You just can put your query as a subquery: SELECT avg(count) FROM ( SELECT COUNT (*) AS Count FROM Table T WHERE T.Update_time = (SELECT MAX (B.Update_time ) FROM Table B WHERE (B.Id = T.Id)) GROUP BY T.Grouping ) as counts Edit: I think this should be the same: SELECT count(*) / count(distinct T.Grouping) FROM … Read more
This answer based on my good understanding of low-pass filters (“exponential moving average” is really just a single-pole lowpass filter), but my hazy understanding of what you’re looking for. I think the following is what you want: First, you can simplify your equation a little bit (looks more complicated but it’s easier in code). I’m … Read more
You can use the Average overload that takes a collection of long in parameter: double doubleAverageTicks = sourceList.Average(timeSpan => timeSpan.Ticks); long longAverageTicks = Convert.ToInt64(doubleAverageTicks); return new TimeSpan(longAverageTicks);
As numpy.convolve is pretty slow, those who need a fast performing solution might prefer an easier to understand cumsum approach. Here is the code: cumsum_vec = numpy.cumsum(numpy.insert(data, 0, 0)) ma_vec = (cumsum_vec[window_width:] – cumsum_vec[:-window_width]) / window_width where data contains your data, and ma_vec will contain moving averages of window_width length. On average, cumsum is about … Read more
Your solution is essentially the “standard” optimal online solution for keeping a running track of average without storing big sums and also while running “online”, i.e. you can just process one number at a time without going back to other numbers, and you only use a constant amount of extra memory. If you want a … Read more
With Java 8 it is a bit easier: OptionalDouble average = marks .stream() .mapToDouble(a -> a) .average(); Thus your average value is average.getAsDouble() return average.isPresent() ? average.getAsDouble() : 0;
first you need to remove empty values, otherwise average will be not accurate. so $a = array_filter($a); $average = array_sum($a)/count($a); echo $average; DEMO More concise and recommended way $a = array_filter($a); if(count($a)) { echo $average = array_sum($a)/count($a); } See here
If your array arr has a length divisible by 3: np.mean(arr.reshape(-1, 3), axis=1) Reshaping to a higher dimensional array and then performing some form of reduce operation on one of the additional dimensions is a staple of numpy programming.
np.average takes an optional weight parameter. If it is not supplied they are equivalent. Take a look at the source code: Mean, Average np.mean: try: mean = a.mean except AttributeError: return _wrapit(a, ‘mean’, axis, dtype, out) return mean(axis, dtype, out) np.average: … if weights is None : avg = a.mean(axis) scl = avg.dtype.type(a.size/avg.size) else: #code … Read more
A solution I consider more elegant: const sum = times.reduce((a, b) => a + b, 0); const avg = (sum / times.length) || 0; console.log(`The sum is: ${sum}. The average is: ${avg}.`);