Can a lambda function call itself recursively in Python?

Question

The only way I can think of to do this amounts to giving the function a name:

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

or alternately, for earlier versions of python:

fact = lambda x: x == 0 and 1 or x * fact(x-1)

Update: using the ideas from the other answers, I was able to wedge the factorial function into a single unnamed lambda:

>>> map(lambda n: (lambda f, *a: f(f, *a))(lambda rec, n: 1 if n == 0 else n*rec(rec, n-1), n), range(10))
[1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880]

So it's possible, but not really recommended!

without reduce, map, named lambdas or python internals:

(lambda a:lambda v:a(a,v))(lambda s,x:1 if x==0 else x*s(s,x-1))(10)

Contrary to what sth said, you CAN directly do this.

(lambda f: (lambda x: f(lambda v: x(x)(v)))(lambda x: f(lambda v: x(x)(v))))(lambda f: (lambda i: 1 if (i == 0) else i * f(i - 1)))(n)

The first part is the fixed-point combinator Y that facilitates recursion in lambda calculus

Y = (lambda f: (lambda x: f(lambda v: x(x)(v)))(lambda x: f(lambda v: x(x)(v))))

the second part is the factorial function fact defined recursively

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

Y is applied to fact to form another lambda expression

F = Y(fact)

which is applied to the third part, n, which evaulates to the nth factorial

>>> n = 5
>>> F(n)
120

or equivalently

>>> (lambda f: (lambda x: f(lambda v: x(x)(v)))(lambda x: f(lambda v: x(x)(v))))(lambda f: (lambda i: 1 if (i == 0) else i * f(i - 1)))(5)
120

If however you prefer fibs to facts you can do that too using the same combinator

>>> (lambda f: (lambda x: f(lambda v: x(x)(v)))(lambda x: f(lambda v: x(x)(v))))(lambda f: (lambda i: f(i - 1) + f(i - 2) if i > 1 else 1))(5)
8

You can't directly do it, because it has no name. But with a helper function like the Y-combinator Lemmy pointed to, you can create recursion by passing the function as a parameter to itself (as strange as that sounds):

# helper function
def recursive(f, *p, **kw):
   return f(f, *p, **kw)

def fib(n):
   # The rec parameter will be the lambda function itself
   return recursive((lambda rec, n: rec(rec, n-1) + rec(rec, n-2) if n>1 else 1), n)

# using map since we already started to do black functional programming magic
print map(fib, range(10))

This prints the first ten Fibonacci numbers: [1, 1, 2, 3, 5, 8, 13, 21, 34, 55],

Yes. I have two ways to do it, and one was already covered. This is my preferred way.

(lambda v: (lambda n: n * __import__('types').FunctionType(
        __import__('inspect').stack()[0][0].f_code, 
        dict(__import__=__import__, dict=dict)
    )(n - 1) if n > 1 else 1)(v))(5)

This answer is pretty basic. It is a little simpler than Hugo Walter's answer:

>>> (lambda f: f(f))(lambda f, i=0: (i < 10)and f(f, i + 1)or i)
10
>>>

Hugo Walter's answer:

(lambda a:lambda v:a(a,v))(lambda s,x:1 if x==0 else x*s(s,x-1))(10)

def recursive(def_fun):
    def wrapper(*p, **kw):
        fi = lambda *p, **kw: def_fun(fi, *p, **kw)
        return def_fun(fi, *p, **kw)

    return wrapper


factorial = recursive(lambda f, n: 1 if n < 2 else n * f(n - 1))
print(factorial(10))

fibonaci = recursive(lambda f, n: f(n - 1) + f(n - 2) if n > 1 else 1)
print(fibonaci(10))

Hope it would be helpful to someone.