斐波那契数，在Python 3中只有一线吗? [英] Fibonacci numbers, with an one-liner in Python 3?

问题描述

我知道用适当的函数结构编写没有什么问题，但是我想知道如何用最Python的方式用单行找到第n个斐波那契数.

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.

我编写了该代码，但是在我看来，这并不是最好的方法:

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

怎么会更好，更简单?

推荐答案

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

(这将维护一个从[a，b]映射到[b，a + b]的元组，初始化为[0,1]，迭代N次，然后获取第一个元组元素)

(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

(请注意，在此编号中，fib(0)= 0，fib(1)= 1，fib(2)= 1，fib(3)= 2，依此类推)

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

(还请注意:reduce是Python 2.7中的内置功能，但不是Python 3中的内置功能；您需要在Python 3中执行from functools import reduce.)

(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.)

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