The actual factorial is nearly as you'd anticipate it. You gather that the a is... the factorial function. b is the real parameter.
<factorial> = lambda a, b: b*a(a, b-1) if b > 0 else 1
This is actually a application of the factorial
<factorial-application> = (lambda a, b: a(a, b))(<factorial>, b)
Here, a is simply the factorial function. It accepts itself as its first argument, and the evaluation point will be the second. This can be summed up to recursive_lambda as long as you wouldn't fret a(a, b - 1) rather than a(b - 1):
recursive_lambda = (lambda func: lambda *args: func(func, *args))
print(recursive_lambda(lambda self, x: x * self(self, x - 1) if x > 0 else 1)(6))
# Or, using the function verbatim:
print(recursive_lambda(lambda a, b: b*a(a, b-1) if b > 0 else 1)(6))
So, the outer part:
(lambda b: <factorial-application>)(num)
As you can see all the caller needs to pass is the evaluation point.
In the event that you really needed to have a recursive lambda, you could simply name the lambda:
fact = lambda x: 1 if x == 0 else x * fact(x-1)
If not, you can utilize a basic helper function. You'll see that ret is a lambda that can r refer to itself, not at all like in the past code where no lambda could refer to itself.
return func(ret, *args)
print(recursive_lambda(lambda factorial, x: x * factorial(x - 1) if x > 1 else 1)(6)) # 720
The two different ways you don't need to turn to ridiculous methods for passing the lambda to itself.
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