Split a string to even sized chunks

Question

How would I be able to take a string like \'aaaaaaaaaaaaaaaaaaaaaaa\' and split it into 4 length tuples like (aaaa,aaaa,aaaa

Answer 1

Another solution using regex:

>>> s = 'aaaaaaaaaaaaaaaaaaaaaaa'
>>> import re
>>> re.findall('[a-z]{4}', s)
['aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaaa']
>>>

Answer 2

Use textwrap.wrap:

>>> import textwrap
>>> s = 'aaaaaaaaaaaaaaaaaaaaaaa'
>>> textwrap.wrap(s, 4)
['aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaa']

Answer 3

I think this method is simpler. But the message length must be split with split_size. Or letters must be added to the message. Example: message = "lorem ipsum_" then the added letter can be deleted.

message = "lorem ipsum"

array = []

temp = ""

split_size = 3

for i in range(1, len(message) + 1):
    temp += message[i - 1]

    if i % split_size == 0:
        array.append(temp)
        temp = ""

print(array)

Output: ['lor', 'em ', 'ips']

Answer 4

s = 'abcdef'

We need to split in parts of 2

[s[pos:pos+2] for pos,i in enumerate(list(s)) if pos%2 == 0]

Answer:

['ab', 'cd', 'ef']

Answer 5

s = 'abcdefghi'

k - no of parts of string

k = 3

parts - list to store parts of string

parts = [s[i:i+k] for i in range(0, len(s), k)]

---

parts --> ['abc', 'def', 'ghi']

Answer 6

You could use the grouper recipe, zip(*[iter(s)]*4):

In [113]: s = 'aaaaaaaaaaaaaaaaaaaaaaa'

In [114]: [''.join(item) for item in zip(*[iter(s)]*4)]
Out[114]: ['aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaaa']

Note that textwrap.wrap may not split s into strings of length 4 if the string contains spaces:

In [43]: textwrap.wrap('I am a hat', 4)
Out[43]: ['I am', 'a', 'hat']

The grouper recipe is faster than using textwrap:

In [115]: import textwrap

In [116]: %timeit [''.join(item) for item in zip(*[iter(s)]*4)]
100000 loops, best of 3: 2.41 µs per loop

In [117]: %timeit textwrap.wrap(s, 4)
10000 loops, best of 3: 32.5 µs per loop

And the grouper recipe can work with any iterator, while textwrap only works with strings.