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Can I calculate the average of these numbers?

Tags:

math

average

I was wondering if it's possible to calculate the average of some numbers if I have this:

int currentCount = 12;
float currentScore = 6.1123   (this is a range of 1 <-> 10).

Now, if I receive another score (let's say 4.5), can I recalculate the average so it would be something like:

int currentCount now equals 13
float currentScore now equals ?????

or is this impossible and I still need to remember the list of scores?

like image 773
Pure.Krome Avatar asked Dec 04 '08 12:12

Pure.Krome


3 Answers

The following formulas allow you to track averages just from stored average and count, as you requested.

currentScore = (currentScore * currentCount + newValue) / (currentCount + 1)
currentCount = currentCount + 1

This relies on the fact that your average is currently your sum divided by the count. So you simply multiply count by average to get the sum, add your new value and divide by (count+1), then increase count.

So, let's say you have the data {7,9,11,1,12} and the only thing you're keeping is the average and count. As each number is added, you get:

+--------+-------+----------------------+----------------------+
| Number | Count |   Actual average     | Calculated average   |
+--------+-------+----------------------+----------------------+
|      7 |     1 | (7)/1           =  7 | (0 * 0 +  7) / 1 = 7 |
|      9 |     2 | (7+9)/2         =  8 | (7 * 1 +  9) / 2 = 8 |
|     11 |     3 | (7+9+11)/3      =  9 | (8 * 2 + 11) / 3 = 9 |
|      1 |     4 | (7+9+11+1)/4    =  7 | (9 * 3 +  1) / 4 = 7 |
|     12 |     5 | (7+9+11+1+12)/5 =  8 | (7 * 4 + 12) / 5 = 8 |
+--------+-------+----------------------+----------------------+
like image 98
paxdiablo Avatar answered Oct 27 '22 15:10

paxdiablo


I like to store the sum and the count. It avoids an extra multiply each time.

current_sum += input;
current_count++;
current_average = current_sum/current_count;
like image 20
John with waffle Avatar answered Oct 27 '22 15:10

John with waffle


It's quite easy really, when you look at the formula for the average: A1 + A2 + ... + AN/N. Now, If you have the old average and the N (numbers count) you can easily calculate the new average:

newScore = (currentScore * currentCount + someNewValue)/(currentCount + 1)
like image 26
Kasprzol Avatar answered Oct 27 '22 17:10

Kasprzol