Small circular queue buffer with median.
Usecase: Sometimes you just need to get the median value out of 5-10 entries on low resource, Arduino type of system. Most often, those values are simple uint8_t or uint16_t and there is no need for bigger or more complex solutions.
Functions:
| function | description |
|---|---|
| `T push()` | puts data in |
| `T pop()` | pops data out (the oldest one) |
| `T peek()` | reads data (the oldest one) |
| `unsignedT` peekTime() | reads timestamp of the oldest data entry |
| `clear()` | clears all |
| `bool isFull()` | returns if buffer is full |
| `bool isEmpty()` | returns if buffer is empty |
| `uint8_t getCount()` | returns count of items in buffer |
| `uint8_t getPushCount()` | returns count of all puts (255 max, then overflows) |
| `resetPushCount()` | resets push counter |
| `bool deleteOld()` | deletes older then (now - interval) |
| `T maxValue()` | gets max value |
| `T minValue()` | gets min value |
| `T range()` | max - min |
| `uint8_t occurenceOfValue()` | for a given value, how many time it appears |
| `resultingT frequencyOfValue()` | for a given value, 1 / occurenceOfValue |
| `meanAbsoluteDeviationAroundAverage()` | average of abs(each value-average) |
| `meanAbsoluteDeviationAroundMedianAverage()` | average of abs(each value-median) |
| `T average()` | gets average of all items |
| `T median()` | gets median, original entry no matter what |
| `T medianAverage()` | gets median, average of values around median at max distance from |
| `T averageInterval()` | gets average interval of time |
| `T averageRateOfChange()` | 1 / averageInterval |
| `T medianInterval()` | gets median time interval between sequential items |
| `T medianAverageInterval()` | gets medianAverage time interval between sequential items |
| `T medianRateOfChange()` | 1 / medianInterval (usefull for count/second measurements) |
| `T medianAverageRateOfChange()` | 1 / medianAverageInterval (usefull for count/second measurements) |
Note: Median is often expressed as one of two following equations. The latter is used here.
(double)(a[(n - 1) / 2] + a[n / 2]) / 2.0
(double)(a[(n / 2) - 1] + a[n / 2]) / 2.0