Can you assign variables in a lambda?

Question

I was using a lambda statement to perform math, and happened to repeatedly use one certain value. Therefore I was wondering if it was possible to assign and use

Answer 1

Nope, you can't. Only expressions allowed in lambda:

lambda_expr        ::=  "lambda" [parameter_list]: expression
lambda_expr_nocond ::=  "lambda" [parameter_list]: expression_nocond

You could, however, define a second lambda inside the lambda and immediately call it with the parameter you want. (Whether that's really better might be another question.)

>>> a = lambda n: ((3+2*n), n*(3+2*n))  # for reference, with repetition
>>> a(42)
(87, 3654)
>>> a2 = lambda n: (lambda b: (b, n*b))(3+2*n)  # lambda inside lambda
>>> a2(42)
(87, 3654)
>>> a3 = lambda n: (lambda b=3+2*n: (b, n*b))()  # using default parameter
>>> a3(42)
(87, 3654)

Of course, both the outer and the inner lambda can have more than one parameter, i.e. you can define multiple "variables" at once. The benefit of this approach over, e.g., defining a second lambda outside of the first is, that you can still also use the original parameters (not possible if you invoked a with b pre-calculated) and you have to do the calculation for b only once (other than repeatedly invoking a function for the calculation of b within a).

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Also, inspired by the top answer to the linked question, you could also define one or more variables as part of a list comprehension or generator within the lambda, and then get the next (first and only) result from that generator or list:

>>> a4 = lambda n: next((b, n*b) for b in [3+2*n])
>>> a4(42)
(87, 3654)

However, I think the intent behind the lambda-in-a-lambda is a bit clearer. Finally, keep in mind that instead of a one-line lambda, you could also just use a much clearer three-line def statement...

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Also, starting with Python 3.8, there will be assignment expressions, which should make it possible to write something like this. (Note that I could not try/verify this as I do not have Python 3.8 yet.)

>>> a5 = lambda n: ((b := 3+2*n), n*b))