Split string in Python into chunks with constant length, but right-aligned

Question

Basically, I have a string like "12345678" and need a list containing this information, but split into substrings of length 3. The problem is, that I need it to be right-aligned, so the output must be ['12', '345', '678'] and NOT ['123', '456', '78'].
How do I best achieve with few code lines and preferably without additional imports?

Answer 1

It's easy enough to adapt the top answer from How do I split a list into equally-sized chunks?:

def chunks_rightaligned(l, n):
    orphan = len(l) % n
    if orphan:
        yield l[: orphan]         
    for i in range(orphan, len(l), n):
        yield l[i : i + n]

This yields a chunk of the remainder length first, then iterates over the indices in chunk-size steps starting from the orphan size rather than 0.

Demo:

>>> def chunks_rightaligned(l, n):
...     orphan = len(l) % n
...     if orphan:
...         yield l[: orphan]         
...     for i in range(orphan, len(l), n):
...         yield l[i : i + n]
... 
>>> list(chunks_rightaligned("12345678", 3))
['12', '345', '678']
>>> list(chunks_rightaligned("1234567", 3))
['1', '234', '567']
>>> list(chunks_rightaligned("123456", 3))
['123', '456']

Answer 2

If you want to try regular expressions, you can use the re.split() function

>>> re.split(r"(...)(?=(?:\d\d\d)+$)","12345678")
['12', '345', '678']

>>> re.split(r"(...)(?=(?:\d\d\d)+$)","123")
['123']

---

EDIT

A better solution would be to use re.findall()

>>> re.findall(r"\d{1,3}(?=(?:\d{3})*$)", "12345")
['12', '345']

>>> re.findall(r"\d{1,3}(?=(?:\d{3})*$)", "123456")
['123', '456']

>>> re.findall(r"\d{1,3}(?=(?:\d{3})*$)", "1234567")
['1', '234', '567']

What it does?

• \d{1,3} Matches maximum of 3 characters, and minimum of 1 character.

• (?=(?:\d{3})*$) Positive look ahead. Ensures that the matched characters is followed by a multiple of 3 digits.

  • (?:\d{3}) matches 3 digits.

---

You can use variables in the regex string to generate variable chunk of data.

Example

>>> $limit=4
>>> regex = r"\d{1,%d}(?=(?:\d{%d})*$)" %(limit,limit)
>>> re.findall(regex, "1234567")
['123', '4567']

>>> limit=3
>>> regex = r"\d{1,%d}(?=(?:\d{%d})*$)" %(limit,limit)
>>> re.findall(regex, "1234567")
['1', '234', '567']