Split string into strings by length?

Question

Is there a way to take a string that is 4*x characters long, and cut it into 4 strings, each x characters long, without knowing the length of the s

Answer 1

Got an re trick:

In [28]: import re

In [29]: x = "qwertyui"

In [30]: [x for x in re.split(r'(\w{2})', x) if x]
Out[30]: ['qw', 'er', 'ty', 'ui']

Then be a func, it might looks like:

def split(string, split_len):
    # Regex: `r'.{1}'` for example works for all characters
    regex = r'(.{%s})' % split_len
    return [x for x in re.split(regex, string) if x]

Answer 2

In Split string every nth character?, "the wolf" gives the most concise answer:

>>> import re
>>> re.findall('..','1234567890')
['12', '34', '56', '78', '90']

Answer 3

• :param s: str; source string
• :param w: int; width to split on

Using the textwrap module:

PyDocs-textwrap

import textwrap
def wrap(s, w):
    return textwrap.fill(s, w)

:return str:

Inspired by Alexander's Answer

PyDocs-data structures

def wrap(s, w):
    return [s[i:i + w] for i in range(0, len(s), w)]

• :return list:

Inspired by Eric's answer

PyDocs-regex

import re
def wrap(s, w):    
    sre = re.compile(rf'(.{{{w}}})')
    return [x for x in re.split(sre, s) if x]

• :return list:

Complete Code Examples/Alternative Methods

Answer 4

The string splitting is required in many cases like where you have to sort the characters of the string given, replacing a character with an another character etc. But all these operations can be performed with the following mentioned string splitting methods.

The string splitting can be done in two ways:

1. Slicing the given string based on the length of split.

2. Converting the given string to a list with list(str) function, where characters of the string breakdown to form the the elements of a list. Then do the required operation and join them with 'specified character between the characters of the original string'.join(list) to get a new processed string.

Answer 5

l = 'abcdefghijklmn'

def group(l,n):
    tmp = len(l)%n
    zipped = zip(*[iter(l)]*n)
    return zipped if tmp == 0 else zipped+[tuple(l[-tmp:])]

print group(l,3)

Answer 6

My solution

   st =' abs de fdgh  1234 556 shg shshh'
   print st

   def splitStringMax( si, limit):
    ls = si.split()
    lo=[]
    st=''
    ln=len(ls)
    if ln==1:
        return [si]
    i=0
    for l in ls:
        st+=l
        i+=1
        if i <ln:
            lk=len(ls[i])
            if (len(st))+1+lk < limit:
                st+=' '
                continue
        lo.append(st);st=''
    return lo

   ############################

   print  splitStringMax(st,7)
   # ['abs de', 'fdgh', '1234', '556', 'shg', 'shshh']
    print  splitStringMax(st,12)

   # ['abs de fdgh', '1234 556', 'shg shshh']