Question

I’m not even able to properly search google for it, but here goes:

a = {}
b = {}
c = [a, b]
for d in c:
    d['ID'] = d
print c

returns:

[{'ID': {...}}, {'ID': {...}}]

why isn’t it:

[{'ID': a}, {'ID': b}]

Answer 1

Let’s step through this:

a = {}
b = {}
c = [a, b]

So far, so good.

for d in c:
    d['ID'] = d

We can unroll this to:

d = c[0]
d['ID'] = d
d = c[1]
d['ID'] = 1

And expand that to:

d = a
d['ID'] = d
d = b
d['ID'] = d

Now substitute:

a['ID'] = a
b['ID'] = a

So, let’s forget about the loop for a second and look at what that does:

>>> a = {}
>>> a['ID'] = a
>>> a
{'ID': {...}}

In other words, you’re making each dict recursively contain a copy of itself, under the key ID. How would you expect this to be printed?

So, the obvious thing to do is to try to print the whole dictionary:

{'ID': {'ID': {'ID': { …

But this would be an infinitely-long string, and Python would run out of stack space before reaching infinity. So it needs to truncate it somehow.

It can’t print this:

{'ID': a}

Because a is just a name that happens to be bound to the dict, just like d is at the time. In fact, the loop doesn’t even know that a is bound to that at the time; it knows that d is. But even if it did know, the result would be wrong. Think about this:

>>> e = a
>>> a = 0
>>> e
???

So, the obvious answer is to use an ellipsis (kind of like I did in the human-readable version) to represent “and so on”.