Factorial using lambda in python

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))

It is this simple:

n = input()

print reduce(lambda x, y: x * y, range(1, n + 1))

Let's peel this one liner open like an onion.

print(lambda b: (Y))(num)

We are making an anonymous function (the keyword lambda means we're about to type a series of parameter names, then a colon, then a function that uses those parameters) and then pass it num to satisfy its one parameter.

   (lambda a, b: a(a, b))(X, b)

Inside of the lambda, we define another lambda. Call this lambda Y. This one takes two parameters, a and b. a is called with a and b, so a is a callable that takes itself and one other parameter

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

For 1, it's nothing more than:

def f(a, b):
   if b > 0:
   b * a(a, b - 1)
else :
   1

For 2, this b

(lambda b: (lambda a, b: a(a, b))(lambda a, b: b * a(a, b - 1) if b > 0
   else 1, b))(num)
(this one)

is actually this b:

(lambda b: (lambda a, b: a(a, b))(lambda a, b: b * a(a, b - 1) if b > 0
   else 1, b))(num)
(this one)

We can use the below lambda expression

     fact = lambda n: 1
     if n == 0
     else n * fact(n - 1)
     print(fact(5) >>>
           120

the generalized def of recursive lambda is as below

recursive_lambda = (lambda func: lambda * args: func(func, * args))

Suggestion : 2

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Method 1: Using Lambda Expression

x = lambda num: 1
if num <= 1
else num * x(num - 1)

number = int(input('Enter number: '))

print('%d != %d' % (number, x(number)))

Output

Enter number: 23
23 != 25852016738884976640000

Method 2: Using Reduce Function & Lambda Expression

from functools
import reduce

number = int(input('Enter number: '))

factorial = reduce(lambda x, y: x * y, range(1, n + 1))

print('%d != %d' % (number, factorial))

Suggestion : 3

In the following example, we defined a lambda function that can calculate the factorial of a given number, using recursion.,In the following example, we define a lambda function with no arguments, and returns 6 when called.,We already mentioned that a Python Lambda function can accept multiple arguments. In the following example, we define a lambda function with two arguments, say a and b, and we shall multiplication of a and b.,Concluding this tutorial of Python Examples, we learned how to define a lambda function, and use it in your program constructively, with the help of example programs.

Following is the syntax of Lambda Function.

lambda arguments: expression

You can assign this lambda function to a variable. Something like in below code snippet.

x = lambda arguments: expression

And you can call the lambda function with the variable name x. Following is an example code snippet to call the lambda function.

Suggestion : 4

This recipe implements the recursive definition of the factorial function as a lambda form. Since lambda forms can only be expressions, this is slightly tricky, since if/else is a statement, and therefore not allowed inside an expression. Still, a short-circuiting form of a Python idiom for a conditional (ternary) operator takes care of that (see Recipe 17.6 for other ways to simulate the ternary operator, both with and without short-circuiting). ,The real issue, of course, is that since lambda’s forte is making anonymous functions, how then do we recurse? This question is what makes this recipe’s subject a good bet to win a drink from your Python-using friends and acquaintances who are misguided enough that they have not yet read this book cover to cover. ,You want to write a recursive function, such as a factorial, using lambda (you probably made a bet about whether it could be done). ,Just make sure the terms of the bet mention only lambda and do not specify that the resulting function will be left unnamed. Some might consider this cheating, but we Python programmers are a bunch of pragmatists. Thus, we simply bind a name to the lambda form with an assignment statement, and in the body of the lambda itself, we use the name to which we will assign the lambda. Since the body executes only when the lambda is called (not at the time it’s created), the name will be bound by the time we use it. And the bet is won!

f = lambda n: n - 1 + abs(n - 1) and f(n - 1) * n or 1

Suggestion : 5

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

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)

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))

Suggestion : 6

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. ,These are the parameters to Y. The first one is a lambda function, call it X. We can see that X is the factorial function, and that the second parameter will become its number.,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):,The factorial itself is almost as you'd expect it. You infer that the a is... the factorial function. b is the actual parameter.

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)

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)