是否有可能分割字符串每n个字符?
例如,假设我有一个包含以下内容的字符串:
'1234567890'
我怎样才能让它看起来像这样:
['12','34','56','78','90']
关于列表的相同问题,请参见如何将列表分割为大小相等的块?。同样的技术通常适用,尽管有一些变化。
是否有可能分割字符串每n个字符?
例如,假设我有一个包含以下内容的字符串:
'1234567890'
我怎样才能让它看起来像这样:
['12','34','56','78','90']
关于列表的相同问题,请参见如何将列表分割为大小相等的块?。同样的技术通常适用,尽管有一些变化。
当前回答
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)))
其他回答
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)))
>>> 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']
more_itertools。切片之前提到过。下面是more_itertools库中的另外四个选项:
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)]
后面的每个选项都会产生以下输出:
['12', '34', '56', '78', '90']
所讨论选项的文档:grouper, chunked, windosed, split_after
你可以使用itertools中的grouper()方法:
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)
这些函数是内存高效的,并且适用于任何可迭代对象。
使用groupby的解决方案:
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)
输出:
['www', 'orl', 'dgg', 'ggr', 'eat', 'tec', 'che', 'mgg', 'pww', 'wza', 'z']