Split a string into N equal parts?

Question

I have a string I would like to split into N equal parts.

For example, imagine I had a string with length 128 and I want to split it in to 4 chunks of length 32 each; i.e., first 32 chars, then the second 32 and so on.

How can I do this?

Answer 1

import textwrap
print(textwrap.wrap("123456789", 2))
#prints ['12', '34', '56', '78', '9']

Note: be careful with whitespace etc - this may or may not be what you want.

"""Wrap a single paragraph of text, returning a list of wrapped lines.

    Reformat the single paragraph in 'text' so it fits in lines of no
    more than 'width' columns, and return a list of wrapped lines.  By
    default, tabs in 'text' are expanded with string.expandtabs(), and
    all other whitespace characters (including newline) are converted to
    space.  See TextWrapper class for available keyword args to customize
    wrapping behaviour.
    """

Answer 2

You may use a simple loop:

parts = [your_string[i:i+n] for i in range(0, len(your_string), n)]

Answer 3

Another common way of grouping elements into n-length groups:

>>> s = '1234567890'
>>> list(map(''.join, zip(*[iter(s)]*2)))
['12', '34', '56', '78', '90']

This method comes straight from the docs for zip().

Answer 4

Recursive way:

def split_str(seq, chunk, skip_tail=False):
    lst = []
    if chunk <= len(seq):
        lst.extend([seq[:chunk]])
        lst.extend(split_str(seq[chunk:], chunk, skip_tail))
    elif not skip_tail and seq:
        lst.extend([seq])
    return lst

Demo:

seq = "123456789abcdefghij"

print(split_str(seq, 3))
print(split_str(seq, 3, skip_tail=True))

# ['123', '456', '789', 'abc', 'def', 'ghi', 'j']
# ['123', '456', '789', 'abc', 'def', 'ghi']

Answer 5

You can treat a string similarly to a list in many cases. There are lots of answers here: Splitting a list of into N parts of approximately equal length

for example you could work out the chunk_size = len(my_string)/N

Then to access a chunk you can go my_string[i: i + chunk_size] (and then increment i by chunk_size) - either in a for loop or in a list comprehension.

Answer 6

I like iterators!

def chunk(in_string,num_chunks):
    chunk_size = len(in_string)//num_chunks
    if len(in_string) % num_chunks: chunk_size += 1
    iterator = iter(in_string)
    for _ in range(num_chunks):
        accumulator = list()
        for _ in range(chunk_size):
            try: accumulator.append(next(iterator))
            except StopIteration: break
        yield ''.join(accumulator)

## DEMO
>>> string = "a"*32+"b"*32+"c"*32+"d"*32
>>> list(chunk(string,4))
['aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa', 'bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb', 'cccccccccccccccccccccccccccccccc', 'dddddddddddddddddddddddddddddddd']
>>> string += "e" # so it's not evenly divisible
>>> list(chunk(string,4))
['aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab', 'bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbcc', 'ccccccccccccccccccccccccccccccddd', 'ddddddddddddddddddddddddddddde']

Also demonstrably faster than textwrap.wrap, although almost certainly less "good"

>>> timeit.timeit(lambda: list(chunk(string,4)),number=500)
0.047726927170444355
>>> timeit.timeit(lambda: textwrap.wrap(string,len(string)//4),number=500)
0.20812756575945457

And pretty easy to hack to work with any iterable (just drop the str.join and yield accumulator unless isinstance(in_string,str))

# after a petty hack
>>> list(chunk([1,2,3,4,5,6,7,8,9,10,11,12],4))
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]]