将十六进制字符串拆分为Python中的列表 - 如何？

Question

如果我有这个字符串:

hexstring = '001122334455'

如何将其拆分为一个列表,结果如下:

hexlist = ['00','11','22','33','44','55']

我想不出一个好的,pythonic方式来做到这一点:/

Answer 1

>>> [hexstring[i:i+2] for i in range(0,len(hexstring), 2)]
['00', '11', '22', '33', '44', '55']

Answer 2

或者:

>>> hexstring = "01234567"
>>> it=iter(hexstring); [a+b for a,b in zip(it, it)]
['01', '23', '45', '67']

如果你的目标是Python 2.x,请使用itertools.izip而不是zip.

此方法是grouperitertools配方中的特定版本.

一些微观基准:

$ python2.6 -m timeit -s 'hexstring = "01234567"*500' '[hexstring[i:i+2] for i in xrange(0,len(hexstring), 2)]'
1000 loops, best of 3: 409 usec per loop

$ python2.6 -m timeit -s 'hexstring = "01234567"*500' '[hexstring[i:i+2] for i in range(0,len(hexstring), 2)]'
1000 loops, best of 3: 438 usec per loop

$ python2.6 -m timeit -s 'hexstring = "01234567"*500' 'it=iter(hexstring); [a+b for a,b in zip(it, it)]'
1000 loops, best of 3: 526 usec per loop

$ python2.6 -m timeit -s 'hexstring = "01234567"*500; from itertools import izip' 'it=iter(hexstring); [a+b for a,b in izip(it, it)]'
1000 loops, best of 3: 406 usec per loop

$ python2.6 -m timeit -s 'hexstring = "01234567"*500; import re; r=re.compile(".{1,2}"); f=r.findall' 'f(hexstring)'
1000 loops, best of 3: 458 usec per loop

$ python3.1 -m timeit -s 'hexstring = "01234567"*500' '[hexstring[i:i+2] for i in range(0,len(hexstring), 2)]'
1000 loops, best of 3: 756 usec per loop

$ python3.1 -m timeit -s 'hexstring = "01234567"*500' 'it=iter(hexstring); [a+b for a,b in zip(it, it)]'
1000 loops, best of 3: 414 usec per loop

$ python3.1 -m timeit -s 'hexstring = "01234567"*500; import re; r=re.compile(".{1,2}"); f=r.findall' 'f(hexstring)'
1000 loops, best of 3: 865 usec per loop


$ python2.6 -m timeit -s 'hexstring = "01234567"' '[hexstring[i:i+2] for i in xrange(0,len(hexstring), 2)]'
1000000 loops, best of 3: 1.52 usec per loop

$ python2.6 -m timeit -s 'hexstring = "01234567"' '[hexstring[i:i+2] for i in range(0,len(hexstring), 2)]'
1000000 loops, best of 3: 1.76 usec per loop

$ python2.6 -m timeit -s 'hexstring = "01234567"' 'it=iter(hexstring); [a+b for a,b in zip(it, it)]'
100000 loops, best of 3: 3.78 usec per loop

$ python2.6 -m timeit -s 'hexstring = "01234567"; from itertools import izip' 'it=iter(hexstring); [a+b for a,b in izip(it, it)]'
100000 loops, best of 3: 2.39 usec per loop

$ python2.6 -m timeit -s 'hexstring = "01234567"; import re; r=re.compile(".{1,2}"); f=r.findall' 'f(hexstring)'
1000000 loops, best of 3: 1.45 usec per loop

$ python3.1 -m timeit -s 'hexstring = "01234567"' '[hexstring[i:i+2] for i in range(0,len(hexstring), 2)]'
100000 loops, best of 3: 2.46 usec per loop

$ python3.1 -m timeit -s 'hexstring = "01234567"' 'it=iter(hexstring); [a+b for a,b in zip(it, it)]'
1000000 loops, best of 3: 1.84 usec per loop

$ python3.1 -m timeit -s 'hexstring = "01234567"; import re; r=re.compile(".{1,2}"); f=r.findall' 'f(hexstring)'
100000 loops, best of 3: 2.07 usec per loop

观察:

• 使用Python 2.6上的长字符串,@ Sharphost和我的方法是最快的.当然你需要使用懒惰的迭代器,例如xrange和izip.
• 使用Python 2.6上的短字符串,@ Nick的正则表达式是最快的.
• 在Python 3.1上,我的方法在两种情况下都是最快的,但我相信这是因为Python 3.x的优化程度较低.
• 当然,过早的优化是邪恶的等等.