Timeline for Incremental averaging
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 28, 2019 at 14:23 | comment | added | Hydrargyrum | By the way, I have a practical application for this formula. I'm doing moving average calculations in a real-time system where I want to incrementally update the moving average as old values expire out of the averaging window, without recalculating the entire average each time a new value arrives. | |
| Aug 28, 2019 at 14:21 | comment | added | Hydrargyrum | @MichalWolodzko This seems like a very dumb objection, but I'm just working through the equations in my head. Let's say that for the first equation, we want the average from n = 7 to m = 5. That gives three terms in the numerator (a_5, a_6, and a_7) and a denominator of 7 - 5 = 2. This doesn't seem right - surely the denominator must be the number of samples being averaged. Am I missing something obvious? | |
| Feb 12, 2014 at 16:09 | comment | added | Michal Wolodzko | Yes. You need at least to know how many elements you include in the average, the first element in the average and the average from the previous step. Note please, that $n-m$, that is the total count of elements in the average can be a constant number. So at least one piece of information you have given "for free" - you don't have to calculate it when new data arrives. | |
| Feb 10, 2014 at 16:20 | comment | added | Ali | So, what happens in the case you are going one step at a time and don't want to store any past information, like you have the previous average, now you have one new element? what else you need to keep, total count of elements and the first one? | |
| Feb 10, 2014 at 16:18 | comment | added | Michal Wolodzko | I have to mention, that I didn't manage to find any practical appliaction for this avarage. I thought about using it for calculating complexity-efficient avarage with big ammount of data (let sat n-m = 1000), but in the end I gave up. If you know how to store $n-m$th element (for $a_{m}$) without using $n-m$th element array, please let me know. | |
| S Feb 10, 2014 at 13:52 | review | Late answers | |||
| Feb 10, 2014 at 13:54 | |||||
| S Feb 10, 2014 at 13:52 | review | First posts | |||
| Feb 10, 2014 at 14:03 | |||||
| Feb 10, 2014 at 13:33 | history | answered | Michal Wolodzko | CC BY-SA 3.0 |