每隔第n个字符拆分字符串？

Question

是否有可能每隔第n个字符拆分一个python字符串？

例如,假设我有一个包含以下内容的字符串:

'1234567890'

我怎么能让它看起来像这样:

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

Answer 1

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

Answer 2

为了完成,您可以使用正则表达式执行此操作:

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

正如评论中指出的那样,你可以这样做:

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

您还可以执行以下操作,以简化较长块的正则表达式:

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

re.finditer如果字符串很长,你可以使用chunk来生成chunk.

Answer 3

为此,python中已经有一个内置函数.

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

这就是包装的docstring所说的:

>>> 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.
'''

Answer 4

将元素分组为n长度组的另一种常用方法:

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

这个方法直接来自docs zip().

Answer 5

我认为这比itertools版本更短,更易读:

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)))

Answer 6

我喜欢这个解决方案:

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

Answer 7

使用PyPI中的more-itertools:

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

Answer 8

您可以使用以下grouper()配方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)

这些函数具有内存效率,适用于任何迭代.

Answer 9

我被困在同一个场景中。

这对我有用

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

输出

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

Answer 10

试试下面的代码：

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)))

Answer 11

这可以通过一个简单的 for 循环来实现。

a = '1234567890a'
result = []

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

输出看起来像 ['12', '34', '56', '78', '90', 'a']

Answer 12

>>> 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']

Answer 13

一如既往，对于那些喜欢单衬的人

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])]

Answer 14

试试这个:

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

输出:

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