Is it possible to change a function's default parameters in Python?
Just use functools.partial
multiplyNumbers = functools.partial(multiplyNumbers, y = 42)
One problem here: you will not be able to call it as multiplyNumbers(5, 7, 9);
you should manually say y=7
If you need to remove default arguments I see two ways:
Store original function somewhere
oldF = f f = functools.partial(f, y = 42) //work with changed f f = oldF //restore
use
partial.func
f = f.func //go to previous version.
Technically, it is possible to do what you ask… but it's not a good idea. RiaD's answer is the Pythonic way to do this.
In Python 3:
>>> def f(x=1, y=2, z=3):
... print(x, y, z)
>>> f()
1 2 3
>>> f.__defaults__ = (4, 5, 6)
4 5 6
As with everything else that's under the covers and hard to find in the docs, the inspect
module chart is the best place to look for function attributes.
The details are slightly different in Python 2, but the idea is the same. (Just change the pulldown at the top left of the docs page from 3.3 to 2.7.)
If you're wondering how Python knows which defaults go with which arguments when it's just got a tuple… it just counts backward from the end (or the first of *
, *args
, **kwargs
—anything after that goes into the __kwdefaults__
dict instead). f.__defaults = (4, 5)
will set the defaults to y
and z
to 4
and 5
, and with default for x
. That works because you can't have non-defaulted parameters after defaulted parameters.
There are some cases where this won't work, but even then, you can immutably copy it to a new function with different defaults:
>>> f2 = types.FunctionType(f.__code__, f.__globals__, f.__name__,
... (4, 5, 6), f.__closure__)
Here, the types
module documentation doesn't really explain anything, but help(types.FunctionType)
in the interactive interpreter shows the params you need.
The only case you can't handle is a builtin function. But they generally don't have actual defaults anyway; instead, they fake something similar in the C API.