Python 如何找到最近的斐波那契数列?
我的下一步是,如果输入不在斐波那契数列中,程序必须给出一个与输入最近的数列中的数字的输出。我不知道如何进行,有人能帮我吗Python 如何找到最近的斐波那契数列?,python,fibonacci,Python,Fibonacci,我的下一步是,如果输入不在斐波那契数列中,程序必须给出一个与输入最近的数列中的数字的输出。我不知道如何进行,有人能帮我吗 def fibs(): a,b = 0,1 yield a yield b while True: a,b = b,a+b yield b n = int(input("please, enter a number: ")) for fib in fibs(): if n == fib:
def fibs():
a,b = 0,1
yield a
yield b
while True:
a,b = b,a+b
yield b
n = int(input("please, enter a number: "))
for fib in fibs():
if n == fib:
print("Yes! Your number is a Fibonacci number!")
break
if fib > n:
print("No! Your number is not a Fibonacci number!")
break
如果您不介意进行额外的生成器调用,那么保留上一个斐波那契函数是没有必要的 首先将生成器存储在变量中
gen = fibs()
n = int(input("please, enter a number: "))
for fib in gen:
if n == fib:
print("Yes! Your number is a Fibonacci number!")
break
if fib > n:
print("No! Your number is not a Fibonacci number!")
next_fib = next(gen)
prev = next_fib - fib
closest = prev if n - prev < fib - n else fib # Search for Python ternary operator
# If you are a stranger to this line of code.
print("The closest fibonacci number to your entry is %s" % closest)
break
gen=fibs()
n=int(输入(“请输入一个数字:”)
对于fib in gen:
如果n==fib:
打印(“是的!您的数字是斐波那契数!”)
打破
如果fib>n:
打印(“不!您的数字不是斐波那契数字!”)
next_fib=next(gen)
上一个=下一个
如果n-prev
Edit:我第一次使用
gen.next()
来获取收益率的下一个值,但是我忘记了在Python 3中,它被重命名为gen.next()
。请小心使用next(gen)
是两个Python版本的预期用法。如果您不介意进行额外的生成器调用,则不必保留以前的斐波那契
首先将生成器存储在变量中
gen = fibs()
n = int(input("please, enter a number: "))
for fib in gen:
if n == fib:
print("Yes! Your number is a Fibonacci number!")
break
if fib > n:
print("No! Your number is not a Fibonacci number!")
next_fib = next(gen)
prev = next_fib - fib
closest = prev if n - prev < fib - n else fib # Search for Python ternary operator
# If you are a stranger to this line of code.
print("The closest fibonacci number to your entry is %s" % closest)
break
gen=fibs()
n=int(输入(“请输入一个数字:”)
对于fib in gen:
如果n==fib:
打印(“是的!您的数字是斐波那契数!”)
打破
如果fib>n:
打印(“不!您的数字不是斐波那契数字!”)
next_fib=next(gen)
上一个=下一个
如果n-prev
Edit:我第一次使用
gen.next()
来获取收益率的下一个值,但是我忘记了在Python 3中,它被重命名为gen.next()
。请小心使用next(gen)
是两个Python版本的预期用法。您可以将fibs本身压缩为:
n = int(input("please, enter a number: "))
for fib, next_fib in itertools.izip(fibs(), itertools.islice(fibs(), 1, None)):
if n == fib:
print("Yes! Your number is a Fibonacci number!")
break
if next_fib > n:
closest = fib if n - fib < next_fib - n else next_fib
print("The closest Fibonacci number is {}".format(closest))
break
n=int(输入(“请输入一个数字:”)
对于fib,itertools.izip中的下一个fib(fibs(),itertools.islice(fibs(),1,None)):
如果n==fib:
打印(“是的!您的数字是斐波那契数!”)
打破
如果next_fib>n:
如果n-fib
您可以使用itertools.tee
对其进行一些优化。您可以将fibs本身压缩为:
n = int(input("please, enter a number: "))
for fib, next_fib in itertools.izip(fibs(), itertools.islice(fibs(), 1, None)):
if n == fib:
print("Yes! Your number is a Fibonacci number!")
break
if next_fib > n:
closest = fib if n - fib < next_fib - n else next_fib
print("The closest Fibonacci number is {}".format(closest))
break
n=int(输入(“请输入一个数字:”)
对于fib,itertools.izip中的下一个fib(fibs(),itertools.islice(fibs(),1,None)):
如果n==fib:
打印(“是的!您的数字是斐波那契数!”)
打破
如果next_fib>n:
如果n-fib
您可以使用itertools.tee
对其进行一些优化。这里有一个简单的方法,可以使用生成器测试小数值
def fibs():
a,b = 0,1
yield a
yield b
while True:
a,b = b,a+b
yield b
def nearest_fib(n):
''' If n is a Fibonacci number return True and n
Otherwise, return False and the nearest Fibonacci number
'''
for fib in fibs():
if fib == n:
return True, n
elif fib < n:
prev = fib
else:
# Is n closest to prev or to fib?
if n - prev < fib - n:
return False, prev
else:
return False, fib
# Test
for i in range(35):
print(i, nearest_fib(i))
更新
这里有一个更有效的方法,它首先用来近似y:F(y)=n。然后,它使用一对与相关的恒等式(可以在O(log(n))时间内计算F(n))递归地找到与n最近的斐波那契数。递归非常快,因为它使用缓存来保存已经计算过的值。在没有缓存的情况下,该算法的速度与Rockybilly算法大致相同
from math import log, sqrt
def fast_fib(n, cache={0: 0, 1: 1}):
if n in cache:
return cache[n]
m = (n + 1) // 2
a, b = fast_fib(m - 1), fast_fib(m)
fib = a * a + b * b if n & 1 else (2 * a + b) * b
cache[n] = fib
return fib
logroot5 = log(5) / 2
logphi = log((1 + 5 ** 0.5) / 2)
def nearest_fib(n):
if n == 0:
return 0
# Approximate by inverting the large term of Binet's formula
y = int((log(n) + logroot5) / logphi)
lo = fast_fib(y)
hi = fast_fib(y + 1)
return lo if n - lo < hi - n else hi
for i in range(35):
print(i, nearest_fib(i))
请注意,fast\u fib
使用a作为缓存,但这里没有问题,因为我们希望缓存记住它以前的内容
在我的速度测试中,默认可变参数缓存比任何其他形式的缓存都快,但它的缺点是无法从函数外部清除缓存,如果向函数添加逻辑以清除缓存,则在不想清除缓存时,会影响大多数调用的性能
更新
实际上,可以从函数外部清除默认的可变参数缓存。我们可以通过函数的。\uuuu default\uuuu
属性访问函数的默认参数(或者在Python2的旧版本中是.func\u defaults
;在Python2.6中是。\uuu default\uuuuuu
工作的,但在2.5中不是)
例如
下面是一些代码(在Python2和Python3上运行),它们对为这个问题提交的一些算法执行计时测试。Rockybilly的与我的第一个版本非常相似,只是它避免了保存以前的值。我还使OP的fibs
生成器更加紧凑
Douglas的代码适用于小数值,或者当参数实际上是斐波那契数时,但对于大的非斐波那契数,由于逐个搜索的速度很慢,所以它变得非常慢。通过避免重新计算各种数量,我已经能够对其进行一些优化,但这对运行速度没有太大影响
在这个版本中,myfast\u fib()
函数使用一个全局缓存,以便在测试之间清除它,从而使计时更加公平
#!/usr/bin/env python3
""" Find the nearest Fibonacci number to a given integer
Test speeds of various algorithms
See https://stackoverflow.com/questions/40682947/fibonacci-in-python
Written by PM 2Ring 2016.11.19
Incorporating code by Rockybilly and Douglas
"""
from __future__ import print_function, division
from math import log, sqrt
from time import time
def fibs():
a, b = 0, 1
while True:
yield a
a, b = b, a + b
def nearest_fib_Rocky(n):
''' Find the nearest Fibonacci number to n '''
fibgen = fibs()
for fib in fibgen:
if fib == n:
return n
elif fib > n:
next_fib = next(fibgen)
return next_fib - fib if 2 * n < next_fib else fib
def nearest_fib_Doug(n):
a = 5 * n * n
if sqrt(a + 4)%1 == 0 or sqrt(a - 4)%1 == 0:
return n
c = 1
while True:
m = n + c
a = 5 * m * m
if sqrt(a + 4)%1 == 0 or sqrt(a - 4)%1 == 0:
return m
m = n - c
a = 5 * m * m
if sqrt(a + 4)%1 == 0 or sqrt(a - 4)%1 == 0:
return m
c += 1
cache={0: 0, 1: 1}
def fast_fib(n):
if n in cache:
return cache[n]
m = (n + 1) // 2
a, b = fast_fib(m - 1), fast_fib(m)
fib = a * a + b * b if n & 1 else (2 * a + b) * b
cache[n] = fib
return fib
logroot5 = log(5) / 2
logphi = log((1 + 5 ** 0.5) / 2)
def nearest_fib_PM2R(n):
if n == 0:
return 0
# Approximate by inverting the large term of Binet's formula
y = int((log(n) + logroot5) / logphi)
lo = fast_fib(y)
hi = fast_fib(y + 1)
return lo if n - lo < hi - n else hi
funcs = (
nearest_fib_PM2R,
nearest_fib_Rocky,
nearest_fib_Doug,
)
# Verify that all the functions return the same result
def verify(lo, hi):
for n in range(lo, hi):
a = [f(n) for f in funcs]
head, tail = a[0], a[1:]
if not all(head == u for u in tail):
print('Error:', n, a)
return False
else:
print('Ok')
return True
def time_test(lo, hi):
print('lo =', lo, 'hi =', hi)
for f in funcs:
start = time()
for n in range(lo, hi):
f(n)
t = time() - start
print('{0:18}: {1}'.format(f.__name__, t))
print()
verify(0, 1000)
cache={0: 0, 1: 1}
time_test(0, 1000)
funcs = funcs[:-1]
cache={0: 0, 1: 1}
time_test(1000, 50000)
这些时间是在Linux上运行Python 3.6的旧2GHz 32位机器上进行的。Python2.6给出了类似的计时
FWIW,Rockybilly和我的代码都可以轻松处理非常大的数字。以下是时间测试(10**1000,10**1000+1000)的定时输出:
这里有一个使用生成器的简单方法,可以用来测试小数字
def fibs():
a,b = 0,1
yield a
yield b
while True:
a,b = b,a+b
yield b
def nearest_fib(n):
''' If n is a Fibonacci number return True and n
Otherwise, return False and the nearest Fibonacci number
'''
for fib in fibs():
if fib == n:
return True, n
elif fib < n:
prev = fib
else:
# Is n closest to prev or to fib?
if n - prev < fib - n:
return False, prev
else:
return False, fib
# Test
for i in range(35):
print(i, nearest_fib(i))
更新
这里有一个更有效的方法,它首先用来近似y:F(y)=n。然后它使用pa
Ok
lo = 0 hi = 1000
nearest_fib_PM2R : 0.005465507507324219
nearest_fib_Rocky : 0.02432560920715332
nearest_fib_Doug : 0.45461463928222656
lo = 1000 hi = 50000
nearest_fib_PM2R : 0.26880311965942383
nearest_fib_Rocky : 1.266334056854248
nearest_fib_PM2R : 0.011492252349853516
nearest_fib_Rocky : 7.556792497634888
from math import *
n = int(input("Enter a number:"))
if sqrt(5*n**2+4)%1==0 or sqrt(5*n**2-4)%1==0:
print("Your number is a Fibonacci number!")
else:
print("Your number is not a Fibonacci number.")
c = 0
while 1:
c += 1
if sqrt(5*(n+c)**2+4)%1==0 or sqrt(5*(n+c)**2-4)%1==0:
print("%s is the closest Fibonacci number to your entry." % str(n+c))
break
if sqrt(5*(n-c)**2+4)%1==0 or sqrt(5*(n-c)**2-4)%1==0:
print("%s is the closest Fibonacci number to your entry." % str(n-c))
break
Enter a number: 9999999999
Your number is not a Fibonacci number.
9999816735 is the closest Fibonacci number to your entry.
Enter a number: 9999816735
Your number is a Fibonacci number!