python syntax: how to return 0 instead of False when evaluating 0

Question

For an assignment we were asked to define a fibonacci function, which I accomplished with this: ``` def fibonacci(n): if n < 2: return n return fibonacci(n-1) + fibonacci(n-2) ``` However, I have seen recursive functions, such as the factorial function, defined in a one line return statement like so: ``` def factorial(n): return n > 1 and n * factorial(n-1) or 1 ``` So, I attempted to apply the same to my fibonacci function. After several attempts, I got it to work for all tested cases except when s = 0, in which case it returns False when it should return 0. Here is where I am: ``` def fibonacci(n): return ((n == 0 or n == 1) and n) or (n > 1 and (fibonacci(n-1) + fibonacci(n-2))) ``` I understand that python evaluates 0 to False, so how would I have python return zero instead of False when n is 0 while maintaining the current length/structure of the code? Is that even possible? Also, is this style of creating a function (recursive or otherwise) more or less desirable/pythonic than t…

Accepted Answer

The `x and y or z` idiom doesn't work if `y` is falsy. You can swap the condition to make it work nonetheless:

```
def fibonacci(n):
    return n >= 2 and fibonacci(n-1) + fibonacci(n-2) or n
```

However, as of Python 2.5 (released 6 years ago), we have proper conditional expressions and don't need the `and/or` hack any longer:

```
def fibonacci(n):
    return n if n < 2 else fibonacci(n-1) + fibonacci(n-2)
```

Now this has an exponential runtime complexity. If you want to be efficient, use the `O(n)` algorithm:

```
def fibonacci(n):
    a, b = 0, 1
    for _ in range(n):
        a, b = b, a + b
    return a
```

Or even write a generator to yield all the numbers and only take as much as you need.

python syntax: how to return 0 instead of False when evaluating 0

Solution

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