Screen Link:
https://app.dataquest.io/m/306/the-weighted-mean-and-the-median/5/distributions-with-even-number-of-values
My Code:
houses_median = houses['TotRms AbvGrd'].copy()
houses_median = houses_median.replace('10 or more' , 10)
houses_median = houses_median.astype(int)
houses_median.sort_values(ascending=True, inplace=True)
# Find the median
length = len(houses_median)
if length // 2 != 0:
median = houses_median[length//2]
else:
median = (houses_median[length//2] + length[(length//2)-1])/2
What I expected to happen:
Our task is to first check the length of the series and based on whether its odd or even calculate the median. My code does exactly that.
When I analysed the code provided by DQ, I understand the code works fine when the length of the series is even(which is the case with our example). But how would the code work correctly when the length of the series is odd? The variable ‘middle_indices’ is going to provide two indices no matter the whether the length of the series is odd or even. But that should not be the case.
Please provide your thoughts.
The code provided by DQ:
# Find the median
middle_indices = [int((len(rooms_sorted) / 2) - 1),
int((len(rooms_sorted) / 2))
] # len - 1 and len because Series use 0-indexing
middle_values = rooms_sorted.iloc[middle_indices] # make sure you don't use loc[]
median = middle_values.mean()
