Create lambdas that use *args in a loop

Asked Dec 22 '20 at 19:56

Active Dec 22 '20 at 19:56

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I've found some other questions that deal with the problem of creating lambdas inside loops, and the last value in the iteration being assigned to all lambda functions. While the solution is generally using lambda x=x: instead of lambda x:, or using functools.partial, what should I do when the lambda expects *args?

>>> keys = ['a', 'b', 'c']
>>> for func in [lambda *args: print(x, args) for x in keys]:
...     func(1, 2, 3)
... 
c (1, 2, 3)
c (1, 2, 3)
c (1, 2, 3)

Expected result:

a (1, 2, 3)
b (1, 2, 3)
c (1, 2, 3)

The actual code is:

for event in emitter_events:
    self.ws.on(event, lambda *args: logging.debug(f"'{event}': {repr(args)}"))

asked Dec 22 '20 at 19:56

admirabilis

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`lambda x=x:` is a hack. In general, the "correct" solution is to force the closure to refer to a variable that does not change by creating another enclosing scope. One way, `lambda event: (lambda event: lambda *args: logging.debug(f"'{event}': {repr(args)}"))(event)` – juanpa.arrivillaga Dec 22 '20 at 20:33
1

However, it is better to create a factory function: `def make_func(event): return lambda *args: logging.debug(...)` and then use *that* in your loop: `self.ws.on(event, make_func(event))` – juanpa.arrivillaga Dec 22 '20 at 20:34
Note, all of this isn't really unique to lambda expressions. – juanpa.arrivillaga Dec 22 '20 at 20:40
@juanpa.arrivillaga Thanks, that works! Why is the second option better, though? – admirabilis Dec 22 '20 at 21:43
Because it is much cleaner. – juanpa.arrivillaga Dec 22 '20 at 21:44

Create lambdas that use *args in a loop

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