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I had a function in Python 2 which I used to generate a 22 length random string from a UUID....

def make_base64_string():
    return uuid.uuid4().bytes.encode("base64")[:22]

I have since started to test in Python 3 and just finished watching the Pragmatic Unicode presentation, most of which went way over my head. Anyway, I didn't assume this function would now work in Python 3.4 and I was right....

So next, I tried what I was hoping was the solution giving that bytes.encode from my understating is gone... (treat everything as a byte, right?).

base64.b64encode(bytes(uuid.uuid4(), 'utf-8'))

but this gives me the following error...

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: encoding or errors without a string argument

Why? I would like to understand this more.

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  • 1
    This is what I suspect, but what I did not confirm: uuid.uuid4() does not return bytes, but UUID object. Thus the encoder cannot properly handle it as this object might not have default string presentation anymore. Commented May 22, 2015 at 12:32
  • You haven't given enough of a traceback for the first error to be useful Commented May 22, 2015 at 12:32
  • Some useful answers that really helped my understanding thank you. Commented May 22, 2015 at 12:39
  • Helpful reference: python-future.org/… Commented Aug 19, 2016 at 23:42

2 Answers 2

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base64.b64encode takes bytes as argument, no need to decode-re-encode.

>>> base64.b64encode(uuid.uuid4().bytes)
b'58jvz9F7QXaulSScqus0NA=='
>>> base64.b64encode(uuid.uuid4().bytes)
b'gLV2vn/1RMSSckMd647jUg=='
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7 Comments

but this would not take the actual UUID but the byte representation, right?
@OrbiterFleet yes, which is the actual UUID. Also your original code in python2 did that (and the slice at 22. byte).
Yep sorry the code did do that with uuid.uuid4().bytes. I understand now!
The confusion stems usually from the fact that str in python2 is bytes, aka strings in python2 are byte strings, and you have to do extra to get unicode, like u"stuff" literals and unicode(stuff) calls. It takes some practice to understand the difference, but it's worth it, at least in my opinion :P (since we have to handle weird chars like ä and å all the time).
Yes, it is an UUID object wrapping the actual byte representation, like @Mohammadhzp points out. It offers helpful methods for inspecting and using the value.
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1 
>>> type(uuid.uuid4())
<class 'uuid.UUID'>

This shows that uuid.uuid4() is not a string and you need to pass an string to bytes, so here is the working code:

base64.b64encode(bytes(str(uuid.uuid4()), 'utf-8'))
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5 Comments

OK good that makes sense. But why do we need bytes( if as in @Ilja answer b64encode takes bytes as argument. I would have guess that base64.b64encode(str(uuid.uuid4()), 'utf-8') should have worked but I get the error 'str' does not support the buffer interface
Your question was about why this code is not working and I want to understand it more, I answered it's because you are passing an object instead of a string to bytes(), the other answer is certainly better but my answer is the correct reply for your question,base64::b64encode() expect parameter one to be bytes not a string and does not take utf-8 for second parameter so base64.b64encode(str(uuid.uuid4()), 'utf-8') must not work
I voted this answer up, not sue why others voted it down. maybe someone else could reply I'm interested also.
the second part of your answer tho takes a string uuid and gives a different result to mine using uuid.uuid4().bytes. I could be wrong but its taking the string value and retuning that into a byte string, right?
No, byte string is different, in this case, bytes() knows this is a string which turns into bytes, bytes() is useful for ASCII characters which means it's ok in here, you get different result because byte order in uuid.UUID().bytes is big-endian(my best guess)

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