How to add and subtract values from an average?

Question

Say I have 100 numbers that are averaged:

number of values = 100
total sum of values = 2000
mean = 2000 / 100 => 20

If I want to add a value and find out the new average:

total sum of values = 2000 + 100
mean = 2100 / 101 => 20.79

If I want to subtract a value and find out the new average:

total sum of values = 2100 - 100
mean = 2000 / 100 => 20

It seems to work, but is the above correct?

Is this the proper way to add/subtract values from a average without having to re-sum all the 100 numbers first?

score 83 · Answer 1 · edited Sep 12 '24 at 14:08

83

I know that's an old thread but I had the same problem. I want to add a value to an existing average without having to calculate the total sum again.

to add a value to an exisitng average we only must know for how many values the average was calculated for: $$ average_{new} = average_{old} + \frac{ value_{new} - average_{old}}{size_{new}} $$

edited Sep 12 '24 at 14:08

Peter Petigru

279

answered Oct 03 '14 at 23:08

501 - not implemented

1,005

26

I may be a smart guy, but the accepted answer above was incomprehensible to me. This answer works perfectly, and I can understand it. Thanks! – Ty H. Nov 13 '14 at 16:37
3

@501 - not implemented thank you! – ThelmaJay Jan 13 '15 at 14:34
1

This answer was exactly what I needed. Elegant! – visc Nov 16 '16 at 23:02
-1 Careful this formula works only for adding 1 new value! Please work out the correct equation using this answer and you'll see! https://math.stackexchange.com/a/1153800/503549 – kouton Nov 16 '17 at 04:25
Is it possible to calculate median in a similar way? – Przemysław Niedziela Mar 09 '20 at 22:14
3

@kouton why would you downvote for an answer that only works for adding one value when the requirement was "...to add A value..." (emphasis added). If I ask you to run a mile and you run a mile, I'm not going to penalize you for not running two miles. – Jamie Apr 05 '22 at 00:14

score 38 · Accepted Answer · edited Jun 05 '20 at 12:41

38

$s=\frac{a_1+...+a_n}{n}$.

If you want the average of $a_1,...,a_n$ and $a_{n+1}$, then $s'=\frac{a_1+...+a_n+a_{n+1}}{n+1}=\frac{ns+a_{n+1}}{n+1} = \frac{(n+1)s+a_{n+1}}{n+1} - \frac{s}{n+1} = s + \frac{a_{n+1}-s}{n+1}$

If you want the average of $a_1,...,a_{n-1}$ then $s''=\frac{a_1+...+a_{n-1}}{n-1}=\frac{ns-a_n}{n-1}= \frac{(n-1)s-a_n}{n-1} + \frac{s}{n-1}=s+\frac{s-a_n}{n-1}$.

edited Jun 05 '20 at 12:41

John St

3

answered Feb 16 '11 at 13:19

Weltschmerz

7,148

3

And $ns$ is the old sum, as you can derive from the first equation. – Ross Millikan Feb 16 '11 at 13:21
2

I'm tempted to downvote as ns is not explained in the answer – Celeritas Aug 22 '14 at 20:32
At first I commented that @501's answer was clearer but realized I didn't fully understand why the question worked. This answer, while denser, actually explains the result. – jds Dec 10 '14 at 23:05
4

@Celeritas $ns = n \cdot s$, $n$ and $s$ are defined quantities in the post. – Axoren Jan 13 '15 at 15:13
This is the working answer. @501's answer doesn't work when increasing n by more than 1. – kouton Nov 16 '17 at 04:26

score 27 · Answer 3 · edited Jun 17 '24 at 17:10

27

To put it programmatically, and since the question was about how to both add and subtract:

Add a value:

average = average + ((value - average) / (nValues + 1))

Subtract a value:

average = (average * nValues - value) / (nValues - 1)

edited Jun 17 '24 at 17:10

9 Guy

103

answered Dec 09 '15 at 10:35

Damien

511

12

Oh thank god there's another programmer on here. Those math equations make my head hurt. – Matt Jan 05 '19 at 06:38
1

@MattD: see also this answer on StackOverflow for replacing a number in a running average. – Dan Dascalescu Jan 13 '20 at 05:03
3

Shouldn't the first formula end with / nValues + 1) rather than / nValues)? – Magnus Jun 18 '21 at 19:56

score 4 · Answer 4 · edited May 08 '21 at 22:07

4

Another formula can be

$$\text{NewAvg} = \frac{( \text{OldAvg} \cdot \text{OldSize} ) + \text{NewValue} } { \text{NewSize}}$$

edited May 08 '21 at 22:07

656475

5,473

answered May 08 '21 at 20:11

Sushrut Thakur

49

1

I was looking for this answer. It's intuitive because newAvg = (sum + newValue) / newSize (by definition of average). And also by definition sum = avg * size. – dekuShrub Mar 03 '22 at 06:57
And also easy for multiple values: newAvg = (sum1 + sum2) / (size1 + size2). Or in the other format newAvg = (avg1 * size1 + avg2 * size2) / (size1 + size2). – dekuShrub Mar 03 '22 at 07:00

How to add and subtract values from an average?

4 Answers4