Understanding __call__ and list.sort(key)

Question

I have the following code I am trying to understand:

>>> class DistanceFrom(object):
        def __init__(self, origin):
            self.origin = origin
        def __call__(self, x):
            return abs(x - self.origin)  

>>> nums = [1, 37, 42, 101, 13, 9, -20]
>>> nums.sort(key=DistanceFrom(10))
>>> nums
[9, 13, 1, 37, -20, 42, 101]

Can anyone explain how this works? As far as I have understood, __call__ is what is called when object() is called - calling the object as a function.

What I don't understand is how nums.sort(key=DistanceFrom(10)). How does this work? Can anyone please explain this line?

Thanks!

Answer 1

__call__ in python allows a class to be run as if it's a function. You can try this out manually:

>>> dis = DistanceFrom(10)
>>> print dis(10), dis(5), dis(0)
0 5 10
>>> 

What sort does is call that function for every item in your list and uses the returned value as sort key. In this example you'll get a list back with the items closest to 10 first, and the one further away more towards the end.

Answer 2

Here I have defined a function DistanceFrom() which can be used in a similar way to your class, but might be easier to follow

>>> def DistanceFrom(origin):
...     def f(x):
...         retval = abs(x - origin)
...         print "f(%s) = %s"%(x, retval)
...         return retval
...     return f
... 
>>> nums = [1, 37, 42, 101, 13, 9, -20]
>>> nums.sort(key=DistanceFrom(10))
f(1) = 9
f(37) = 27
f(42) = 32
f(101) = 91
f(13) = 3
f(9) = 1
f(-20) = 30
>>> nums
[9, 13, 1, 37, -20, 42, 101]

So you see that the object returned by DistanceFrom is called once for each item of nums and then nums is returned sorted in accordance with the returned values