Calculating average in interchangable range?

Question

I know it's because of n, but n is supposed to be any variable, and left as n, this is what I have:

def average(n):
    if n >= 0:
        avg = sum((range(1:int(n)))/float(len(range(1:int(n)))))
    print avg

how do I fix it?

Answer 1

If your range is always 1:n, why don't you just use this:

avg = sum((range(1:int(n)))/float(n))

Or maybe I am not understanding your question...

Answer 2

I may be wrong but range(1:int(n)) doesn't look like syntactically correct and parenthesis don't match. You may want to calculate the average of numbers in the range of 0 to n. In that case, I would replace your code like this:

def average(n):
if n >= 0:
    avg = sum((range(int(n))))/float(n)
print avg

Answer 3

The summation of x from 1 to n is simply (n + 1) * (n / 2). The number of elements being summed is n . Do a little simplification and your new function is

def average(n):
    return (n + 1) / 2.0

You'll have to adjust this if you actually wanted Python's behavior of an exclusive upper-bound for range() (i.e., having average(10) return the average of the sum of values 1 - 9 instead of 1 - 10).