Fibonacci numbers, with an one-liner in Python 3?

I know there is nothing wrong with writing with proper function structure, but I would like to know how can I find nth fibonacci number with most Pythonic way with a one-line.

I wrote that code, but It didn't seem to me best way:

>>> fib = lambda n:reduce(lambda x, y: (x[0]+x[1], x[0]), [(1,1)]*(n-2))[0]
>>> fib(8)
13

How could it be better and simplier?

Solution 1:

fib = lambda n:reduce(lambda x,n:[x[1],x[0]+x[1]], range(n),[0,1])[0]

(this maintains a tuple mapped from [a,b] to [b,a+b], initialized to [0,1], iterated N times, then takes the first tuple element)

>>> fib(1000)
43466557686937456435688527675040625802564660517371780402481729089536555417949051
89040387984007925516929592259308032263477520968962323987332247116164299644090653
3187938298969649928516003704476137795166849228875L

(note that in this numbering, fib(0) = 0, fib(1) = 1, fib(2) = 1, fib(3) = 2, etc.)

(also note: reduce is a builtin in Python 2.7 but not in Python 3; you'd need to execute from functools import reduce in Python 3.)

Solution 2:

A rarely seen trick is that a lambda function can refer to itself recursively:

fib = lambda n: n if n < 2 else fib(n-1) + fib(n-2)

By the way, it's rarely seen because it's confusing, and in this case it is also inefficient. It's much better to write it on multiple lines:

def fibs():
    a = 0
    b = 1
    while True:
        yield a
        a, b = b, a + b