Python 2 vs 3：从字节字符串中获取字节的结果一致

Question

Is there any simple way to get consistent results in both Python 2 and Python 3 for operatioIn like "give me N-th byte in byte string"? Getting either byte-as-integer or byte-as-character will do for me, as long as that will be consistent.

Ie given

s = b"123"

Naïve approach yields:

s[1] # => Python 2: '2', <type 'str'>
s[1] # => Python 3: 50, <class 'int'>

Wrapping that in ord(...) yields an error in Python 3:

ord(s[1]) # => Python 2: 50, <type 'int'> 
ord(s[1]) # => Python 3: TypeError: ord() expected string of length 1, but int found

I can think of a fairly complicated compat solution:

ord(s[1]) if (type(s[1]) == type("str")) else s[1] # 50 in both Python 2 and 3

... but may be there's an easier way which I just don't notice?

Answer 1

A length-1 slice will be also be a byte-sequence in either 2.x or 3.x:

s = b'123'
s[1:2] # 3.x: b'2'; 2.x: '2', which is the same thing but the repr() rules are different.

Answer 2

What about something like this?

import sys

if sys.version_info.major == 3:
    def index(s, n):
        return s[n]
elif sys.version_info.major == 2:
    def index(s, n):
        return ord(s[n])
else:
    raise NotImplementedError

Answer 3

If you use (converting if needed) the bytearray type, behavior will be indentical on both version, always matching Python 3 behavior. That's because bytearray is actually a distinct type on Python 2 (with Python 3 behavior), where bytes is just an alias for str there.

The more typical solution would be to use the six compatibility library, which provides six.indexbytes , so on either version of Python, you could do:

>>> six.indexbytes(s, 1)
50

Answer 4

Prefix your string with u and you'll get consistency across Python versions.

# Python 2
>>> ord(u"123"[0])
49

# Python 3
>>> ord(u"123"[0])
49