# Python lambda function to calculate factorial of a number

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I have recently begun learning python. I went over lambda functions. On one of the issues, the writer requested to compose a one liner lambda function for factorial of a number.

Look at the solution which is 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)

But, I can't understand this syntax. Can anyone please explain this?

Thanks

## 1 Answer

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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.

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

The two different ways you don't need to turn to ridiculous methods for passing the lambda to itself.

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