In the question "What's the numerically best way to calculate the average" it was suggested, that calculating a rolling mean, i.e.
mean = a[n]/n + (n-1)/n * mean
might be numerically more stable than calculating the sum and then dividing by the total number of elements. This was questioned by a commenter. I can not tell which one is true - can someone else? The advantage of the rolling mean is, that you keep the mean small (i.e. at roughly the same size of all vector entries). Intuitively this should keep the error small. But the commenter claims:
Part of the issue is that 1/n introduces errors in the least significant bits, so n/n != 1, at least when it is performed as a three step operation (divide-store-multiply). This is minimized if the division is only performed once, but you'd be doing it over GB of data.
So I have multiple questions:
Is the rolling mean more precise than summing and then dividing?
Does that depend on the question whether 1/n is calculated first and then multiplied?
If so, do computers implement a one step division? (I thought so, but I am unsure now)
If yes, is it more precise than Kahan summation and then dividing?
If compareable - which one is faster? In both cases we have additional calculations.
If more precise, could you use this for precise summation?
In many circumstances, yes. Consider a sequence of all positive terms, all on the same order of magnitude. Adding them all generates a large intermediate sum, to which we add small terms, which might round precisely to the intermediate sum. Using the rolling sum, you get terms on the same order of magnitude, and in addition, the sum is much harder to overflow. However, this is not open and shut: Adding the terms and then dividing allows us to use AVX instructions, which are significantly faster than the subtract/divide/add instructions of the rolling loop. In addition, there are distributions which cause one or the other to be more accurate. This has been examined in:
Robert F Ling. Comparison of several algorithms for computing sample means and variances. Journal of the American Statistical Association, 69(348): 859–866, 1974
Kahan summation is an orthogonal issue. You can apply Kahan summation to the sequence x[n] = (x[n-1]-mu)/n; this is very accurate.
I'm currently writing a UCB1 algorithm for a game. The algorithm as I'm using it is:
average(i) + sqrt( (2 * ln(totalcount)) / count(i) )
Where averagei is the average score of arm i, count(i) is the count for arm i, and totalcount is the total samples of all arms. This is the equation for the score for a given arm i. The arm with the highest score, max(i), is chosen to be sampled. The algorithm then repeats this equation with the new data it got from that sample, ad infinitum, or until it runs out of thinking time.
I have an assignment which tells me to "modify the exploration constant" for the algorithm. I also notice that it exploits rather than explores almost all the time -- it hardly ever tries arms it has only visited once. However, I don't see any exploration constant. Am I missing some part of the algorithm?
The 2 is the exploration constant. The larger it is, the more the algorithm favors exploration over exploitation.
Also beware that this formula makes sense only when the payoffs are in [0,1] range, otherwise a large payoff (say 1000) will nullify the influence of the "exploration" part of the formula, effectively making it exploitation-only.
I'm having sort of an issue trying to figure out how to tune the parameters for my perceptron algorithm so that it performs relatively well on unseen data.
I've implemented a verified working perceptron algorithm and I'd like to figure out a method by which I can tune the numbers of iterations and the learning rate of the perceptron. These are the two parameters I'm interested in.
I know that the learning rate of the perceptron doesn't affect whether or not the algorithm converges and completes. I'm trying to grasp how to change n. Too fast and it'll swing around a lot, and too low and it'll take longer.
As for the number of iterations, I'm not entirely sure how to determine an ideal number.
In any case, any help would be appreciated. Thanks.
Start with a small number of iterations (it's actually more conventional to count 'epochs' rather than iterations--'epochs' refers to the number of iterations through the entire data set used to train the network). By 'small' let's say something like 50 epochs. The reason for this is that you want to see how the total error is changing with each additional training cycle (epoch)--hopefully it's going down (more on 'total error' below).
Obviously you are interested in the point (the number of epochs) where the next additional epoch does not cause a further decrease in total error. So begin with a small number of epochs so you can approach that point by increasing the epochs.
The learning rate you begin with should not be too fine or too coarse, (obviously subjective but hopefully you have a rough sense for what is a large versus small learning rate).
Next, insert a few lines of testing code in your perceptron--really just a few well-placed 'print' statements. For each iteration, calculate and show the delta (actual value for each data point in the training data minus predicted value) then sum the individual delta values over all points (data rows) in the training data (i usually take the absolute value of the delta, or you can take the square root of the sum of the squared differences--doesn't matter too much. Call that summed value "total error"--just to be clear, this is total error (sum of the error across all nodes) per epoch.
Then, plot the total error as a function of epoch number (ie, epoch number on the x axis, total error on the y axis). Initially of course, you'll see the data points in the upper left-hand corner trending down and to the right and with a decreasing slope
Let the algorithm train the network against the training data. Increase the epochs (by e.g., 10 per run) until you see the curve (total error versus epoch number) flatten--i.e., additional iterations doesn't cause a decrease in total error.
So the slope of that curve is important and so is its vertical position--ie., how much total error you have and whether it continues to trend downward with more training cycles (epochs). If, after increasing epochs, you eventually notice an increase in error, start again with a lower learning rate.
The learning rate (usually a fraction between about 0.01 and 0.2) will certainly affect how quickly the network is trained--i.e., it can move you to the local minimum more quickly. It can also cause you to jump over it. So code a loop that trains a network, let's say five separate times, using a fixed number of epochs (and a the same starting point) each time but varying the learning rate from e.g., 0.05 to 0.2, each time increasing the learning rate by 0.05.
One more parameter is important here (though not strictly necessary), 'momentum'. As the name suggests, using a momentum term will help you get an adequately trained network more quickly. In essence, momentum is a multiplier to the learning rate--as long as the the error rate is decreasing, the momentum term accelerates the progress. The intuition behind the momentum term is 'as long as you traveling toward the destination, increase your velocity'.Typical values for the momentum term are 0.1 or 0.2. In the training scheme above, you should probably hold momentum constant while varying the learning rate.
About the learning rate not affecting whether or not the perceptron converges - That's not true. If you choose a learning rate that is too high, you will probably get a divergent network. If you change the learning rate during learning, and it drops too fast (i.e stronger than 1/n) you can also get a network that never converges (That's because the sum of N(t) over t from 1 to inf is finite. that means the vector of weights can only change by a finite amount).
Theoretically it can be shown for simple cases that changing n (learning rate) according to 1/t (where t is the number of presented examples) should work good, but I actually found that in practice, the best way to do this, is to find good high n value (the highest value that doesn't make your learning diverge) and low n value (this one is tricker to figure. really depends on the data and problem), and then let n change linearly over time from high n to low n.
The learning rate depends on the typical values of data. There is no rule of thumb in general. Feature scaling is a method used to standardize the range of independent variables or features of data. In data processing, it is also known as data normalization and is generally performed during the data preprocessing step.
Normalizing the data to a zero-mean, unit variance or between 0-1 or any other standard form can help in selecting a value of learning rate. As doug mentioned, learning rate between 0.05 and 0.2 generally works well.
Also this will help in making the algorithm converge faster.
Source: Juszczak, P.; D. M. J. Tax, and R. P. W. Dui (2002). "Feature scaling in support vector data descriptions". Proc. 8th Annu. Conf. Adv. School Comput. Imaging: 95–10.
Is there an algorithm to estimate the median, mode, skewness, and/or kurtosis of set of values, but that does NOT require storing all the values in memory at once?
I'd like to calculate the basic statistics:
mean: arithmetic average
variance: average of squared deviations from the mean
standard deviation: square root of the variance
median: value that separates larger half of the numbers from the smaller half
mode: most frequent value found in the set
skewness: tl; dr
kurtosis: tl; dr
The basic formulas for calculating any of these is grade-school arithmetic, and I do know them. There are many stats libraries that implement them, as well.
My problem is the large number (billions) of values in the sets I'm handling: Working in Python, I can't just make a list or hash with billions of elements. Even if I wrote this in C, billion-element arrays aren't too practical.
The data is not sorted. It's produced randomly, on-the-fly, by other processes. The size of each set is highly variable, and the sizes will not be known in advance.
I've already figured out how to handle the mean and variance pretty well, iterating through each value in the set in any order. (Actually, in my case, I take them in the order in which they're generated.) Here's the algorithm I'm using, courtesy http://en.wikipedia.org/wiki/Algorithms_for_calculating_variance#On-line_algorithm:
Initialize three variables: count, sum, and sum_of_squares
For each value:
Increment count.
Add the value to sum.
Add the square of the value to sum_of_squares.
Divide sum by count, storing as the variable mean.
Divide sum_of_squares by count, storing as the variable mean_of_squares.
Square mean, storing as square_of_mean.
Subtract square_of_mean from mean_of_squares, storing as variance.
Output mean and variance.
This "on-line" algorithm has weaknesses (e.g., accuracy problems as sum_of_squares quickly grows larger than integer range or float precision), but it basically gives me what I need, without having to store every value in each set.
But I don't know whether similar techniques exist for estimating the additional statistics (median, mode, skewness, kurtosis). I could live with a biased estimator, or even a method that compromises accuracy to a certain degree, as long as the memory required to process N values is substantially less than O(N).
Pointing me to an existing stats library will help, too, if the library has functions to calculate one or more of these operations "on-line".
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.
I would love to see an incremental mode estimator of a similar form...
UPDATE (2011-09-19)
I just 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.
UPDATE (2021-11-19)
For further details about the median estimator described here, I'd like to highlight this paper linked in the comments below: Bylander & Rosen, 1997, A Perceptron-Like Online Algorithm for Tracking the Median. Here is a postscript version from the author's website.
Skewness and Kurtosis
For the on-line algorithms for Skewness and Kurtosis (along the lines of the variance), see in the same wiki page here the parallel algorithms for higher-moment statistics.
Median
Median is tough without sorted data. If you know, how many data points you have, in theory you only have to partially sort, e.g. by using a selection algorithm. However, that doesn't help too much with billions of values. I would suggest using frequency counts, see the next section.
Median and Mode with Frequency Counts
If it is integers, I would count
frequencies, probably cutting off the highest and lowest values beyond some value where I am sure that it is no longer relevant. For floats (or too many integers), I would probably create buckets / intervals, and then use the same approach as for integers. (Approximate) mode and median calculation than gets easy, based on the frequencies table.
Normally Distributed Random Variables
If it is normally distributed, I would use the population sample mean, variance, skewness, and kurtosis as maximum likelihood estimators for a small subset. The (on-line) algorithms to calculate those, you already now. E.g. read in a couple of hundred thousand or million datapoints, until your estimation error gets small enough. Just make sure that you pick randomly from your set (e.g. that you don't introduce a bias by picking the first 100'000 values). The same approach can also be used for estimating mode and median for the normal case (for both the sample mean is an estimator).
Further comments
All the algorithms above can be run in parallel (including many sorting and selection algorithm, e.g. QuickSort and QuickSelect), if this helps.
I have always assumed (with the exception of the section on the normal distribution) that we talk about sample moments, median, and mode, not estimators for theoretical moments given a known distribution.
In general, sampling the data (i.e. only looking at a sub-set) should be pretty successful given the amount of data, as long as all observations are realizations of the same random variable (have the same distributions) and the moments, mode and median actually exist for this distribution. The last caveat is not innocuous. For example, the mean (and all higher moments) for the Cauchy Distribution do not exist. In this case, the sample mean of a "small" sub-set might be massively off from the sample mean of the whole sample.
I implemented the P-Square Algorithm for Dynamic Calculation of Quantiles and Histograms without Storing Observations in a neat Python module I wrote called LiveStats. It should solve your problem quite effectively. The library supports every statistic that you mention except for mode. I have not yet found a satisfactory solution for mode estimation.
Ryan, I'm afraid you are not doing the mean and variance right... This came up a few weeks ago here. And one of the strong points of the online version (which actually goes by the name of Welford's method) is the fact that it is specially accurate and stable, see the discussion here. One of the strong points is the fact that you do not need to store the total sum or total sum of squares...
I can't think of any on-line approach to the mode and median, which seem to require considering the whole list at once. But it may very well be that a similar approach than the one for the variance and mean will work also for the skewness and kurtosis...
The Wikipedia article quoted in the question contains the formulas for calcualting skewness and kurtosis on-line.
For mode - I believe - there is no way doing this on-line. Why? Assume that all values of your input are different besides the last one that duplicates a previous one. In this case you have to remember all values allready seen in the input to detect that the last value duplicates a value seen befor and makes it the most frequent one.
For median it is almost the same - up to the last input you don't know what value will become the median if all input values are different because it could be before or after the current median. If you know the length of the input, you can find the median without storing all values in memory, but you will still have to store many of them (I guess around the half) because a bad input sequence could shift the median heavily in the second half possibly making any value from the first half the median.
(Note that I am refering to exact calculation only.)
If you have billions of data points, then it's not likely that you need exact answers, as opposed to close answers. Generally, if you have billions of data points the underlying process which generates them will likely obey some kind of statistical stationarity / ergodicity / mixing property. Also it may matter whether you expect the distributions to be reasonably continuous or not.
In these circumstances, there exist algorithms for on-line, low memory, estimation of quantiles (the median is a special case of 0.5 quantile), as well as modes, if you don't need exact answers. This is an active field of statistics.
quantile estimation example: http://www.computer.org/portal/web/csdl/doi/10.1109/WSC.2006.323014
mode estimation example: Bickel DR. Robust estimators of the mode and skewness of continuous data. Computational Statistics and Data Analysis. 2002;39:153–163. doi: 10.1016/S0167-9473(01)00057-3.
These are active fields of computational statistics. You are getting into the fields where there isn't any single best exact algorithm, but a diversity of them (statistical estimators, in truth), which have different properties, assumptions and performance. It's experimental mathematics. There are probably hundreds to thousands of papers on the subject.
The final question is whether you really need skewness and kurtosis by themselves, or more likely some other parameters which may be more reliable at characterizing the probability distribution (assuming you have a probability distribution!). Are you expecting a Gaussian?
Do you have ways of cleaning/preprocessing the data to make it mostly Gaussianish? (for instance, financial transaction amounts are often somewhat Gaussian after taking logarithms). Do you expect finite standard deviations? Do you expect fat tails? Are the quantities you care about in the tails or in the bulk?
Everyone keeps saying that you can't do the mode in an online manner but that is simply not true. Here is an article describing an algorithm to do just this very problem invented in 1982 by Michael E. Fischer and Steven L. Salzberg of Yale University. From the article:
The majority-finding algorithm uses one of its registers for temporary
storage of a single item from the stream; this item is the current
candidate for majority element. The second register is a counter
initialized to 0. For each element of the stream, we ask the algorithm
to perform the following routine. If the counter reads 0, install the
current stream element as the new majority candidate (displacing any
other element that might already be in the register). Then, if the
current element matches the majority candidate, increment the counter;
otherwise, decrement the counter. At this point in the cycle, if the
part of the stream seen so far has a majority element, that element is
in the candidate register, and the counter holds a value greater than
0. What if there is no majority element? Without making a second pass through the data—which isn't possible in a stream environment—the
algorithm cannot always give an unambiguous answer in this
circumstance. It merely promises to correctly identify the majority
element if there is one.
It can also be extended to find the top N with more memory but this should solve it for the mode.
Ultimately if you have no a priori parametric knowledge of the distribution I think you have to store all the values.
That said unless you are dealing with some sort of pathological situation, the remedian (Rousseuw and Bassett 1990) may well be good enough for your purposes.
Very simply it involves calculating the median of batches of medians.
median and mode can't be calculated online using only constant space available. However, because median and mode are anyway more "descriptive" than "quantitative", you can estimate them e.g. by sampling the data set.
If the data is normal distributed in the long run, then you could just use your mean to estimate the median.
You can also estimate median using the following technique: establish a median estimation M[i] for every, say, 1,000,000 entries in the data stream so that M[0] is the median of the first one million entries, M[1] the median of the second one million entries etc. Then use the median of M[0]...M[k] as the median estimator. This of course saves space, and you can control how much you want to use space by "tuning" the parameter 1,000,000. This can be also generalized recursively.
I would tend to use buckets, which could be adaptive. The bucket size should be the accuracy you need. Then as each data point comes in you add one to the relevant bucket's count.
These should give you simple approximations to median and kurtosis, by counting each bucket as its value weighted by its count.
The one problem could be loss of resolution in floating point after billions of operations, i.e. adding one does not change the value any more! To get round this, if the maximum bucket size exceeds some limit you could take a large number off all the counts.
OK dude try these:
for c++:
double skew(double* v, unsigned long n){
double sigma = pow(svar(v, n), 0.5);
double mu = avg(v, n);
double* t;
t = new double[n];
for(unsigned long i = 0; i < n; ++i){
t[i] = pow((v[i] - mu)/sigma, 3);
}
double ret = avg(t, n);
delete [] t;
return ret;
}
double kurt(double* v, double n){
double sigma = pow(svar(v, n), 0.5);
double mu = avg(v, n);
double* t;
t = new double[n];
for(unsigned long i = 0; i < n; ++i){
t[i] = pow( ((v[i] - mu[i]) / sigma) , 4) - 3;
}
double ret = avg(t, n);
delete [] t;
return ret;
}
where you say you can already calculate sample variance (svar) and average (avg)
you point those to your functions for doin that.
Also, have a look at Pearson's approximation thing. on such a large dataset it would be pretty similar.
3 (mean − median) / standard deviation
you have median as max - min/2
for floats mode has no meaning. one would typically stick them in bins of a sginificant size (like 1/100 * (max - min)).
This problem was solved by Pebay et al:
https://prod-ng.sandia.gov/techlib-noauth/access-control.cgi/2008/086212.pdf
Median
Two recent percentile approximation algorithms and their python implementations can be found here:
t-Digests
https://arxiv.org/abs/1902.04023
https://github.com/CamDavidsonPilon/tdigest
DDSketch
https://arxiv.org/abs/1908.10693
https://github.com/DataDog/sketches-py
Both algorithms bucket data. As T-Digest uses smaller bins near the tails the
accuracy is better at the extremes (and weaker close to the median). DDSketch additionally provides relative error guarantees.
for j in range (1,M):
y=np.zeros(M) # build the vector y
y[0]=y0
#generate the white noise
eps=npr.randn(M-1)*np.sqrt(var)
#increment the y vector
for k in range(1,T):
y[k]=corr*y[k-1]+eps[k-1]
yy[j]=y
list.append(y)
Assume we're frequently sampling a particular value and want to keep statistics on the samples. The simplest approach is to store every sample so we can calculate whatever stats we want, but this requires unbounded storage. Using a constant amount of storage, we can keep track of some stats like minimum and maximum values. What else can we track using only constant storage? I am thinking of percentiles, standard deviation, and any other useful statistics.
That's the theoretical question. In my actual situation, the samples are simply millisecond timings: profiling information for a long-running application. There will be millions of samples but not much more than a billion or so. So what stats can be kept for the samples using no more than, say, 10 variables?
Minimum, maximum, average, total count, variance are all easy and useful. That's 5 values.
Usually you'll store sum and not average, and when you need the average you can just divide the sum by the count.
So, in your loop
maxVal=max(x, maxVal);
minVal=min(x, minVal);
count+=1;
sum+=x;
secondorder+=x*x;
later, you may print any of these stats. Mean and standard deviation can be computed at any time and are:
mean=sum/count;
std=sqrt(secondorder/count - mean*mean);
Median and percentile estimation are more difficult, but possible. The usual trick is to make a set of histogram bins and fill the bins when a sample is found inside them.
You can then estimate median and such by looking at the distribution of those bin populations.
This is only an approximation to the distribution, but often enough. To find the exact median, you must store all samples.
Well, lots of them. Consider, eg, σ. Square root of the variance, which is computed from the average and the data point, repeatedly. At point Xi, you have an estimated variance for all the points up to this one; then add (x-μ)² to that for the next estimate. There's a nice section on that in wikipedia.
What you're looking for, in general, is called "on line methods for calculation".