Function to find factors of a number, optionally including 1 and the number itself I have a function for finding factors of a number in Python

def factors(n, one=True, itself=False):
    def factors_functional():
        # factors below sqrt(n)
        a = filter(lambda x: n % x == 0,
                   xrange(2, int(n**0.5)+1))
        result = a + map(lambda x: n / x, reversed(a))
        return result


factors(28) gives [2, 4, 7, 14]. Now i want this function to also include 1 in the beginning of a list, if one is true, and 28 in the end, if itself is true.

([1] if one else []) + a + map(lambda x: n / x, reversed(a)) + ([n] if itself else [])


This is "clever" but inelegant.

if one:
    result = [1] + result
if itself:
    result = result + [n]


This is straight-forward but verbose.

Is it possible to implement optional one and itself in the output with the most concise yet readable code possible? Or maybe the whole idea should be done some other way - like the function factors can be written differently and still add optional 1 and itself?
What about this?

[1] * one + result + [n] * itself