[Python-ideas] Reference variable in assignment: x = foo(?)

[Steven D'Aprano]

On 12/07/13 07:07, Corey Sarsfield wrote:
> I came up with the idea after having some code on dicts that looked like:
>
> a[b][c] = foo(a[b][c])
>
> So in this case there are twice as many look-ups going on as there need to
> be, even if a[b][c] were to be pulled out into x.

How do you reason that? You need two sets of lookups however you do it: first you call __getitem__ 'b' and 'c', then you call __setitem__. In the general case, you can't cache the reference 1) because that's not how references work in Python, and 2) even if it were, any of the __getitem__ or __setitem__ calls may have side-effects.


> If I were to do:
>
> a[b][c] += 1
>
> Would it be doing the lookups twice behind the scenes?


Easy enough to find out with a helper class:

class Test(list):
     def __getitem__(self, attr):
         print("Looking up item %s" % attr)
         return super(Test, self).__getitem__(attr)
     def __setitem__(self, attr, value):
         print("Setting item %s to %r" % (attr, value))
         return super(Test, self).__setitem__(attr, value)

instance = Test([1, 2, 3])
instance[1] = Test([4, 5, 6])

instance[1][2] += 100

print(instance)



-- 
Steven