Hi Nils,
yes, that's exactly the same problem! Thank you for reminding me. I've
totally forgotten how the "correct" encoding was called.
Best regards,
Simon
On 2020-01-22, Nils Bruin <nbr...@sfu.ca> wrote:
> On Wednesday, January 22, 2020 at 1:35:16 PM UTC-8, Simon King wrote:
>>
>> Here is an example:
>> Unpickling the data initially results in a string, s:
>> sage: s = '\x80\x1f'
>> I want it to be interpreted as the following bytes, b:
>> sage: b = b'\x80\x1f'
>>
>> How can I efficiently transform s into b? The following works, but I
>>
>
> This is exactly what we encountered on #28444, where we found that "latin1"
> decoding has a left-inverse (if decoding acts on the left) of "latin1"
> encoding. It looks like the encoding Py3 used for your py2-string was
> "latin1" (that's what we decided was the smart thing to do most of the time
> for possibly binary data), so
>
> s.encode(encoding="latin1")
>
> should do the trick. That should be pretty efficient.
>
>
>
>> doubt that it is very efficient:
>> sage: import struct
>> sage: struct.pack('{}B'.format(len(s)),*(ord(_) for _ in s)) == b
>> True
>>
>> Note that sage.cpython.string.str_to_bytes does't do what I need:
>> sage: sage.cpython.string.str_to_bytes(s)
>> b'\xc2\x80\x1f'
>>
>> Best regards,
>> Simon
>>
>>
>>
>
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