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Suppose I have a python function that takes two arguments, but I want the second arg to be optional, with the default being whatever was passed as the first argument. So, I want to do something like this:

def myfunc(arg1, arg2=arg1):
    print (arg1, arg2)

Except that doesn't work. The only workaround I can think of is this:

def myfunc(arg1, arg2=None):
    if arg2 is None:
        arg2 = arg1
    print (arg1, arg2)

Is there a better way to do this?

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I'm just rethinking this, it might be possible to achieve with a decorator. It's going to be unclear what it does at first glance so probably a bad idea, but it's an idea. –  marcog Jan 1 '11 at 19:39
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3 Answers

up vote 9 down vote accepted

As @Ignacio says, you can't do this. In your latter example, you might have a situation where None is a valid value for arg2. If this is the case, you can use a sentinel value:

sentinel = object()
def myfunc(arg1, arg2=sentinel):
    if arg2 is sentinel:
        arg2 = arg1
    print (arg1, arg2)

myfunc("foo")           # Prints 'foo foo'
myfunc("foo", None)     # Prints 'foo None'
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@martineau Thanks for the edit, nice spot! –  marcog Jan 1 '11 at 20:39
No problem -- excellent answer, BTW. –  martineau Jan 1 '11 at 20:52
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The second way is fine in my opinion: it is exactly clear what you do, and for all those who will come after you and read the code. While there is a slight overhead in documenting the default behavior, the use of 'None' or any other to make your function generate a default value is very common. Putting logic inside the function calls would certainly not be commendable in my opinion, it could become very complex very soon. So just leave it in the function body, where all the code is.

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Not really. The other argument names don't exist until the function is actually entered.

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But you can tidy it up slightly by changing the if clause in the workaround to: arg2 = arg2 or arg1 –  Michael Dunn Jan 1 '11 at 19:23
@Michael That will fail if arg2 is 0, False or an empty sequence. –  marcog Jan 1 '11 at 19:27
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