将字符串分成一定长度的较小字符串列表

Question

If i have this string:

"0110100001100101011011000110110001101111"

How can I divide it at every eighth character into smaller strings, and put it into a list, so that it looks like this?:

['01101000','01100101','01101100','01101100','01101111']

So far, I can't figure out how this would be possible.

I should mention, since my strings are in binary so the length is always a multiple of 8.

Answer 1

>>> mystr = "0110100001100101011011000110110001101111"
>>> [mystr[i:i+8] for i in range(0, len(mystr), 8)]
['01101000', '01100101', '01101100', '01101100', '01101111']

The solution uses a so called list comprehension ( this seems to be a pretty decent tutorial) and is doing basically the same as Aleph's answer, just in one line.

Answer 2

t=[]
for i in range(len(yourstring)/8):
    t.append(yourstring[i*8: (i+1)*8])

Answer 3

You can also use re

import re

In [21]: s="0110100001100101011011000110110001101111"

In [22]: re.findall('.{8}',s)
Out[22]: ['01101000', '01100101', '01101100', '01101100', '01101111']

As your strings only contain digits we can use . to match any character.

You can also explicitly use the \\d{8} which matches 8 digits in a low.

In [23]: In [22]: re.findall('\d{8}',s)
Out[23]: ['01101000', '01100101', '01101100', '01101100', '01101111']