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 following:

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

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

a = fib()
a.next()
0
for i in range(a):
    print a.next(),

Pilfer

I think 'while True' is more concise – Gannes

How could the 2nd iteration understand "a=1"? Shouldn't we have to write a = a+b rather than "a, b=b, a+b" – Nordstrom

@Nordstrom That expression is a two-part expression. First, it makes a tuple with (b,a+b) using the current values of a and b, then it unpacks that tuple when assigning it to a,b. (Technically speaking Python can do some optimization, but I'm ignoring this for now). – Epps

Answer 1

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)))

Speculate

fib(0) doesn't yield anything, which can be fixed using range(n+1) instead. – Rimester

I rewrote this by using a=0, b=1, a=b and b=a+b because I was uncomfortable with that notation but it gave completely different answers. Am I missing something? – Petrosal

As The Amateur Coder says below, should be yield b not yield a. So it begins 1,1,2,3, etc... No? – Hydrophilic

Answer 2

You are giving a too many meanings:

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

vs.

a = fib()       

You won't run into the problem (as often) if you give your variables more descriptive names (3 different uses of the name a in 10 lines of code!):

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

and change range(a) to range(amount).

Schofield

Answer 3

Since you are writing a generator, why not use two yields, to save doing the extra shuffle?

import itertools as it

num_iterations = int(raw_input('How many? '))
def fib():
    a,b = 0,1
    while True:
        yield a
        b = a+b
        yield b
        a = a+b

for x in it.islice(fib(), num_iterations):
    print x

.....

Francyne

Answer 4

Really simple with generator:

def fin(n):
    a, b = 0, 1

    for i in range(n):
        yield a
        a, b = b, a + b


ln = int(input('How long? '))
print(list(fin(ln))) # [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...]

Yoder

So this is how yield works! BTW, it should be yield b instead of yield a as it repeats the elements twice (a becomes b). – Wentz

Answer 5

Python is a dynamically typed language. the type of a variable is determined at runtime and it can vary as the execution is in progress. Here at first, you have declared a to hold an integer type and later you have assigned a function to it and so its type now became a function.

you are trying to apply 'a' as an argument to range() function which expects an int arg but you have in effect provided a function variable as argument.

the corrected code should be

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

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

b = fib()
b.next()

for i in range(a):
    print b.next(),

this will work

Retiform

I am gettting error on this File "D:\Projects\Python\fibonacci.py", line 18, in <module> b.next() AttributeError: 'generator' object has no attribute 'next' – Impregnable

I found it for Python 3, it is b.__next__(). Sorry :) and Thanks – Impregnable

Answer 6

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

Stampede

Answer 7

def genFibanocciSeries():

    a=0
    b=1
    for x in range(1,10):
        yield a
        a,b = b, a+b

for fib_series in genFibanocciSeries():
    print(fib_series)

Phocis

Answer 8

To get the fibonacci numbers till any number (100 in this case) with generator, you can do this.

def getFibonacci():
    a, b = 0, 1

    while True:
        yield b
        b = a + b
        a = b - a

for num in getFibonacci():
    if num > 100:
        break
    print(num)

Surfing

Answer 9

Your a is a global name so-to-say.

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

Whenever Python sees an a, it thinks you are talking about the above one. Calling it something else (elsewhere or here) should help.

Softshoe

Answer 10

Also you can use enumerate infinite generator:

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

Pennebaker

Answer 11

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)

Modestine

Answer 12

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`

Lapstrake

Answer 13

Simple way to print Fibonacci series till n number

def Fib(n):
    i=a=0
    b=1
    while i<n:
        print (a)
        i=i+1
        c=a+b
        a=b
        b=c




Fib(input("Please Enter the number to get fibonacci series of the Number :  "))

Ern

Answer 14

a = 3 #raw_input

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

fs = fib_gen()
next(fs)
for i in range(a):
    print (next(fs))

Betty

only code answers are discouraged here. It is encouraged to add some details to how this code will fix the problem. – Brierwood

Answer 15

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.

Videlicet

Answer 16

It looks like you are using the a twice. Try changing that to a different variable name.

The following seems to be working great for me.

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

f = fib()
for x in range(100):
    print(f.next())

Prevention

Answer 17

i like this version:

array = [0,1]

for i in range(20):
   x = array[0]+array[1]   
   print(x)
   array[0] = array[1]
   array[1] = x

Elegy

Answer 18

Below are two solution for fiboncci generation:

def fib_generator(num):
    '''
    this will works as generator function and take yield into account.
    '''
    assert num > 0
    a, b = 1, 1
    while num > 0:
        yield a
        a, b = b, a+b
        num -= 1


times = int(input('Enter the number for fib generaton: '))
fib_gen = fib_generator(times)
while(times > 0):
    print(next(fib_gen))
    times = times - 1


def fib_series(num):
    '''
    it collects entires series and then print it.
    '''
    assert num > 0
    series = []
    a, b = 1, 1
    while num > 0:
        series.append(a)
        a, b = b, a+b
        num -= 1
    print(series)


times = int(input('Enter the number for fib generaton: '))
fib_series(times)

Dyestuff

Answer 19

Why do you go for complex here is one of my snippet to work on!!

n = int(input('Enter your number..: '))
a = 0
b = 1
c = 0
print(a)
print(b)
for i in range(3, n+1):
    c = a+b
    print(c)
    a,b=b,c 

check out my git - rohith-sreedharan

Dereism

Answer 20

I recently discovered this type of construct (many thanks to @KellyBundy for this), so I couldn't resist:

n = 20
fib = (a for a,b in [(0,1)] for _ in range(n) for a,b in [(b,a+b)])

print(list(fib))
# [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765]

Kreit

Answer 21

We can use it in a short way without through 'yield and 'next' statements

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

Marlomarlon

Yeah but the question is asking to print the first n numbers, not the nth number... – Noyes