Is there a way to pass optional parameters to a function?

Question

Is there a way in Python to pass optional parameters to a function while calling it and in the function definition have some code based on “only if the optional parameter is passed”

Asked By: user1795998

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Answer 1

def my_func(mandatory_arg, optional_arg=100):
    print(mandatory_arg, optional_arg)

http://docs.python.org/2/tutorial/controlflow.html#default-argument-values

I find this more readable than using **kwargs.

To determine if an argument was passed at all, I use a custom utility object as the default value:

MISSING = object()

def func(arg=MISSING):
    if arg is MISSING:
        ...

Answered By: warvariuc

Answer 2

If you want give some default value to a parameter assign value in (). like (x =10). But important is first should compulsory argument then default value.

eg.

(y, x =10)

but

(x=10, y) is wrong

Answered By: sumit

Answer 3

The Python 2 documentation, 7.6. Function definitions gives you a couple of ways to detect whether a caller supplied an optional parameter.

First, you can use special formal parameter syntax *. If the function definition has a formal parameter preceded by a single *, then Python populates that parameter with any positional parameters that aren’t matched by preceding formal parameters (as a tuple). If the function definition has a formal parameter preceded by **, then Python populates that parameter with any keyword parameters that aren’t matched by preceding formal parameters (as a dict). The function’s implementation can check the contents of these parameters for any “optional parameters” of the sort you want.

For instance, here’s a function opt_fun which takes two positional parameters x1 and x2, and looks for another keyword parameter named “optional”.

>>> def opt_fun(x1, x2, *positional_parameters, **keyword_parameters):
...     if ('optional' in keyword_parameters):
...         print 'optional parameter found, it is ', keyword_parameters['optional']
...     else:
...         print 'no optional parameter, sorry'
... 
>>> opt_fun(1, 2)
no optional parameter, sorry
>>> opt_fun(1,2, optional="yes")
optional parameter found, it is  yes
>>> opt_fun(1,2, another="yes")
no optional parameter, sorry

Second, you can supply a default parameter value of some value like None which a caller would never use. If the parameter has this default value, you know the caller did not specify the parameter. If the parameter has a non-default value, you know it came from the caller.

Answered By: Jim DeLaHunt

Answer 4

def op(a=4,b=6):
    add = a+b
    print add

i)op() [o/p: will be (4+6)=10]
ii)op(99) [o/p: will be (99+6)=105]
iii)op(1,1) [o/p: will be (1+1)=2]
Note:
 If none or one parameter is passed the default passed parameter will be considered for the function.

Answered By: Sanyal

Answer 5

You can specify a default value for the optional argument with something that would never passed to the function and check it with the is operator:

class _NO_DEFAULT:
    def __repr__(self):return "<no default>"
_NO_DEFAULT = _NO_DEFAULT()

def func(optional= _NO_DEFAULT):
    if optional is _NO_DEFAULT:
        print("the optional argument was not passed")
    else:
        print("the optional argument was:",optional)

then as long as you do not do func(_NO_DEFAULT) you can be accurately detect whether the argument was passed or not, and unlike the accepted answer you don’t have to worry about side effects of ** notation:

# these two work the same as using **
func()
func(optional=1)

# the optional argument can be positional or keyword unlike using **
func(1) 

#this correctly raises an error where as it would need to be explicitly checked when using **
func(invalid_arg=7)

Answered By: Tadhg McDonald-Jensen

Is there a way to pass optional parameters to a function?

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