How to find the average of the differences between all the numbers of a Python List

Question:

I have a python list like this,

arr = [110, 60, 30, 10, 5] 

What I need to do is actually find the difference of every number with all the other numbers and then find the average of all those differences.

So, for this case, it would first find the difference between 110 and then all the remaining elements, i.e. 60, 30, 10, 5, and then it will find the difference of 60 with the remaining elements, i.e. 30, 10, 5 and etc.

After which, it will compute the Average of all these differences.

Now, this can easily be done with two For Loops but in O(n^2) time complexity and also a little bit of "messy" code. I was wondering if there was a faster and more efficient way of doing this same thing?

Asked By: Asad Hussain

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Answers:

I’ll just give the formula first:

n = len(arr)
out = np.sum(arr * np.arange(n-1, -n, -2) ) / (n*(n-1) / 2)
# 52

Explanation: You want to find the mean of

a[0] - a[1], a[0] - a[2],..., a[0] - a[n-1]
             a[1] - a[2],..., a[1] - a[n-1]
                         ...

there, your

`a[0]` occurs `n-1` times with `+` sign, `0` with `-` -> `n-1` times
`a[1]` occurs `n-2` times with `+` sign, `1` with `-` -> `n-3` times
... and so on 
Answered By: Quang Hoang

Not as brilliant as @QuangHoang’s answer, this answer uses numpy broadcasting to calculate the differences matrix, then average the upper triangle values to get the answer.

import numpy as np

a = np.array([110, 60, 30, 10, 5])

dif = np.triu(a.reshape(len(a),-1) - a)
out = np.mean(dif[dif != 0])
52.0
Answered By: AboAmmar

Now, this can easily be done with two For Loops but in O(n^2) time complexity and also a little bit of messy code.

No need for the code to be messy.

from statistics import mean

arr = [110, 60, 30, 10, 5]

m = mean(arr[i] - arr[j] for i in range(len(arr)) for j in range(i+1, len(arr)))

print(m)
# 52

If the array is not sorted, replace arr[i] - arr[j] with abs(arr[i] - arr[j]).

Answered By: Stef
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