What is this doing (Python)

Question

I came across a piece of code today that looks like this:

class ClassName(object):
     def __init__(self):
         self._vocabulary = None

     def vocabulary(self):
         self._vocabulary = self._vocabulary or self.keys()
         return self._vocabulary

What exactly is the line self._vocabulary = self._vocabulary or self.keys() doing?

The first part of your code snippet is invalid Python syntax. — Sven Marnach, May 04 '11 at 19:25

Santiago Alessandri · Accepted Answer · 2011-05-05T10:54:55.530

8

A line like this:

self._vocabulary = self._vocabulary or self.keys()

Is what's called lazy initialization, when you are retrieving the value for the first time it initializes. So if it has never been initialized self._vocabulary will be None (because the __init__ method had set this value) resulting in the evaluation of the second element of the or so self.keys() will be executed, assigning the returning value to self._vocabulary, thus initializing it for future requests.

When a second time vocabulary is called self._vocabulary won't be None and it will keep that value.

edited May 05 '11 at 10:54

answered May 04 '11 at 19:28

Santiago Alessandri

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Upvoted, but it would be good to clarify that it is the explicit line in `__init__` that sets `self._vocabulary` to `None`, it doesn't happen automatically. – ncoghlan May 05 '11 at 06:26
@ncoghlan Thanks for the vote and for the advice. I edited the post, is it clear now? – Santiago Alessandri May 05 '11 at 10:56

score 2 · Answer 2 · answered May 04 '11 at 19:27

2

In a nutshell, if self._vocabulary evaluates to a logical false (e.g. if it's None, 0, False, etc), then it will be replaced with self.keys().

The or operator in this case, returns whichever value evaluates to a logical true.

Also, your code should look more like this:

class Example(object):
     def __init__(self):
         self._vocabulary = None

     def vocabulary(self):
         self._vocabulary = self._vocabulary or self.keys()
         return self._vocabulary

     def keys(self):
         return ['a', 'b']

ex = Example()
print ex.vocabulary()
ex._vocabulary = 'Hi there'
print ex.vocabulary()

answered May 04 '11 at 19:27

Joe Kington

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`The or operator in this case, returns whichever value evaluates to a logical true.` - This is a little misleading. If `self._vocabulary` evaluates to `False` then `self.keys()` is executed and used for the assignment, whether or not it evaluates as `True` or `False`. See http://stackoverflow.com/questions/1452489/evaluation-of-boolean-expressions-in-python/1452500#1452500 for explanation of how things evaluate in boolean expressions. – MattH May 04 '11 at 19:37
@dave - No, lots of other languages have similar operators. For example, the `||` operator does the same thing in ruby. The equivalent would be `vocabulary ||= keys()` or `vocabulary = (vocabulary || keys())` in ruby. It's common of plenty of other languages besides ruby and python, as well. – Joe Kington May 04 '11 at 19:39
@MattH - I used slightly poor phrasing, but that's _exactly_ what I was saying. I.e. logical testing, not identity testing whether it `is False`. (Thus the lowercase "logical false" and not the object `False`) Edit: whoops, I misread your comment. Good point... – Joe Kington May 04 '11 at 19:41

score 0 · Answer 3 · answered May 04 '11 at 19:28

Hard to say, the code will not run for a number of reasons. To guess though, I'd say it looks like it would be evaluated as a logical expression, self._vocabulary would be evaluated False by python as type None, and self.keys() is a method that would (hopefully) return something to be evaluated as well. Then it's just a logical OR between the two and the result gets put into self._vocabulary.

What is this doing (Python)

3 Answers3