bug: regression: slow to compute name of deeply nested expressions

[NickCrews]

What happened?

import ibis

x = ibis.literal(1)
for i in range(100):
    x = x + 1
    # if I hardcode the name, this is instantaneous
    # x = x.name(f"step_{i}")
x

If I ctrl-c this after a few seconds, I see see eg

File ~/code/ibis/ibis/expr/operations/core.py:89, in Value.name(self)
     86 @property
     87 def name(self) -> str:
     88     names = (arg.name for arg in self.__args__ if hasattr(arg, "name"))
---> 89     return f"{self.__class__.__name__}({', '.join(names)})"

File ~/code/ibis/ibis/expr/operations/core.py:88, in <genexpr>(.0)
     86 @property
     87 def name(self) -> str:
---> 88     names = (arg.name for arg in self.__args__ if hasattr(arg, "name"))
     89     return f"{self.__class__.__name__}({', '.join(names)})"

    [... skipping similar frames: <genexpr> at line 88 (97 times), Value.name at line 89 (97 times)]

File ~/code/ibis/ibis/expr/operations/core.py:89, in Value.name(self)
     86 @property
     87 def name(self) -> str:
     88     names = (arg.name for arg in self.__args__ if hasattr(arg, "name"))
---> 89     return f"{self.__class__.__name__}({', '.join(names)})"

File ~/code/ibis/ibis/expr/operations/core.py:88, in <genexpr>(.0)
     86 @property
     87 def name(self) -> str:
---> 88     names = (arg.name for arg in self.__args__ if hasattr(arg, "name"))
     89     return f"{self.__class__.__name__}({', '.join(names)})"

What version of ibis are you using?

main 09b6ada

What backend(s) are you using, if any?

NA

Relevant log output

No response

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[cpcloud]

Thanks!

If this is noticeably slow then it's probably recursing infinitely.

[cpcloud]

Ok, it's not infinite: reducing the number of iterations to 10 returns quickly.