Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
452 views
in Technique[技术] by (71.8m points)

python - Dictionary class attribute that refers to other class attributes in the definition

While there are numerous ways around this, because of a personality fault I can't let it go until I understand the nature of the failure.

Attempting:

class OurFavAnimals(object):
    FAVE = 'THATS ONE OF OUR FAVORITES'
    NOTFAVE = 'NAH WE DONT CARE FOR THAT ONE'
    UNKNOWN = 'WHAT?'
    FAVES = defaultdict(lambda: UNKNOWN, {x:FAVE for x in ['dog', 'cat']})
    FAVES['Crab'] = NOTFAVE 

Fails with:

      3     NOTFAVE = 'NAH WE DONT CARE FOR THAT ONE'
      4     UNKNOWN = 'WHAT?'
----> 5     FAVES = defaultdict(lambda: UNKNOWN, {x:FAVE for x in ['dog', 'cat']})
      6     FAVES['Crab'] = NOTFAVE

NameError: global name 'FAVE' is not defined

Why? Why can it find UNKNOWN but not FAVE? Is it because it's in a dictionary comprehension?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Yes, it's because it's in a dictionary comprehension. Note that it's not "finding" UNKNOWN either; it's just not looking for it yet, because UNKNOWN is only referenced in a lambda. If you replace your dict comprehension with something else to allow the class definition to succeed, you'll get an error later if you try to access a nonexistent key (because then it will try to call that lambda). So if you change it to

FAVES = defaultdict(lambda: UNKNOWN, {'a': 1})

You'll get:

>>> OurFavAnimals.FAVES['x']
Traceback (most recent call last):
  File "<pyshell#4>", line 1, in <module>
    OurFavAnimals.FAVES['x']
  File "<pyshell#2>", line 5, in <lambda>
    FAVES = defaultdict(lambda: UNKNOWN, {'a': 1})
NameError: global name 'UNKNOWN' is not defined

In both cases, the reason is that variables defined in the class scope are not available in nested scopes. In other words, it's the same reason this fails:

class Foo(object):
    something = "Hello"
    def meth(self):
        print(something)

Both the lambda and the dictionary comprehension create function scopes that are nested in the class scope, so they don't have access to the class variables directly. See also this related question.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...