单一陈述斐波那契 [英] Single Statement Fibonacci
问题描述
可能重复:
斐波那契数，在Python 3中有一个内胆?a>
Possible Duplicate:
Fibonacci numbers, with an one-liner in Python 3?
这可能很容易，但是我对Python还是很陌生.我想出了这句话斐波那契.
It may be very easy thing, but I am very new to Python. I came up with this single statement Fibonacci.
[fibs.append(fibs[-2]+fibs[-1]) for i in xrange(1000)]
不过，并不是真正的单一陈述.在触发此语句之前，我需要初始化列表 fibs ，即 fibs = [0，1] .
Not really single statement, though. I need to initialise the list, fibs, before firing this statement i.e. fibs = [0, 1].
现在，我有2个问题，
-
我们如何摆脱此列表初始化语句
fibs = [0，1]，以便使其真正成为单个语句?
How can we get rid of this list initialisation statement,
fibs = [0, 1], in order to make it really single statement?
原始语句打印 None n 次；其中 n 是在 xrange()中传递的数字.有什么办法可以完全避免这种情况?如果语句可以打印系列，则更好.然后，我们无需显式打印 fibs .
Original statement prints None n times; where n is the number passed in xrange(). Is there any way to avoid that altogether? Or better if the statement can print the series, instead. Then we don't need to print fibs explicitly.
[已编辑]
还是我们可以使用 list.append()的替代方法，该方法返回附加到其上的 list ?
Or do we have any alternative to list.append() which returns the list it appends to?
推荐答案
这有效:
for n in range(1000):
print(0.4472135954999579392818347337462552470881236719223051448541*(pow(1.6180339887498948482045868343656381177203091798057628621354,n) - pow(-0.618033988749894848204586834365638117720309179805762862135,n)))
这是Binet公式的实现. http://en.wikipedia.org/wiki/Fibonacci_number#Relation_to_the_golden_ratio
This is an implementation of Binet's Formula. http://en.wikipedia.org/wiki/Fibonacci_number#Relation_to_the_golden_ratio
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