Question

What does any = lambda v: v mean? It seems v is only v itself.

class Object(object):                   
    """Common base class supporting automatic kwargs->attributes handling,
    and cloning."""
    attrs = ()

    def __init__(self, *args, **kwargs):
        any = lambda v: v
        for name, type_ in self.attrs:  
            value = kwargs.get(name)    
            if value is not None:       
                setattr(self, name, (type_ or any)(value))
            else:                       
                try:                    
                    getattr(self, name) 
                except AttributeError:  
                    setattr(self, name, None)

Answer 1

lambda v: v creates an identity function, which just returns its argument unchanged. Assigning it to a local variable is equivalent to defining a local function like this:

def any(v):
    return v

It can be useful as a fallback for code that wants to call a function to do some processing on the argument, for the cases where the real function is not available, or the processing is undesired.

In the code you posted, type_ can be logically false (most likely None), which means that it is not to be called, so it’s replaced with an identity function. The author could also have used a more explicit if to skip the function call in that case, at the price of additional clutter in the loop.

BTW any is a bad name for a local variable because it shadows the built-in function with the same name and a completely different meaning.