Dynamic instance attributes

Question

Say i have a class:

class Foo(object):
 def __init__(self,d):
 self.d=d
d={'a':1,'b':2}
inst=Foo(d)
inst.d
Out[315]: {'a': 1, 'b': 2}

Is there a way to dyamically create n attributes where each attribute would be a dict key, so inst.a would return 1 and so on.

• nneonneo: Yes, but you wouldn't be able to use them because inst.0 is illegal syntax.
• user395760: Yes, but why would you do that instead of accessing items by index, either straight from the list, or by overriding __getitem__ and friends.
• root: @ delnan -- it's just an example, not a very practical solution, I would just like to learn how to dynamically create attributes.

Answer 1

class Foo(object):
 def __init__(self, attributes):
 self.__dict__.update(attributes)

That would do it.

>>>foo = Foo({'a': 42, 'b': 999})
>>>foo.a
42
>>>foo.b
999

You can also use the setattr built-in method:

class Foo(object):
 def __init__(self, attributes):
 for attr, value in attributes.iteritems():
 setattr(self, attr, value)

Answer 2

use setattr():

>>> class foo(object):
 def __init__(self, d):
 self.d = d
 for x in self.d:
 setattr(self, x, self.d[x])
>>> d = {'a': 1, 'b': 2}
>>> l = foo(d)
>>> l.d
{'a': 1, 'b': 2}
>>> l.a
1
>>> l.b
2
>>> 

• Ashwini Chaudhary: @root yes, just move the for-loop inside the __init__().see the edited solution.

Answer 3

Here is a solution even more outlandish than the one offered by pythonm:

class Foo(object):
 def __init__(self, d):
 self.__dict__ = d

Instead of using inst.d, use inst.__dict__ directly. An added benefit is that new keys added to d automatically become attributes. That's as dynamic as it gets.

Answer 4

You could do something like this:

class Foo(object):
 def __init__(self, **kwdargs):
 self.__dict__.update(kwdargs)
d = {'a':1,'b':2}
foo = Foo(**d)
foo2 = Foo(a=1, b=2)

Answer 5

You can also use __getattr__.

class Foo(object):
 def __init__(self, d):
 self.d = d
 def __getattr__(self, name):
 return self.d[name]