Moving elements of a python dictionary to another one under certain condition using less code

OK, so this is more a code "optimization" exercise.

I have to move all elements from a python dictionary to another, under certain condition, while emptying the source dict at the same time. (No matter I find matching elements or not, the source dict must be empty by the end)

Let's have:

pie={"A1":{2000,2001,2002},"A2":{2003,2004,2005},"A3":{2000,2004,2007}} ; slices={}
condition_check=2000

As I said, I needed to move the elements to the destination dict only when they meet a certain condition. Match is made on keys and/or values while emptying the source dict, and thus I started with this code (in this case the condition is on values):

while bool(pie):
    temp = pie.popitem()
    if (condition_check in temp[1]): slices.update({temp[0]:temp[1]})

Now the question is: is there a way to do without the temp = pie.popitem() line? (I thought perhaps it would be possible using an assignment expression to put it within the if condition maybe?)

P.S. I know I could go with

for s in [{k:v} for k,v in pie.items() if (condition_check in v)]: slices.update(s)
pie.clear()

or, alternatively

s = {k:v for k,v in pie.items() if (condition_check in v)} ; slices.update(s)
pie.clear()

Finally, I think slices.update({k:v for k,v in pie.items() if (condition_check in v)}) ; pie.clear() could suit well

Yet, I'd still like optimizing the first code in such a way to "move/merge" the temp = pie.popitem() assignment inside the if condition somehow... And I'm just curious to know if is it possible to achieve something like that anyway tbh

EDIT:

As @KellyBundy noted on comments : indeed, the thing to really assure here was that the source dictionary had to be empty by the end before going onward. Within the while loop this is done step by step, using popitem(). But also a clear() as very first instruction immediately after the for comprehension is fine for this

For clean code, you just want something like:

>>> slices = {}
>>> slices.update(item for item in pie.items() if condition_check in item[1])
>>> pie.clear()
>>> pie
{}
>>> slices
{'A1': {2000, 2001, 2002}, 'A3': {2000, 2004, 2007}}
>>> condition_check
2000

That, of course, is assuming that slices is already defined, if you were just going to define it:

slices = {k:v for k, v in pie.items() if condition_check in v}
pie.clear()

You could use an assignment expression in your original construction, but that doesn't really optimize as much as obfuscate. Stop trying to put everything on one line, stick to PEP 8 style guidelines and use indentation after conditions, etc. But just to show you how:

>>> pie = {"A1":{2000,2001,2002},"A2":{2003,2004,2005},"A3":{2000,2004,2007}} ; slices={}
>>> slices = {}
>>> while pie:
...     if condition_check in (item := pie.popitem())[1]:
...         slices[item[0]] = item[1]
...
>>> pie
{}
>>> slices
{'A3': {2000, 2004, 2007}, 'A1': {2000, 2001, 2002}}

Also note, there's no need for while bool(pie): ... because bool() will be called implicitly in boolean contexts like the condition of a while loop.

But the above code is ugly and unpythonic. The following is so much more readable and cleaner:

while pie:
    key, values = pie.popitem()
    if condition_check in values:
        slices[key] = values

More lines are good. They help make your code more readable. But again, I would just use one of the first two instead of this while loop.