Python Fibonacci Generator

Question

I need to make a program that asks for the amount of Fibonacci numbers printed and then prints them like 0, 1, 1, 2... but I can\'t get it to work. My code looks the followi

Answer 1

def fibonacci(n):
    fn = [0, 1,]
    for i in range(2, n):
        fn.append(fn[i-1] + fn[i-2])
    return fn

Answer 2

I've build this a while ago:

a = int(raw_input('Give amount: '))

fab = [0, 1, 1]
def fab_gen():
    while True:
        fab.append(fab[-1] + fab[-2])
        yield fab[-4]

fg = fab_gen()
for i in range(a): print(fg.next())

No that fab will grow over time, so it isn't a perfect solution.

Answer 3

You had the right idea and a very elegant solution, all you need to do fix is your swapping and adding statement of a and b. Your yield statement should go after your swap as well

a, b = b, a + b #### should be a,b = a+b,a #####

`###yield a`

Answer 4

I would use this method:

Python 2

a = int(raw_input('Give amount: '))

def fib(n):
    a, b = 0, 1
    for _ in xrange(n):
        yield a
        a, b = b, a + b

print list(fib(a))

Python 3

a = int(input('Give amount: '))

def fib(n):
    a, b = 0, 1
    for _ in range(n):
        yield a
        a, b = b, a + b

print(list(fib(a)))

Answer 5

Also you can use enumerate infinite generator:

for i,f  in enumerate(fib()):
    print i, f
    if i>=n: break

Answer 6

Also you can try the closed form solution (no guarantees for very large values of n due to rounding/overflow errors):

root5 = pow(5, 0.5)
ratio = (1 + root5)/2

def fib(n):
    return int((pow(ratio, n) - pow(1 - ratio, n))/root5)