How to get the concrete class name as a string?

Question

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?

This 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, Jun 17 '18 at 18:59

score 333 · Accepted Answer · answered Feb 06 '09 at 18:20

333

 instance.__class__.__name__

example:

>>> class A():
    pass
>>> a = A()
>>> a.__class__.__name__
'A'

answered Feb 06 '09 at 18:20

SilentGhost

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77

I prefer: `type(instance).__name__` – Chris May 08 '11 at 18:58
5

What if you don't want to instantiate the class? can you just get the name of the class? (without the object) – Gabriel Oct 15 '13 at 23:17
48

For class A, use the following: `A.__name__` – bk0 Nov 13 '13 at 02:04
2

Is there a way to get it with the import path? For example, `str(type(foo)) == ""` but `type(foo).__name__ == "youup"`. How do I get `never.gonna.give.youup`? – Martin Thoma Apr 29 '19 at 08:00
1

For path you can use type(foo).__module__ – xyxel Oct 06 '20 at 12:47

score 32 · Answer 2 · answered Feb 06 '09 at 18:23

32

<object>.__class__.__name__

answered Feb 06 '09 at 18:23

Morendil

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Javier · Answer 3 · 2009-02-06T19:22:27.563

9

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

edited Feb 06 '09 at 19:22

answered Feb 06 '09 at 19:15

Javier

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+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. – bobince Feb 06 '09 at 19:19
Sadly, this falls flat on subclasses, but +1 for the hint to "real" OOP and polymorphism. – Torsten Marek Feb 06 '09 at 19:25

How to get the concrete class name as a string?

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