Considering this problem : Having a vector of 1000 real positive numbers,find the optim partition of the 1000 elements in 7 parts so that the sum of parts have aproximative(close) values.
How would you make the chromosome representation, operators (mutation,crossover), fitness function, selection.. so that you solve the problem in the most efficent & optimized way ?
My idea is to give each number a index (the lowest number has index 1, the highest has index 1000 for example)... but I don't think this is the most efficent way? Any suggestions are welcome !
Since its a partition problem I think you need to have the whole set in a single chromosome. Say you have an array of length 1000 that can have values from 1 to 7. The fitness function can calculate the difference of the sum of every partition (less is better). Crossover can be done with single point or double point. Then mutation can randomly change an individual gene from its value to a random value, say position 102 is 4, then mutates to 1. With this solution you guarantee that every chromosome is a valid solution, altough possibly a bad one, so you don't have to check after every iteration for chromosomes that do not follow the problem rules (a problem you would have if you choose to have one chromosome per partition). As usual the criteria for crossover and the likelihood of mutation needs exploration and tunning before achiving best performance.
There is a very expensive computation I must make frequently.
The computation takes a small array of numbers (with about 20 entries) that sums to 1 (i.e. the histogram) and outputs something that I can store pretty easily.
I have 2 things going for me:
I can accept approximate answers
The "answers" change slowly. For example: [.1 .1 .8 0] and [.1
.1 .75 .05] will yield similar results.
Consequently, I want to build a look-up table of answers off-line. Then, when the system is running, I can look-up an approximate answer based on the "shape" of the input histogram.
To be precise, I plan to look-up the precomputed answer that corresponds to the histogram with the minimum Earth-Mover-Distance to the actual input histogram.
I can only afford to store about 80 to 100 precomputed (histogram , computation result) pairs in my look up table.
So, how do I "spread out" my precomputed histograms so that, no matter what the input histogram is, I'll always have a precomputed result that is "close"?
Finding N points in M-space that are a best spread-out set is more-or-less equivalent to hypersphere packing (1,2) and in general answers are not known for M>10. While a fair amount of research has been done to develop faster methods for hypersphere packings or approximations, it is still regarded as a hard problem.
It probably would be better to apply a technique like principal component analysis or factor analysis to as large a set of histograms as you can conveniently generate. The results of either analysis will be a set of M numbers such that linear combinations of histogram data elements weighted by those numbers will predict some objective function. That function could be the “something that you can store pretty easily” numbers, or could be case numbers. Also consider developing and training a neural net or using other predictive modeling techniques to predict the objective function.
Building on #jwpat7's answer, I would apply k-means clustering to a huge set of randomly generated (and hopefully representative) histograms. This would ensure that your space was spanned with whatever number of exemplars (precomputed results) you can support, with roughly equal weighting for each cluster.
The trick, of course, will be generating representative data to cluster in the first place. If you can recompute from time to time, you can recluster based on the actual data in the system so that your clusters might get better over time.
I second jwpat7's answer, but my very naive approach was to consider the count of items in each histogram bin as a y value, to consider the x values as just 0..1 in 20 steps, and then to obtain parameters a,b,c that describe x vs y as a cubic function.
To get a "covering" of the histograms I just iterated through "possible" values for each parameter.
e.g. to get 27 histograms to cover the "shape space" of my cubic histogram model I iterated the parameters through -1 .. 1, choosing 3 values linearly spaced.
Now, you could change the histogram model to be quartic if you think your data will often be represented that way, or whatever model you think is most descriptive, as well as generate however many histograms to cover. I used 27 because three partitions per parameter for three parameters is 3*3*3=27.
For a more comprehensive covering, like 100, you would have to more carefully choose your ranges for each parameter. 100**.3 isn't an integer, so the simple num_covers**(1/num_params) solution wouldn't work, but for 3 parameters 4*5*5 would.
Since the actual values of the parameters could vary greatly and still achieve the same shape it would probably be best to store ratios of them for comparison instead, e.g. for my 3 parmeters b/a and b/c.
Here is an 81 histogram "covering" using a quartic model, again with parameters chosen from linspace(-1,1,3):
edit: Since you said your histograms were described by arrays that were ~20 elements, I figured fitting parameters would be very fast.
edit2 on second thought I think using a constant in the model is pointless, all that matters is the shape.
Say that you have a list of scores with binary labels (for simplicity, assume no ties), and that we've used the labels to compute the area under the associated receiver operating characteristic (ROC) curve. For a set of n scores, this calculation is straightforward to do in O(n log n) time -- you simply sort the list, then traverse the list in sorted order, keeping a running total of the number of positively labeled examples you've seen so far. Every time you see a negative label, you add the number of positives, and at the end you divide the resulting sum by the product of the number of positives times the number of negatives.
Now, having done that calculation, say that someone comes along and flips exactly one label (from positive to negative or vice versa). The scores themselves do not change, so you don't need to re-sort. It's straightforward to calculate the new area under the curve (AUC) in O(n) time by re-traversing the sorted list. My question is, is it possible to compute the new AUC in something better than O(n)? I.e., do I have to re-traverse the entire sorted list to get the new AUC?
I think I can do the re-calculation in O(1) time by storing a count, at each position in the ranked list, to the number of positives and negatives above this position. But I am going to need to repeatedly calculate the AUC as more labels get flipped. And I think that if I rely on those stored values, then updating them for the next time will be O(n).
Yes, it is possible to compute AUC in O(log(n)). You need two sets of scores, one for positives and one for negatives, that provide the following operations:
Querying the number of items with higher (or lower) score than a given value (score of the label being flipped).
Inserting and removing the elements.
Knowing the number of positives above/below given position lets you update AUC efficiently as you already mentioned. After that you have to remove the item from the set of positives/negatives and insert to negatives/positives, respectively.
Balanced search trees can do both operations in O(log(n)).
Furthermore, actual values of scores do not matter, only position is relevant. This leads to very simple and efficient implementation using binary indexed tree. See http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=binaryIndexedTrees for explanation.
Also, you don't really need to maintain two sets. Since you already know the total number of positives and negatives above given position, single set is enough.
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)