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I solved this by making the two scenarios into variables; even and odd. Each variable finds their respective midpoints, which makes them easy to average and call in the control flow. There’s probably a more elegant way to solve this, though, haha.
def middle_element(lst): even = int(len(lst)/2) - 1 odd = int(len(lst)/2) if len(lst) % 2 != 0: return lst[odd] else: return (lst[even]+lst[odd])/2
Rather than doing floating point arithmetic and rounding later, stick with integer operations:
# don't: int(a / b) # do: a // b
Ah! Wasn’t aware of that operator. Thanks!
Would be worth the ten minutes or so it will take to research it further while it is fresh in your mind.
It’s important to note that floats have floors, too, and if one or both operands are a float, the result will be a float.
10.3 // 2 => 5.0
Only when both operands are integer, will the quotient be an integer.
I couldn’t remember how to fool python into accounting for odd or even (that was a few excercises ago), so I hit the “see solution” button. It wrote this:
#Write your function here def middle_element(lst): if len(lst) % 2 == 0: sum = lst[int(len(lst)/2)] + lst[int(len(lst)/2) - 1] return sum / 2 else: return lst[int(len(lst)/2)] #Uncomment the line below when your function is done print(middle_element([5, 2, -10, -4, 4, 5]))
and when I hit solve it printed out: -7.0
I thought the program was supposed to show the middle 2 elements when even. -7.0 isn’t even the middle 2 added or subtracted from each other. So what the H? Did I fully misunderstand the exercise, or was the solution wrong?
Hello, @bit2995930909, and welcome to the forums.
It would seem you misunderstood. The exercise directs us to return the average of the two middle values when there are an even number of values.
We don’t attempt to fool Python. We just write code. We write it, and the computer executes what we wrote.
Ok, I missed the avg line. Thank you. Though I do feel like I’m trying to fool something that can add but doesn’t know what an odd number is. That is a joke. I mean, my joke, not the program.