How to default to the default argument in Python

Question

So lets say I've got a function:

def print_something(a, b, c='something', d='something else'):
    print '{}/{}/{}/{}'.format(a, b, c, d)

And I've got this method:

def print_method(self, a, b):
    if self.c and self.b:
        print_something(a, b, self.c, self.d)
    elif self.c:
        print_something(a, b, c=self.c)
    elif self.d:
        print_something(a, b, d=self.d)
    else:
        print_something(a, b)

Is there a way to get the same functionality as print_method() without having to provide an if for each possible combination of self.c and self.d?

They could be many things. Maybe strings that are initially empty but may be appended to. — chmod 777 j
– chmod 777 j, Commented Feb 17, 2015 at 22:16

Peter DeGlopper · Accepted Answer · 2015-02-17 21:59:35Z

4

You can pass a dictionary to the function call containing keyword argument names and values. So something like this should work:

kwargs = dict((k, getattr(self, k)) for k in ('c', 'd') if getattr(self, k))
print_something(a, b, **kwargs)

answered Feb 17, 2015 at 21:59

Peter DeGlopper

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Comments

Padraic Cunningham · Accepted Answer · 2015-02-17 22:40:54Z

2

You could change your other function setting values to empty strings so the default will be used unless we get non empty strings passed in:

def print_something(a, b, c="", d=""):
    c = 'something' if not c else c
    d = 'something else' if not d else d
    print('{}/{}/{}/{}'.format(a, b, c, d))

def print_method(self, a, b):
        print_something(a, b, self.c, self.d)

Or more concisely as you consider any falsey value to be false in the if checks in your class we can use c = c or default

def print_something(a, b, c="", d=""):
    c = c or 'something' 
    d = d or 'something else' 
    print('{}/{}/{}/{}'.format(a, b, c, d))

edited Feb 17, 2015 at 22:40

answered Feb 17, 2015 at 22:22

Padraic Cunningham

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4 Comments

acushner Over a year ago

i always like the the more concise c = c or 'something' formulation.

Padraic Cunningham Over a year ago

@acushner, yes I have a bit of a mental block when it comes to using that but yes it will work , I will add it

Andrew Gorcester Over a year ago

I feel the x if x else y construct is more idiomatic.

Padraic Cunningham Over a year ago

@AndrewGorcester, I feel more comfortable using it too but I suppose in this case they are equivalent as we are not checking ` if x is not None` etc..

Kevin · Accepted Answer · 2015-02-18 03:36:18Z

-1

You could use **args here:

def print_method(self, a, b):
    extra_args = {}
    if self.c:
        extra_args['c'] = self.c
    if self.d:
        extra_args['d'] = self.d
    print_something(a, b, **extra_args)

This calls print_something() with however many extra arguments you have.

edited Feb 18, 2015 at 3:36

answered Feb 17, 2015 at 21:59

Kevin

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1 Comment

jgritty Over a year ago

This won't do the right thing. Because if self.d is set, but self.c is not, positionally, c will be assigned self.d, which seems counter to what the questioner desires.

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