CODEWITHSUNDEEP
pythonpython-3.x
Coder
Coder

Reputation: 455

Call function with multiple optional arguments of different types

I have already checked this post and this post, but couldn't find a good way of sorting the problem of my code out.

I have a code as follows:

class foo:
    def __init__(self, foo_list, str1, str2):

        self.foo_list = foo_list
        self.str1 = str1
        self.str2 = str2

    def fun(self, l=None, s1=None, s2=None):

        if l is None:
            l = self.foo_list

        if s1 is None:
            s1 = self.str1

        if s2 is None:
            s2 = self.str2

        result_list = [pow(i, 2) for i in l]

        return result_list, s1[-1], len(s2)

Then I create "f" and call "fun" function:

f = foo([1, 2, 3, 4], "March", "June")
print(f.fun())

The output is:

([1, 4, 9, 16], 'h', 4)

which is correct, but if I do:

print(f.fun("April"))

I get the following error:

TypeError: unsupported operand type(s) for ** or pow(): 'str' and 'int'

Apparently, python confuses the string argument "April" with the list, how do I fix it?

Upvotes: 5

Views: 1886

Answers (1)

Niko Fohr
Niko Fohr

Reputation: 33988

By default, the first argument passed to the function will be assigned to the first parameter. If you want to assign the first argument to the second (or n:th) parameter, you must give it as keyword argument. See, for example

In [19]: def myfunc(x='X', y=5):
    ...:     print(x,y)
    ...:
    ...:

# No arguments -> Using default parameters
In [20]: myfunc()
X 5

# Only one positional argument -> Assigned to the first parameter, which is x
In [21]: myfunc(100)
100 5

# One keyword argument -> Assigned by name to parameter y
In [22]: myfunc(y=100)
X 100

The type of the arguments do not matter, but the order you used in the function definition.

Notes on terminology

  • By parameter, I mean the variable in the function definition
  • By argument, I mean the actual value passed to the function.

Upvotes: 5

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