How to reverse string with stride via Python String slicing

Question

I need to reverse an interleaved string, means i have 2-pairs which shouldt get messed up like this:

>>> interleaved = "123456"

reversing

>>> print interleaved[::-1]
654321

but what i actually want is

563412

is there a string slice operation for this?

Answer 1

For even length strings this should do it:

>>> s = "123456"
>>> it = reversed(s)
>>> ''.join(next(it) + x for x in it)
'563412'

For odd length strings, you need to prepend the first character separately:

>>> s = "7123456"
>>> it = reversed(s)
>>> (s[0] if len(s)%2 else '') + ''.join(next(it) + x for x in it)
'7563412'

---

Using slicing and zip:

>>> s = "7123456"
>>> (s[0] if len(s)%2 else '') + ''.join(x+y for x, y in zip(s[-2::-2], s[::-2]))
'7563412'

Answer 2

The shortest way as far as I know would be to use a regex:

import re
''.join(re.findall('..?', '123456', flags=re.S)[::-1])

• Input: '123456'
• Output: '563412'

This also works for odd-length strings without having to implement separate logic for them.

Answer 3

You can bring several ideas to split a string in pieces and then reverse each piece and reassemble (join) the list reversed too.

E.g. (using satomacoto answer in a not-so-readable way...)

''.join([a[::-1][i:i+2][::-1] for i in range(0, len(a), 2)]) 

or (using F.J. answer)

''.join(map(''.join, zip(*[iter(a)]*2))[::-1])

and so on. (Being a your string).