Python中递归函数的斐波那契数列

Question

Hello I am trying to generate Fibonacci series by using a recursive function in python.

Here is my code

def fibolist(n):
    list1 = [1, 1]
    if n in (1,2) :
        return list1
    else:
        fibolist(n-1).append(sum(fibolist(n-1)[n-3:]))
        return list1

but when I enter any number as an argument, the result is [1, 1] Could you please help me?!

Answer 1

You start with

list1 = [1, 1]

You never change that value, and then you return it to the calling routine.

Each invocation of fibolist has a local variable named list1 ; appending to one does not change the list1 value in the calling program. You need to explicitly do that. Try

else:
    return fibolist(n-1) + [sum(fibolist(n-1)[n-3:])]

Answer 2

Just to fix your code:

def fibolist(n):
    if n in (0,1) :
        return [1,1]
    else:
        return fibolist(n-1)+[sum(fibolist(n-1)[n-2:])]

Few notes:

lists in python have starting index=0, so it's better start with it (unless you want to put start return to [0,1,1] for n in (1,2) ).

Also - as already mentioned you shouldn't return local variable, which you preassign in each go.

Answer 3

Your code is not updating the list1 variable that it returns upon coming back from the recursion. Doing fibolist(n-1).append(...) updates the list returned by the next level but that is a separate list so list1 is not affected.

You could also make your function much simpler by making it pass the last two values to itself:

def fibo(n,a=1,b=1): return [a] if n==1 else [a] + fibo(n-1,b,a+b)

BTW, the modern interpretation of the fibonacci sequence starts at 0,1 not 1,1 so the above signature should be def fibo(n,a=0,b=1) .

Output:

print(fibo(5))
#[1, 1, 2, 3, 5]