I want to avoid calling a lot of isinstance() functions, so I'm looking for a way to get the concrete class name for an instance variable as a string.
Any ideas?
- 2This is NOT a duplicate of a question that got me here. I'm reading the "Concrete Objects Layer" documentation for Python 3. That documentation describes C-level mappings with types that are lower-level than the regular class name. So, for example, the documentation describes a "PyLongObject". I'd like to know how to get this low-level name given an arbitrary object.Tom Stambaugh– Tom Stambaugh2018-06-17 18:59:06 +00:00Commented Jun 17, 2018 at 18:59
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3 Answers
instance.__class__.__name__ example:
>>> class A(): pass >>> a = A() >>> a.__class__.__name__ 'A' 
5 Comments
Chris
I prefer:
type(instance).__name__
Gabriel
What if you don't want to instantiate the class? can you just get the name of the class? (without the object)
bk0
For class A, use the following:
A.__name__
Martin Thoma
Is there a way to get it with the import path? For example,
str(type(foo)) == "<class 'never.gonna.give.youup'>" but type(foo).__name__ == "youup". How do I get never.gonna.give.youup?
xyxel
For path you can use type(foo).__module__
<object>.__class__.__name__ 
Comments
you can also create a dict with the classes themselves as keys, not necessarily the classnames
typefunc={ int:lambda x: x*2, str:lambda s:'(*(%s)*)'%s } def transform (param): print typefunc[type(param)](param) transform (1) >>> 2 transform ("hi") >>> (*(hi)*) here typefunc is a dict that maps a function for each type. transform gets that function and applies it to the parameter.
of course, it would be much better to use 'real' OOP
2 Comments
bobince
+1. An 'is' test on the class object itself will be quicker than strings and won't fall over with two different classes called the same thing.
Torsten Marek
Sadly, this falls flat on subclasses, but +1 for the hint to "real" OOP and polymorphism.