python - Python - 将字符串拆分为成对字符列表

Question

可能重复：
如何在 Python 中将列表拆分为大小均匀的块？
按字符数拆分字符串

我有一个字符串（十六进制），例如：

717765717777716571a7202020

我需要把它变成以下格式：

0x71,0x77,0x65,0x71,0x77,0x77...

我不确定最好的方法是什么

Answer 1

>>> s = '717765717777716571a7202020'
>>> ['0x' + s[i:i+2] for i in range(0, len(s), 2)]
['0x71', '0x77', '0x65', '0x71', '0x77', '0x77', '0x71', '0x65', '0x71', '0xa7', '0x20', '0x20', '0x20']

如果您想要一个逗号分隔的字符串作为结果，您可以使用以下内容：

>>> ','.join('0x' + s[i:i+2] for i in range(0, len(s), 2))
'0x71,0x77,0x65,0x71,0x77,0x77,0x71,0x65,0x71,0xa7,0x20,0x20,0x20'

Answer 2

其他方式。

s = '717765717777716571a7202020'

print ','.join('0x'+''.join(d) for d in zip( *[iter(s)]*2 ))

输出：

0x71,0x77,0x65,0x71,0x77,0x77,0x71,0x65,0x71,0xa7,0x20,0x20,0x20