[python] Is there a way to pass optional parameters to a function?

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"

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

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:
        ...

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)

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.