每n个字符分割一个字符串？

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']

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

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.

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.
'''

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().

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)))

I like this solution:

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

Using more-itertools from PyPI:

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

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.

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)))

>>> 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']

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']

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])]

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)

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']

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

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’]