I am trying to write a Fibonacci numbers function recursively in python

Question

I want to print the whole list up to the element that is given by function,

such as [1,1,2,3,5,8]

def Fib(a):
    b=[ ]
    if a==0:
        b.append(1)
    elif a==1:
        b.append(1)
    else:
        b.append(Fib(a-1)+Fib(a-2))
    print(b)

Fib(6)

I get the error of "unsupported operand type(s) for +: 'NoneType' and 'NoneType'" Thanks in advance, all the helps are appreciated.

use [return](http://stackoverflow.com/questions/7129285/why-would-you-use-the-return-statement-in-python) — syntonym, Feb 23 '16 at 20:19
It is worth mentioning that recursive Fibonacci [is not very efficient](http://stackoverflow.com/questions/360748/computational-complexity-of-fibonacci-sequence) — OneCricketeer, Feb 23 '16 at 20:23
OK, thanks for the comments and advices, I added the return statement, actually changed the line print(b) with return b, and after that I print the function itself outside the function, but right now I have the output as : [[[[[[1, 1], 1], [1, 1]], [[1, 1], 1]], [[[1, 1], 1], [1, 1]]]] — Heyyo, Feb 23 '16 at 20:28

score 1 · Answer 1 · answered Feb 23 '16 at 20:37

1

Could make a generator

def fibGen(n, a=1, b=1):
    while n > 0:
        yield a
        a, b, n = b, a+b, n-1

Run it

>>> [i for i in fibGen(6)]
[1, 1, 2, 3, 5, 8]

answered Feb 23 '16 at 20:37

OneCricketeer

179,855
19
132
245

score 0 · Answer 2 · edited Feb 23 '16 at 23:07

0

The function requires a return statement. Furthermore, After each conditional statements, a return statement should be present

If n ==0:
     return 0
Elif n ==1: 
    return 1
Else: 
    return fib(n-1) + fib(n-2)

edited Feb 23 '16 at 23:07

user229044

232,980
40
330
338

answered Feb 23 '16 at 20:25

Spyke

114
4

1

Welcome to Stack Overflow! When answering questions, it is often a good idea to leave a little explanation. You can edit your posts, to make them more informative, if you want. To do so, click the "edit" link below your post. – S.L. Barth is on codidact.com Feb 23 '16 at 20:43
1

While your answer is "correct", it does not attempt to answer the question. You could explain what needs to be in that return statement to achieve the expected output. I don't think a simple return of the calculated fibonacci value or the `b` list will produce `[1,1,2,3,5,8]` – OneCricketeer Feb 23 '16 at 20:44

YBathia · Answer 3 · 2016-02-23T20:57:27.093

0

result = {}
def fibo(n):
    if n ==0 :
        result[n] = 0
        return 0
    elif n == 1:
        result[n] = 1
        return 1
    else:
        val = fibo(n-1) + fibo(n-2)
        result[n] = val
        return (val)

fibo(10)
print(result.values())

Output:

[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]

edited Feb 23 '16 at 20:57

answered Feb 23 '16 at 20:44

YBathia

894
1
8
18

I am trying to write a Fibonacci numbers function recursively in python

3 Answers3