Python lambda function to calculate factorial of a number

lambda, python, python-2.7

Solution

The factorial itself is almost as you'd expect it. You infer that the `a` is... the factorial function. `b` is the actual parameter.

<factorial> = lambda a, b: b*a(a, b-1) if b > 0 else 1

This bit is the application of the factorial:

<factorial-application> = (lambda a, b: a(a, b))(<factorial>, b)

`a` is the factorial function itself. It takes itself as its first argument, and the evaluation point as the second. This can be generalized to `recursive_lambda` as long as you don't mind `a(a, b - 1)` instead of `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 we have the outer part:

(lambda b: <factorial-application>)(num)

As you see all the caller has to pass is the evaluation point.

If you actually wanted to have a recursive lambda, you could just name the lambda:

fact = lambda x: 1 if x == 0 else x * fact(x-1)

If not, you can use a simple helper function. You'll notice that `ret` is a lambda that can refer to itself, unlike in the previous code where no lambda could refer to itself.

def recursive_lambda(func):
    def ret(*args):
        return func(ret, *args)
    return ret

print(recursive_lambda(lambda factorial, x: x * factorial(x - 1) if x > 1 else 1)(6))  # 720

Both ways you don't have to resort to ridiculous means of passing the lambda to itself.

Problem

I have just started learning python. I came across lambda functions. On one of the problems, the author asked to write a one liner lambda function for factorial of a number. This is the solution that was given: ``` num = 5 print (lambda b: (lambda a, b: a(a, b))(lambda a, b: b*a(a, b-1) if b > 0 else 1,b))(num) ``` I cannot understand the weird syntax. What does a(a,b) mean? Can someone explain? Thanks

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