How to call python function with optional parameters
I have the following function:
def create(self, name, page=5, opt1=False, opt2=False,
opt3=False,
opt4=False,
opt5=False,
opt6=False,
*parameters):
Is it possible to assign only one of the optional parameters and some * parameters? eg.
create('some name', opt4=True, 1, 2, 3) # I need 1, 2, 3 to be assigned to *parameters
Most of the time, I don't need to change the opt1 ... opt6 values, I only need to change one of them and assign another one *parameters
. So I'm looking for a way to avoid installing opt1 ... opt6 if I don't want to change their default.
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In Python 3.x try putting parameters
before any of the default arguments, for example -
def create(self, name , *parameters, page=5, opt1=False,
opt2=False,opt3=False,
opt4=False,opt5=False,
opt6=False):
print(parameters)
print(page)
print(opt4)
In [31]: create('C', 'some name', 1, 2, 3, opt4 = True)
(1, 2, 3)
5
True
Please understand that with this method you can only call arguments with a default value using named arguments.
For Python 2.x you can try using *parameters
and then **kwargs
for default parameters -
>>> def create(self, name, *parameters, **kwargs):
... kwargs.setdefault('opt4',False) #You will need to do for all such parameters.
... print(parameters)
... print(kwargs['opt4'])
...
>>> create('C', 'some name', 1, 2, 3, opt4 = True)
(1, 2, 3)
True
Again, just a way to set values for opt1...opt6
or page
etc. will use named arguments.
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Well, for sure it's not nice, but you can probably work on variations of this
def create(name, page=5, opt1=False, opt2=False,
opt3=False,
opt4=False,
opt5=False,
opt6=False,
*parameters):
print(parameters)
print(page)
print(opt4)
myArgs = dict(zip(['page','opt1','opt2','opt3','opt4','opt5','opt6'],create.func_defaults))
myArgs['opt4']=True
create("MyName",**myArgs)
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