How do you split a string in python with number of characters?

Is it possible to split a string every nth character?

For example, suppose I have a string containing the following:

'1234567890'

How can I get it to look like this:

['12','34','56','78','90']

Georgy

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asked Feb 28, 2012 at 1:48

1

>>> line = '1234567890'
>>> n = 2
>>> [line[i:i+n] for i in range(0, len(line), n)]
['12', '34', '56', '78', '90']

answered Feb 28, 2012 at 2:02

4

Just to be complete, you can do this with a regex:

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

For odd number of chars you can do this:

>>> import re
>>> re.findall('..?', '123456789')
['12', '34', '56', '78', '9']

You can also do the following, to simplify the regex for longer chunks:

>>> import re
>>> re.findall('.{1,2}', '123456789')
['12', '34', '56', '78', '9']

And you can use re.finditer if the string is long to generate chunk by chunk.

Georgy

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answered Feb 28, 2012 at 6:31

the wolfthe wolf

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There is already an inbuilt function in python for this.

>>> from textwrap import wrap
>>> s = '1234567890'
>>> wrap(s, 2)
['12', '34', '56', '78', '90']

This is what the docstring for wrap says:

>>> help(wrap)
'''
Help on function wrap in module textwrap:

wrap(text, width=70, **kwargs)
    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.
'''

answered Feb 19, 2018 at 6:57

10

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

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

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

answered Feb 28, 2012 at 2:25

Andrew ClarkAndrew Clark

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5

I think this is shorter and more readable than the itertools version:

def split_by_n(seq, n):
    '''A generator to divide a sequence into chunks of n units.'''
    while seq:
        yield seq[:n]
        seq = seq[n:]

print(list(split_by_n('1234567890', 2)))

answered Feb 28, 2012 at 1:53

Russell BorogoveRussell Borogove

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2

Using more-itertools from PyPI:

>>> from more_itertools import sliced
>>> list(sliced('1234567890', 2))
['12', '34', '56', '78', '90']

answered Jun 22, 2017 at 10:19

Tim DielsTim Diels

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I like this solution:

s = '1234567890'
o = []
while s:
    o.append(s[:2])
    s = s[2:]

answered Sep 12, 2015 at 23:14

vlkvlk

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You could use the grouper() recipe from itertools:

Python 2.x:

from itertools import izip_longest    

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

Python 3.x:

from itertools import zip_longest

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)

These functions are memory-efficient and work with any iterables.

answered Oct 3, 2015 at 20:16

Eugene YarmashEugene Yarmash

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1

This can be achieved by a simple for loop.

a = '1234567890a'
result = []

for i in range(0, len(a), 2):
    result.append(a[i : i + 2])
print(result)

The output looks like ['12', '34', '56', '78', '90', 'a']

answered May 22, 2020 at 18:02

Kasem777Kasem777

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3

I was stucked in the same scenrio.

This worked for me

x="1234567890"
n=2
list=[]
for i in range(0,len(x),n):
    list.append(x[i:i+n])
print(list)

Output

['12', '34', '56', '78', '90']

answered Nov 28, 2019 at 14:54

StrickStrick

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Try the following code:

from itertools import islice

def split_every(n, iterable):
    i = iter(iterable)
    piece = list(islice(i, n))
    while piece:
        yield piece
        piece = list(islice(i, n))

s = '1234567890'
print list(split_every(2, list(s)))

answered Feb 28, 2012 at 1:52

enderskillenderskill

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1

Try this:

s='1234567890'
print([s[idx:idx+2] for idx,val in enumerate(s) if idx%2 == 0])

Output:

['12', '34', '56', '78', '90']

answered Jul 10, 2018 at 3:46

U12-ForwardU12-Forward

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>>> from functools import reduce
>>> from operator import add
>>> from itertools import izip
>>> x = iter('1234567890')
>>> [reduce(add, tup) for tup in izip(x, x)]
['12', '34', '56', '78', '90']
>>> x = iter('1234567890')
>>> [reduce(add, tup) for tup in izip(x, x, x)]
['123', '456', '789']

answered Feb 28, 2012 at 1:56

ben wben w

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As always, for those who love one liners

n = 2  
line = "this is a line split into n characters"  
line = [line[i * n:i * n+n] for i,blah in enumerate(line[::n])]

answered May 20, 2016 at 20:00

SqripterSqripter

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more_itertools.sliced has been mentioned before. Here are four more options from the more_itertools library:

s = "1234567890"

["".join(c) for c in mit.grouper(2, s)]

["".join(c) for c in mit.chunked(s, 2)]

["".join(c) for c in mit.windowed(s, 2, step=2)]

["".join(c) for c in  mit.split_after(s, lambda x: int(x) % 2 == 0)]

Each of the latter options produce the following output:

['12', '34', '56', '78', '90']

Documentation for discussed options: grouper, chunked, windowed, split_after

answered Feb 9, 2018 at 1:16

pylangpylang

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0

A simple recursive solution for short string:

def split(s, n):
    if len(s) < n:
        return []
    else:
        return [s[:n]] + split(s[n:], n)

print(split('1234567890', 2))

Or in such a form:

def split(s, n):
    if len(s) < n:
        return []
    elif len(s) == n:
        return [s]
    else:
        return split(s[:n], n) + split(s[n:], n)

, which illustrates the typical divide and conquer pattern in recursive approach more explicitly (though practically it is not necessary to do it this way)

answered Oct 22, 2018 at 10:25

englealuzeenglealuze

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A solution with groupby:

from itertools import groupby, chain, repeat, cycle

text = "wwworldggggreattecchemggpwwwzaz"
n = 3
c = cycle(chain(repeat(0, n), repeat(1, n)))
res = ["".join(g) for _, g in groupby(text, lambda x: next(c))]
print(res)

Output:

['www', 'orl', 'dgg', 'ggr', 'eat', 'tec', 'che', 'mgg', 'pww', 'wza', 'z']

answered Jul 23, 2021 at 23:08

TigerTV.ruTigerTV.ru

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These answers are all nice and working and all, but the syntax is so cryptic... Why not write a simple function?

def SplitEvery(string, length):
    if len(string) <= length: return [string]        
    sections = len(string) / length
    lines = []
    start = 0;
    for i in range(sections):
        line = string[start:start+length]
        lines.append(line)
        start += length
    return lines

And call it simply:

text = '1234567890'
lines = SplitEvery(text, 2)
print(lines)

# output: ['12', '34', '56', '78', '90']

answered Jul 22 at 9:12

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