Split string into strings by length?

I tried Alexanders answer but got this error in Python3:

TypeError: ‘float’ object cannot be interpreted as an integer

This is because the division operator in Python3 is returning a float. This works for me:

>>> x = "qwertyui"
>>> chunks, chunk_size = len(x), len(x)//4
>>> [ x[i:i+chunk_size] for i in range(0, chunks, chunk_size) ]
['qw', 'er', 'ty', 'ui']

Notice the // at the end of line 2, to ensure truncation to an integer.

Here is a one-liner that doesn’t need to know the length of the string beforehand:

from functools import partial
from StringIO import StringIO

[l for l in iter(partial(StringIO(data).read, 4), '')]

If you have a file or socket, then you don’t need the StringIO wrapper:

[l for l in iter(partial(file_like_object.read, 4), '')]

And for dudes who prefer it to be a bit more readable:

def itersplit_into_x_chunks(string,x=10): # we assume here that x is an int and > 0
    size = len(string)
    chunksize = size//x
    for pos in range(0, size, chunksize):
        yield string[pos:pos+chunksize]

output:

>>> list(itersplit_into_x_chunks('qwertyui',x=4))
['qw', 'er', 'ty', 'ui']

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']

def split2len(s, n):
    def _f(s, n):
        while s:
            yield s[:n]
            s = s[n:]
    return list(_f(s, n))

Here are two generic approaches. Probably worth adding to your own lib of reusables. First one requires the item to be sliceable and second one works with any iterables (but requires their constructor to accept iterable).

def split_bylen(item, maxlen):
    '''
    Requires item to be sliceable (with __getitem__ defined)
    '''
    return [item[ind:ind+maxlen] for ind in range(0, len(item), maxlen)]
    #You could also replace outer [ ] brackets with ( ) to use as generator.

def split_bylen_any(item, maxlen, constructor=None):
    '''
    Works with any iterables.
    Requires item's constructor to accept iterable or alternatively 
    constructor argument could be provided (otherwise use item's class)
    '''
    if constructor is None: constructor = item.__class__
    return [constructor(part) for part in zip(* ([iter(item)] * maxlen))]
    #OR: return map(constructor, zip(* ([iter(item)] * maxlen)))
    #    which would be faster if you need an iterable, not list

So, in topicstarter’s case, the usage is:

string = 'Baboons love bananas'
parts = 5
splitlen = -(-len(string) // parts) # is alternative to math.ceil(len/parts)

first_method = split_bylen(string, splitlen)
#Result :['Babo', 'ons ', 'love', ' ban', 'anas']

second_method = split_bylen_any(string, splitlen, constructor=''.join)
#Result :['Babo', 'ons ', 'love', ' ban', 'anas']

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.

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)

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]

• :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:

length = 4
string = "abcdefgh"
str_dict = [ o for o in string ]
parts = [ ''.join( str_dict[ (j * length) : ( ( j + 1 ) * length ) ]   ) for j in xrange(len(string)/length  )]

some_string="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
x=3 
res=[some_string[y-x:y] for y in range(x, len(some_string)+x,x)]
print(res)

will produce

['ABC', 'DEF', 'GHI', 'JKL', 'MNO', 'PQR', 'STU', 'VWX', 'YZ']

# spliting a string by the length of the string

def len_split(string,sub_string):
    n,sub,str1=list(string),len(sub_string),')/^0*/-'
    for i in range(sub,len(n)+((len(n)-1)//sub),sub+1):
        n.insert(i,str1)   
    n="".join(n)
    n=n.split(str1)
    return n

x="divyansh_looking_for_intership_actively_contact_Me_here"
sub="four"
print(len_split(x,sub))

# Result-> ['divy', 'ansh', 'tiwa', 'ri_l', 'ooki', 'ng_f', 'or_i', 'nter', 'ship', '_con', 'tact', '_Me_', 'here']

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

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

There is a built in function in python for that

import textwrap

text = "Your Text.... and so on"
width = 5 #

textwrap.wrap(text,width)

Vualla