I know that default arguments are created at the function initialisation time and not every time the function is called.  See the following code:def ook (item, lst=[]):    lst.append(item)    print 'ook', lstdef eek (item, lst=None):    if lst is None: lst = []    lst.append(item)    print 'eek', lstmax = 3for x in xrange(max):    ook(x)for x in xrange(max):    eek(x)What I do not get is why this was implemented this way.  What benefits does this behaviour offers over an initialisation at each call time?
Python mutable default argument: Why?
python
I think the reason is implementation simplicity. Let me elaborate.The default value of the function is an expression that you need to evaluate. In your case it is a simple expression that does not depend on the closure, but it can be something that contains free variables - def ook(item, lst = something.defaultList()). If you are to design Python, you will have a choice - do you evaluate it once when the function is defined or every time when the function is called. Python chooses the first (unlike Ruby, which goes with the second option).There are some benefits for this.First, you get some speed and memory boosts. In most cases you will have immutable default arguments and Python can construct them just once, instead of on every function call. This saves (some) memory and time. Of course, it doesn't work quite well with mutable values, but you know how you can go around.Another benefit is the simplicity. It's quite easy to understand how the expression is evaluated - it uses the lexical scope when the function is defined. If they went the other way, the lexical scope might change between the definition and the invocation and make it a bit harder to debug. Python goes a long way to be extremely straightforward in those cases.