There are multiple methods for find the average of a set of numbers.
First, the sum / count quotient. Add all values and divide them by the number of values.
Second, the moving average. The function I found in another Stack answer is:
New average = old average * (n-1)/n + new value /n
This works as long as each value is added to the average one value at a time.
My concern is that the second method is more calculatively complex for my processor to execute, but I also fear that the first method will result in a loss of resolution for data sets that result in large sums. In a 32 bit system, for example, the resolution of a float value stored is reduced automatically as the magnitude of the number grows.
Does a moving average preserve resolution?
"moving average" does not calculate average over large interval.
It smooth data in such way that newer measurements give larger impact, and older measurements weight becomes smaller and smaller.
If you bother about large sums and want to preserve all possible data "bits", consider special methods like Kahan summation algorithm
I have a set of n complex numbers that move through the complex plane from time step 1 to nsampl . I want to plot those numbers and their trace over time (y-axis shows imaginary part, x-axis the real part). The numbers are stored in a n x nsampl vector. However in each time step the order of the n points is random. Thus in each time step I pick a point in the last time step, find its nearest neighbor in the current time step and put it at the same position as the current point. Then I repeat that for all other n-1 points and go on to the next time step. This way every point in the previous step is associated with exactly one point in the new step (1:1 relation). My current implementation and an example are given below. However my implementation is terribly slow (takes about 10s for 10 x 4000 complex numbers). As I want to increase both, the set size n and the time frames nsampl this is really important to me. Is there a smarter way to implement this to gain some performance?
Example with n=3 and nsampl=2:
%manually create a test vector X
X=zeros(3,2); % zeros(n,nsampl)
X(:,1)=[1+1i; 2+2i; 3+3i];
X(:,2)=[2.1+2i; 5+5i; 1.1+1.1i]; % <-- this is my vector with complex numbers
%vector sort algorithm
for k=2:nsampl
Xlast=[real(X(:,k-1)) imag(X(:,k-1))]; % create vector with x/y-coords from last time step
Xcur=[real(X(:,k)) imag(X(:,k))]; % create vector with x/y-coords from current time step
for i=1:size(X,1) % loop over all n points
idx = knnsearch(Xcur(i:end,:),Xlast(i,:)); %find nearest neighbor to Xlast(i,:), but only use the points not already associated, thus Xcur(i:end,:) points
idx = idx + i - 1;
Xcur([i idx],:) = Xcur([idx i],:); %sort nearest neighbor to the same position in the vector as it was in the last time step
end
X(:,k) = Xcur(:,1)+1i*Xcur(:,2); %revert x/y coordinates to a complex number
end
Result:
X(:,2)=[1.1+1.1i; 2.1+2i; 5+5i];
Can anyone help me to speed up this code?
The problem you are tying to solve is combinatorial optimizartion which is solved by the hungarian algorithm (aka munkres). Luckily there is a implementation for matlab available for download. Download the file and put it either on your search path or next to your function. The code to use it is:
for k=2:size(X,2)
%build up a cost matrix, here cost is the distance between two points.
pairwise_distances=abs(bsxfun(#minus,X(:,k-1),X(:,k).'));
%let the algorithm find the optimal pairing
permutation=munkres(pairwise_distances);
%apply it
X(:,k)=X(permutation,k);
end
I come up with this
n=1;
curAvg = 0;
loop{
curAvg = curAvg + (newNum - curAvg)/n;
n++;
}
I think highlights of this way are:
- It avoids big numbers (and possible overflow if you would sum and then divide)
- you save one register (not need to store sum)
The trouble might be with summing error - but I assume that generally there shall be balanced numbers of round up and round down so the error shall not sum up dramatically.
Do you see any pitfalls in this solution?
Have you any better proposal?
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 slightly optimized solution in terms of numerical accuracy, at the cost of being "online", then assuming your numbers are all non-negative, then sort your numbers first from smallest to largest and then process them in that order, the same way you do now. That way, if you get a bunch of numbers that are really small about equal and then you get one big number, you will be able to compute the average accurately without underflow, as opposed to if you processed the large number first.
I have used this algorithm for many years.
The loop is any kind of loop. Maybe it is individual web sessions or maybe true loop. The point is all you need to track is the current count (N) and the current average (avg). Each time a new value is received, apply this algorithm to update the average. This will compute the exact arithmetic average. It has the additional benefit that it is resistant to overflow. If you have a gazillion large numbers to average, summing them all up may overflow before you get to divide by N. This algorithm avoids that pitfall.
Variables that are stored during the computation of the average:
N = 0
avg = 0
For each new value: V
N=N+1
a = 1/N
b = 1 - a
avg = a * V + b * avg
I am trying to find a function in Matlab, or at least the name of an algorithm that does the following:
Let's say that I am analyzing a time series in real time. I initially start with a threshold of 10 and -10, so that when the time series goes above 10 or below -10, it's considered a 'HIT'. Let's say it initially takes the time series 5 minutes to produce a 'HIT', but I want to adjust the threshold so that, on average, it takes only 1 minute for a 'HIT' to be produced. I know it would look something like start with 10 and -10, if it takes too long, drop it to 5 and -5, then increase the threshold if it's too quick, etc.
I know there's a specific name for this type of algorithm, and there's probably built-in functions for this, but the name is eluding me. Can somebody help?
I don't know what the time resolution of your time series is, or if it's constant, so I'll leave that to you. However here is what you can do in matlab if you have a constant time resolution. First take the absolute value of the values in your time series. Then sort these values in reverse order using the sort() command. Then choose the value whose index in the sorted array gives you the average time resolution that you desire. So e.g. if your time series has size N and the time resolution is 0.1 seconds, and you want an alert on average every 1 second, then after sorting you would choose the threshold at (reverse order) sorted position N/10.
I'm trying to calculate the median of a set of values, but I don't want to store all the values as that could blow memory requirements. Is there a way of calculating or approximating the median without storing and sorting all the individual values?
Ideally I would like to write my code a bit like the following
var medianCalculator = new MedianCalculator();
foreach (var value in SourceData)
{
medianCalculator.Add(value);
}
Console.WriteLine("The median is: {0}", medianCalculator.Median);
All I need is the actual MedianCalculator code!
Update: Some people have asked if the values I'm trying to calculate the median for have known properties. The answer is yes. One value is in 0.5 increments from about -25 to -0.5. The other is also in 0.5 increments from -120 to -60. I guess this means I can use some form of histogram for each value.
Thanks
Nick
If the values are discrete and the number of distinct values isn't too high, you could just accumulate the number of times each value occurs in a histogram, then find the median from the histogram counts (just add up counts from the top and bottom of the histogram until you reach the middle). Or if they're continuous values, you could distribute them into bins - that wouldn't tell you the exact median but it would give you a range, and if you need to know more precisely you could iterate over the list again, examining only the elements in the central bin.
There is the 'remedian' statistic. It works by first setting up k arrays, each of length b. Data values are fed in to the first array and, when this is full, the median is calculated and stored in the first pos of the next array, after which the first array is re-used. When the second array is full the median of its values is stored in the first pos of the third array, etc. etc. You get the idea :)
It's simple and pretty robust. The reference is here...
http://web.ipac.caltech.edu/staff/fmasci/home/astro_refs/Remedian.pdf
Hope this helps
Michael
I use these incremental/recursive mean and median estimators, which both use constant storage:
mean += eta * (sample - mean)
median += eta * sgn(sample - median)
where eta is a small learning rate parameter (e.g. 0.001), and sgn() is the signum function which returns one of {-1, 0, 1}. (Use a constant eta if the data is non-stationary and you want to track changes over time; otherwise, for stationary sources you can use something like eta=1/n for the mean estimator, where n is the number of samples seen so far... unfortunately, this does not appear to work for the median estimator.)
This type of incremental mean estimator seems to be used all over the place, e.g. in unsupervised neural network learning rules, but the median version seems much less common, despite its benefits (robustness to outliers). It seems that the median version could be used as a replacement for the mean estimator in many applications.
Also, I modified the incremental median estimator to estimate arbitrary quantiles. In general, a quantile function tells you the value that divides the data into two fractions: p and 1-p. The following estimates this value incrementally:
quantile += eta * (sgn(sample - quantile) + 2.0 * p - 1.0)
The value p should be within [0,1]. This essentially shifts the sgn() function's symmetrical output {-1,0,1} to lean toward one side, partitioning the data samples into two unequally-sized bins (fractions p and 1-p of the data are less than/greater than the quantile estimate, respectively). Note that for p=0.5, this reduces to the median estimator.
I would love to see an incremental mode estimator of a similar form...
(Note: I also posted this to a similar topic here: "On-line" (iterator) algorithms for estimating statistical median, mode, skewness, kurtosis?)
Here is a crazy approach that you might try. This is a classical problem in streaming algorithms. The rules are
You have limited memory, say O(log n) where n is the number of items you want
You can look at each item once and make a decision then and there what to do with it, if you store it, it costs memory, if you throw it away it is gone forever.
The idea for the finding a median is simple. Sample O(1 / a^2 * log(1 / p)) * log(n) elements from the list at random, you can do this via reservoir sampling (see a previous question). Now simply return the median from your sampled elements, using a classical method.
The guarantee is that the index of the item returned will be (1 +/- a) / 2 with probability at least 1-p. So there is a probability p of failing, you can choose it by sampling more elements. And it wont return the median or guarantee that the value of the item returned is anywhere close to the median, just that when you sort the list the item returned will be close to the half of the list.
This algorithm uses O(log n) additional space and runs in Linear time.
This is tricky to get right in general, especially to handle degenerate series that are already sorted, or have a bunch of values at the "start" of the list but the end of the list has values in a different range.
The basic idea of making a histogram is most promising. This lets you accumulate distribution information and answer queries (like median) from it. The median will be approximate since you obviously don't store all values. The storage space is fixed so it will work with whatever length sequence you have.
But you can't just build a histogram from say the first 100 values and use that histogram continually.. the changing data may make that histogram invalid. So you need a dynamic histogram that can change its range and bins on the fly.
Make a structure which has N bins. You'll store the X value of each slot transition (N+1 values total) as well as the population of the bin.
Stream in your data. Record the first N+1 values. If the stream ends before this, great, you have all the values loaded and you can find the exact median and return it. Else use the values to define your first histogram. Just sort the values and use those as bin definitions, each bin having a population of 1. It's OK to have dupes (0 width bins).
Now stream in new values. For each one, binary search to find the bin it belongs to.
In the common case, you just increment the population of that bin and continue.
If your sample is beyond the histogram's edges (highest or lowest), just extend the end bin's range to include it.
When your stream is done, you find the median sample value by finding the bin which has equal population on both sides of it, and linearly interpolating the remaining bin-width.
But that's not enough.. you still need to ADAPT the histogram to the data as it's being streamed in. When a bin gets over-full, you're losing information about that bin's sub distribution.
You can fix this by adapting based on some heuristic... The easiest and most robust one is if a bin reaches some certain threshold population (something like 10*v/N where v=# of values seen so far in the stream, and N is the number of bins), you SPLIT that overfull bin. Add a new value at the midpoint of the bin, give each side half of the original bin's population. But now you have too many bins, so you need to DELETE a bin. A good heuristic for that is to find the bin with the smallest product of population and width. Delete it and merge it with its left or right neighbor (whichever one of the neighbors itself has the smallest product of width and population.). Done!
Note that merging or splitting bins loses information, but that's unavoidable.. you only have fixed storage.
This algorithm is nice in that it will deal with all types of input streams and give good results. If you have the luxury of choosing sample order, a random sample is best, since that minimizes splits and merges.
The algorithm also allows you to query any percentile, not just median, since you have a complete distribution estimate.
I use this method in my own code in many places, mostly for debugging logs.. where some stats that you're recording have unknown distribution. With this algorithm you don't need to guess ahead of time.
The downside is the unequal bin widths means you have to do a binary search for each sample, so your net algorithm is O(NlogN).
David's suggestion seems like the most sensible approach for approximating the median.
A running mean for the same problem is a much easier to calculate:
Mn = Mn-1 + ((Vn - Mn-1) / n)
Where Mn is the mean of n values, Mn-1 is the previous mean, and Vn is the new value.
In other words, the new mean is the existing mean plus the difference between the new value and the mean, divided by the number of values.
In code this would look something like:
new_mean = prev_mean + ((value - prev_mean) / count)
though obviously you may want to consider language-specific stuff like floating-point rounding errors etc.
I don't think it is possible to do without having the list in memory. You can obviously approximate with
average if you know that the data is symmetrically distributed
or calculate a proper median of a small subset of data (that fits in memory) - if you know that your data has the same distribution across the sample (e.g. that the first item has the same distribution as the last one)
Find Min and Max of the list containing N items through linear search and name them as HighValue and LowValue
Let MedianIndex = (N+1)/2
1st Order Binary Search:
Repeat the following 4 steps until LowValue < HighValue.
Get MedianValue approximately = ( HighValue + LowValue ) / 2
Get NumberOfItemsWhichAreLessThanorEqualToMedianValue = K
is K = MedianIndex, then return MedianValue
is K > MedianIndex ? then HighValue = MedianValue Else LowValue = MedianValue
It will be faster without consuming memory
2nd Order Binary Search:
LowIndex=1
HighIndex=N
Repeat Following 5 Steps until (LowIndex < HighIndex)
Get Approximate DistrbutionPerUnit=(HighValue-LowValue)/(HighIndex-LowIndex)
Get Approximate MedianValue = LowValue + (MedianIndex-LowIndex) * DistributionPerUnit
Get NumberOfItemsWhichAreLessThanorEqualToMedianValue = K
is (K=MedianIndex) ? return MedianValue
is (K > MedianIndex) ? then HighIndex=K and HighValue=MedianValue Else LowIndex=K and LowValue=MedianValue
It will be faster than 1st order without consuming memory
We can also think of fitting HighValue, LowValue and MedianValue with HighIndex, LowIndex and MedianIndex to a Parabola, and can get ThirdOrder Binary Search which will be faster than 2nd order without consuming memory and so on...
Usually if the input is within a certain range, say 1 to 1 million, it's easy to create an array of counts: read the code for "quantile" and "ibucket" here: http://code.google.com/p/ea-utils/source/browse/trunk/clipper/sam-stats.cpp
This solution can be generalized as an approximation by coercing the input into an integer within some range using a function that you then reverse on the way out: IE: foo.push((int) input/1000000) and quantile(foo)*1000000.
If your input is an arbitrary double precision number, then you've got to autoscale your histogram as values come in that are out of range (see above).
Or you can use the median-triplets method described in this paper: http://web.cs.wpi.edu/~hofri/medsel.pdf
I picked up the idea of iterative quantile calculation. It is important to have a good value for starting point and eta, these may come from mean and sigma. So I programmed this:
Function QuantileIterative(Var x : Array of Double; n : Integer; p, mean, sigma : Double) : Double;
Var eta, quantile,q1, dq : Double;
i : Integer;
Begin
quantile:= mean + 1.25*sigma*(p-0.5);
q1:=quantile;
eta:=0.2*sigma/xy(1+n,0.75); // should not be too large! sets accuracy
For i:=1 to n Do
quantile := quantile + eta * (signum_smooth(x[i] - quantile,eta) + 2*p - 1);
dq:=abs(q1-quantile);
If dq>eta
then Begin
If dq<3*eta then eta:=eta/4;
For i:=1 to n Do
quantile := quantile + eta * (signum_smooth(x[i] - quantile,eta) + 2*p - 1);
end;
QuantileIterative:=quantile
end;
As the median for two elements would be the mean, I used a smoothed signum function, and xy() is x^y. Are there ideas to make it better? Of course if we have some more a-priori knowledge we can add code using min and max of the array, skew, etc. For big data you would not use an array perhaps, but for testing it is easier.
On homogeneous random ordered and for big enough list, this pseudo code can work:
# find min on the fly
if minDataPoint > dataPoint:
minDataPoint = dataPoint
# find max on the fly
if maxDataPoint < dataPoint:
maxDataPoint = dataPoint
# estimate median base on the current data
estimate_mid = (maxDataPoint + minDataPoint) / 2
#if **new** dataPoint is closer to the mid? stor it
if abs(midDataPoint - estimate_mid) > abs(dataPoint - estimate_mid):
midDataPoint = dataPoint
Inspired by #lakshmanaraj