Python：有没有办法将一串数字分成每个第三个数字？

Question

例如，如果我有一个字符串a = 123456789876567543我可以有一个像...的列表

123 456 789 876 567 543

Answer 1

>>> a="123456789"
>>> [int(a[i:i+3]) for i in range(0, len(a), 3)]
[123, 456, 789]

Answer 2

来自itertools文档的配方（当长度不是3的倍数时，您可以定义fillvalue）：

from itertools import izip_longest

def grouper(n, iterable, fillvalue=None):
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

s = '123456789876567543'

print [''.join(l) for l in grouper(3, s, '')]


>>> ['123', '456', '789', '876', '567', '543']

Answer 3

>>> import re
>>> a = '123456789876567543'
>>> l = re.findall('.{1,3}', a)
>>> l
['123', '456', '789', '876', '567', '543']
>>> 

Answer 4

s = str(123456789876567543)
l = []
for i in xrange(0, len(s), 3):
    l.append(int(s[i:i+3]))
print l

Answer 5

如果你想要正确对齐：

a='123456789876567543'
format(int(a),',').split(',')
['123', '456', '789', '876', '567', '543']
a='12345'
format(int(a),',').split(',')
['12', '345']