. Problems at the boundaries are avoided with the if clauses.

This solution only works when providing the length of the sequence to address. I don't think there's a way to get around this. If there is, or if there is an easier way, I'm open for suggestions!