In DBSCAN, the core points is defined as having more than MinPts within Eps.
So if MinPts = 4, a points with total 5 points in Eps is definitely a core point.
How about a point with 4 points (including itself) in Eps? Is it a core point, or a border point?
Border points are points that are (in DBSCAN) part of a cluster, but not dense themselves (i.e. every cluster member that is not a core point).
In the followup algorithm HDBSCAN, the concept of border points was discarded.
Campello, R. J. G. B.; Moulavi, D.; Sander, J. (2013).
Density-Based Clustering Based on Hierarchical Density Estimates.
Proceedings of the 17th Pacific-Asia Conference on Knowledge Discovery in Databases, PAKDD 2013. Lecture Notes in Computer Science 7819. p. 160.
doi:10.1007/978-3-642-37456-2_14
which states:
Our new definitions are more consistent with a statistical interpretation of clusters as connected components of a level set of a density [...] border objects do not technically belong to the level set (their estimated density is below the threshold).
Actually, I just re-read the original paper and Definition 1 makes it look like the core point belongs to its own eps neighborhood. So if minPts is 4, then a point needs at least 3 others in its eps neighborhood.
Notice in Definition 1 that they say NEps(p) = {q ∈D | dist(p,q) ≤ Eps}. If the point were excluded from its eps neighborhood, then it would have said NEps(p) = {q ∈D | dist(p,q) ≤ Eps and p != q}. Where != is "not equal to".
This is also reinforced by the authors of DBSCAN in their OPTICS paper in Figure 4. http://fogo.dbs.ifi.lmu.de/Publikationen/Papers/OPTICS.pdf
So I think the SciKit interpretation is correct and the Wikipedia illustration is misleading in http://en.wikipedia.org/wiki/DBSCAN
This largely depends on the implementation. The best way is to just play with the implementation yourself.
In the original DBSCAN1 paper, core point condition is given as N_Eps>=MinPts, where N_Eps is the Epsilon neighborhood of a certain data point, which is excluded from its own N_Eps.
Following your example, if MinPts = 4 and N_Eps = 3 (or 4 including itself as you say), then they don't form a cluster according to the original paper. On the other hand, the scikit-learn2 implementation of DBSCAN works otherwise, meaning it counts the point itself for forming a group. So for MinPts=4, four points are needed in total to form a cluster.
[1] Ester, Martin; Kriegel, Hans-Peter; Sander, Jörg; Xu, Xiaowei (1996). "A density-based algorithm for discovering clusters in large spatial databases with noise."
[2] http://scikit-learn.org
Related
I am not sure about the terminology I should use for my problem, so I will give an example.
I have 2 sets of measurements (6 empirical distributions per set = D1-6) that describe 2 different states of the same system (BLUE & RED). These distributions can be multimodal, skewed, undersampled, and strange in some other unpredictable ways.
BLUE is my reference and I want to make RED distributed as closely as possible to BLUE, for all pairwise distributions. For this, I will play with parameters of my RED system and monitor the RED set of measurements D1-6 trying to make it overlap BLUE perfectly.
I know that I can use Jensen-Shannon or Bhattacharyya distances to evaluate the distance between 2 distributions (i.e. RED-D1 and BLUE-D1, for example). However, I do not know if there exist other metrics that could be applied here to get a global distance between all distributions (i.e. quantify the global mismatch between 2 sets of pairwise distributions). Is that the case ?
I am thinking about building an empirical scoring function that would use all the pairwise Jensen-Shannon distances, but I have no better ideas yet. I believe I can NOT just sum all the JS distances because I would get similar scores in these 2 hypothetical, different cases:
D1-6 are distributed as in my image
RED-D1-5 are a much better fit to BLUE-D1-5, BUT RED-D6 is shifted compared to BLUE-D6
And that would be wrong because I would have missed one important feature of my system. Given these 2 cases, it is better to have D1-6 distributed as in my image (solution 1).
The pairwise match between each distribution is equally important and should be equally weighted (i.e. the match between BLUE-D1 and RED-D1 is as important as the match between BLUE-D2 and RED-D2, etc).
D1-3 has a given range DOM1 of [0, 5] and D4-6 has another range DOM2 of [50, 800]. Diamonds represent the weighted means of BLUE and RED distributions.
Thank you very much for your help!
I ended up using a sum of all pairwise Earth Mover's Distance (EMD, https://en.wikipedia.org/wiki/Earth_mover%27s_distance, also known as Wasserstein metric) as a global metric of distance between all pairwise distributions. This describes the difference or similarity between 2 states of my system in an appropriate way.
EMD is implemented in python in package 'pyemd' or using scipy: https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.wasserstein_distance.html.
Can anyone explain me the similarities and differences, of the Correlation and Convolution ? Please explain the intuition behind that, not the mathematical equation(i.e, flipping the kernel/impulse).. Application examples in the image processing domain for each category would be appreciated too
You will likely get a much better answer on dsp stack exchange but... for starters I have found a number of similar terms and they can be tricky to pin down definitions.
Correlation
Cross correlation
Convolution
Correlation coefficient
Sliding dot product
Pearson correlation
1, 2, 3, and 5 are very similar
4,6 are similar
Note that all of these terms have dot products rearing their heads
You asked about Correlation and Convolution - these are conceptually the same except that the output is flipped in convolution. I suspect that you may have been asking about the difference between correlation coefficient (such as Pearson) and convolution/correlation.
Prerequisites
I am assuming that you know how to compute the dot-product. Given two equal sized vectors v and w each with three elements, the algebraic dot product is v[0]*w[0]+v[1]*w[1]+v[2]*w[2]
There is a lot of theory behind the dot product in terms of what it represents etc....
Notice the dot product is a single number (scalar) representing the mapping between these two vectors/points v,w In geometry frequently one computes the cosine of the angle between two vectors which uses the dot product. The cosine of the angle between two vectors is between -1 and 1 and can be thought of as a measure of similarity.
Correlation coefficient (Pearson)
Correlation coefficient between equal length v,w is simply the dot product of two zero mean signals (subtract mean v from v to get zmv and mean w from w to get zmw - here zm is shorthand for zero mean) divided by the magnitudes of zmv and zmw.
to produce a number between -1 and 1. Close to zero means little correlation, close to +/- 1 is high correlation. it measures the similarity between these two vectors.
See http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient for a better definition.
Convolution and Correlation
When we want to correlate/convolve v1 and v2 we basically are computing a series of dot-products and putting them into an output vector. Let's say that v1 is three elements and v2 is 10 elements. The dot products we compute are as follows:
output[0] = v1[0]*v2[0]+v1[1]*v2[1]+v1[2]*v2[2]
output[1] = v1[0]*v2[1]+v1[1]*v2[2]+v1[2]*v2[3]
output[2] = v1[0]*v2[2]+v1[1]*v2[3]+v1[2]*v2[4]
output[3] = v1[0]*v2[3]+v1[1]*v2[4]+v1[2]*v2[5]
output[4] = v1[0]*v2[4]+v1[1]*v2[5]+v1[2]*v2[6]
output[5] = v1[0]*v2[7]+v1[1]*v2[8]+v1[2]*v2[9]
output[6] = v1[0]*v2[8]+v1[1]*v2[9]+v1[2]*v2[10] #note this is
#mathematically valid but might give you a run time error in a computer implementation
The output can be flipped if a true convolution is needed.
output[5] = v1[0]*v2[0]+v1[1]*v2[1]+v1[2]*v2[2]
output[4] = v1[0]*v2[1]+v1[1]*v2[2]+v1[2]*v2[3]
output[3] = v1[0]*v2[2]+v1[1]*v2[3]+v1[2]*v2[4]
output[2] = v1[0]*v2[3]+v1[1]*v2[4]+v1[2]*v2[5]
output[1] = v1[0]*v2[4]+v1[1]*v2[5]+v1[2]*v2[6]
output[0] = v1[0]*v2[7]+v1[1]*v2[8]+v1[2]*v2[9]
Notice that we have less than 10 elements in the output as for simplicity I am computing the convolution only where both v1 and v2 are defined
Notice also that the convolution is simply a number of dot products. There has been considerable work over the years to be able to speed up convolutions. The sweeping dot products are slow and can be sped up by first transforming the vectors into the fourier basis space and then computing a single vector multiplication then inverting the result, though I won't go into that here...
You might want to look at these resources as well as googling: Calculating Pearson correlation and significance in Python
The best answer I got were from this document:http://www.cs.umd.edu/~djacobs/CMSC426/Convolution.pdf
I'm just going to copy the excerpt from the doc:
"The key difference between the two is that convolution is associative. That is, if F and G are filters, then F*(GI) = (FG)*I. If you don’t believe this, try a simple example, using F=G=(-1 0 1), for example. It is very convenient to have convolution be associative. Suppose, for example, we want to smooth an image and then take its derivative. We could do this by convolving the image with a Gaussian filter, and then convolving it with a derivative filter. But we could alternatively convolve the derivative filter with the Gaussian to produce a filter called a Difference of Gaussian (DOG), and then convolve this with our image. The nice thing about this is that the DOG filter can be precomputed, and we only have to convolve one filter with our image.
In general, people use convolution for image processing operations such as smoothing, and they use correlation to match a template to an image. Then, we don’t mind that correlation isn’t associative, because it doesn’t really make sense to combine two templates into one with correlation, whereas we might often want to combine two filter together for convolution."
Convolution is just like correlation, except that we flip over the filter before correlating
I have a large dataset I am trying to do cluster analysis on using SOM. The dataset is HUGE (~ billions of records) and I am not sure what should be the number of neurons and the SOM grid size to start with. Any pointers to some material that talks about estimating the number of neurons and grid size would be greatly appreciated.
Thanks!
Quoting from the som_make function documentation of the som toolbox
It uses a heuristic formula of 'munits = 5*dlen^0.54321'. The
'mapsize' argument influences the final number of map units: a 'big'
map has x4 the default number of map units and a 'small' map has
x0.25 the default number of map units.
dlen is the number of records in your dataset
You can also read about the classic WEBSOM which addresses the issue of large datasets
http://www.cs.indiana.edu/~bmarkine/oral/self-organization-of-a.pdf
http://websom.hut.fi/websom/doc/ps/Lagus04Infosci.pdf
Keep in mind that the map size is also a parameter which is also application specific. Namely it depends on what you want to do with the generated clusters. Large maps produce a large number of small but "compact" clusters (records assigned to each cluster are quite similar). Small maps produce less but more generilized clusters. A "right number of clusters" doesn't exists, especially in real world datasets. It all depends on the detail which you want to examine your dataset.
I have written a function that, with the data set as input, returns the grid size. I rewrote it from the som_topol_struct() function of Matlab's Self Organizing Maps Toolbox into a R function.
topology=function(data)
{
#Determina, para lattice hexagonal, el número de neuronas (munits) y su disposición (msize)
D=data
# munits: número de hexágonos
# dlen: número de sujetos
dlen=dim(data)[1]
dim=dim(data)[2]
munits=ceiling(5*dlen^0.5) # Formula Heurística matlab
#munits=100
#size=c(round(sqrt(munits)),round(munits/(round(sqrt(munits)))))
A=matrix(Inf,nrow=dim,ncol=dim)
for (i in 1:dim)
{
D[,i]=D[,i]-mean(D[is.finite(D[,i]),i])
}
for (i in 1:dim){
for (j in i:dim){
c=D[,i]*D[,j]
c=c[is.finite(c)];
A[i,j]=sum(c)/length(c)
A[j,i]=A[i,j]
}
}
VS=eigen(A)
eigval=sort(VS$values)
if (eigval[length(eigval)]==0 | eigval[length(eigval)-1]*munits<eigval[length(eigval)]){
ratio=1
}else{
ratio=sqrt(eigval[length(eigval)]/eigval[length(eigval)-1])}
size1=min(munits,round(sqrt(munits/ratio*sqrt(0.75))))
size2=round(munits/size1)
return(list(munits=munits,msize=sort(c(size1,size2),decreasing=TRUE)))
}
hope it helps...
Iván Vallés-Pérez
I don't have a reference for it, but I would suggest starting off by using approximately 10 SOM neurons per expected class in your dataset. For example, if you think your dataset consists of 8 separate components, go for a map with 9x9 neurons. This is completely just a ballpark heuristic though.
If you'd like the data to drive the topology of your SOM a bit more directly, try one of the SOM variants that change topology during training:
Growing SOM
Growing Neural Gas
Unfortunately these algorithms involve even more parameter tuning than plain SOM, but they might work for your application.
Kohenon has written on the issue of selecting parameters and map size for SOM in his book "MATLAB Implementations and Applications of the Self-Organizing Map". In some cases, he suggest the initial values can be arrived at after testing several sizes of the SOM to check that the cluster structures were shown with sufficient resolution and statistical accuracy.
my suggestion would be the following
SOM is distantly related to correspondence analysis. In statistics, they use 5*r^2 as a rule of thumb, where r is the number of rows/columns in a square setup
usually, one should use some criterion that is based on the data itself, meaning that you need some criterion for estimating the homogeneity. If a certain threshold would be violated, you would need more nodes. For checking the homogeneity you would need some records per node. Agai, from statistics you could learn that for simple tests (small number of variables) you would need around 20 records, for more advanced tests on some variables at least 8 records.
remember that the SOM represents a predictive model. So validation is the key, absolutely mandatory. Yet, validation of predictive models (see typeI / II error entry in Wiki) is a subject on its own. And the acceptable risk as well as the risk structure also depend fully on your purpose.
You may test the dynamics of the error rate of the model by reducing its size more and more. Then take the smallest one with acceptable error.
It is a strength of the SOM to allow for empty nodes. Yet, there should not be too much of them. Let me say, less than 5%.
Taken all together, from experience, I would recommend the following criterion a minimum of the absolute number of 8..10 records, but those should not be more than 5% of all clusters.
Those 5% rule is of of course a heuristics, which however can be justified by the general usage of the confidence level in statistical tests. You may choose any percentage from 1% to 5%.
I want to cluster my data with KL-divergence as my metric.
In K-means:
Choose the number of clusters.
Initialize each cluster's mean at random.
Assign each data point to a cluster c with minimal distance value.
Update each cluster's mean to that of the data points assigned to it.
In the Euclidean case it's easy to update the mean, just by averaging each vector.
However, if I'd like to use KL-divergence as my metric, how do I update my mean?
Clustering with KL-divergence may not be the best idea, because KLD is missing an important property of metrics: symmetry. Obtained clusters could then be quite hard to interpret. If you want to go ahead with KLD, you could use as distance the average of KLD's i.e.
d(x,y) = KLD(x,y)/2 + KLD(y,x)/2
It is not a good idea to use KLD for two reasons:-
It is not symmetry KLD(x,y) ~= KLD(y,x)
You need to be careful when using KLD in programming: the division may lead to Inf values and NAN as a result.
Adding a small number may affect the accuracy.
Well, it might not be a good idea use KL in the "k-means framework". As it was said, it is not symmetric and K-Means is intended to work on the euclidean space.
However, you can try using NMF (non-negative matrix factorization). In fact, in the book Data Clustering (Edited by Aggarwal and Reddy) you can find the prove that NMF (in a clustering task) works like k-means, only with the non-negative constrain. The fun part is that NMF may use a bunch of different distances and divergences. If you program python: scikit-learn 0.19 implements the beta divergence, which has a variable beta as a degree of liberty. Depending on the value of beta, the divergence has a different behavour. On beta equals 2, it assumes the behavior of the KL divergence.
This is actually very used in the topic model context, where people try to cluster documents/words over topics (or themes). By using KL, the results can be interpreted as a probabilistic function on how the word-topic and topic distributions are related.
You can find more information:
FÉVOTTE, C., IDIER, J. “Algorithms for Nonnegative Matrix
Factorization with the β-Divergence”, Neural Computation, v. 23, n.
9, pp. 2421– 2456, 2011. ISSN: 0899-7667. doi: 10.1162/NECO_a_00168.
Dis- ponível em: .
LUO, M., NIE, F., CHANG, X., et al. “Probabilistic Non-Negative
Matrix Factorization and Its Robust Extensions for Topic Modeling.”
In: AAAI, pp. 2308–2314, 2017.
KUANG, D., CHOO, J., PARK, H. “Nonnegative matrix factorization for
in- teractive topic modeling and document clustering”. In:
Partitional Clus- tering Algorithms, Springer, pp. 215–243, 2015.
http://scikit-learn.org/stable/modules/generated/sklearn.decomposition.NMF.html
K-means is intended to work with Euclidean distance: if you want to use non-Euclidean similarities in clustering, you should use a different method. The most principled way to cluster with an arbitrary similarity metric is spectral clustering, and K-means can be derived as a variant of this where the similarities are the Euclidean distances.
And as #mitchus says, KL divergence is not a metric. You want the Jensen-Shannon divergence or its square root named as the Jensen-Shannon distance as it has symmetry.
I have a scenario where I have several thousand instances of data. The data itself is represented as a single integer value. I want to be able to detect when an instance is an extreme outlier.
For example, with the following example data:
a = 10
b = 14
c = 25
d = 467
e = 12
d is clearly an anomaly, and I would want to perform a specific action based on this.
I was tempted to just try an use my knowledge of the particular domain to detect anomalies. For instance, figure out a distance from the mean value that is useful, and check for that, based on heuristics. However, I think it's probably better if I investigate more general, robust anomaly detection techniques, which have some theory behind them.
Since my working knowledge of mathematics is limited, I'm hoping to find a technique which is simple, such as using standard deviation. Hopefully the single-dimensioned nature of the data will make this quite a common problem, but if more information for the scenario is required please leave a comment and I will give more info.
Edit: thought I'd add more information about the data and what I've tried in case it makes one answer more correct than another.
The values are all positive and non-zero. I expect that the values will form a normal distribution. This expectation is based on an intuition of the domain rather than through analysis, if this is not a bad thing to assume, please let me know. In terms of clustering, unless there's also standard algorithms to choose a k-value, I would find it hard to provide this value to a k-Means algorithm.
The action I want to take for an outlier/anomaly is to present it to the user, and recommend that the data point is basically removed from the data set (I won't get in to how they would do that, but it makes sense for my domain), thus it will not be used as input to another function.
So far I have tried three-sigma, and the IQR outlier test on my limited data set. IQR flags values which are not extreme enough, three-sigma points out instances which better fit with my intuition of the domain.
Information on algorithms, techniques or links to resources to learn about this specific scenario are valid and welcome answers.
What is a recommended anomaly detection technique for simple, one-dimensional data?
Check out the three-sigma rule:
mu = mean of the data
std = standard deviation of the data
IF abs(x-mu) > 3*std THEN x is outlier
An alternative method is the IQR outlier test:
Q25 = 25th_percentile
Q75 = 75th_percentile
IQR = Q75 - Q25 // inter-quartile range
IF (x < Q25 - 1.5*IQR) OR (Q75 + 1.5*IQR < x) THEN x is a mild outlier
IF (x < Q25 - 3.0*IQR) OR (Q75 + 3.0*IQR < x) THEN x is an extreme outlier
this test is usually employed by Box plots (indicated by the whiskers):
EDIT:
For your case (simple 1D univariate data), I think my first answer is well suited.
That however isn't applicable to multivariate data.
#smaclell suggested using K-means to find the outliers. Beside the fact that it is mainly a clustering algorithm (not really an outlier detection technique), the problem with k-means is that it requires knowing in advance a good value for the number of clusters K.
A better suited technique is the DBSCAN: a density-based clustering algorithm. Basically it grows regions with sufficiently high density into clusters which will be maximal set of density-connected points.
DBSCAN requires two parameters: epsilon and minPoints. It starts with an arbitrary point that has not been visited. It then finds all the neighbor points within distance epsilon of the starting point.
If the number of neighbors is greater than or equal to minPoints, a cluster is formed. The starting point and its neighbors are added to this cluster and the starting point is marked as visited. The algorithm then repeats the evaluation process for all the neighbors recursively.
If the number of neighbors is less than minPoints, the point is marked as noise.
If a cluster is fully expanded (all points within reach are visited) then the algorithm proceeds to iterate through the remaining unvisited points until they are depleted.
Finally the set of all points marked as noise are considered outliers.
There are a variety of clustering techniques you could use to try to identify central tendencies within your data. One such algorithm we used heavily in my pattern recognition course was K-Means. This would allow you to identify whether there are more than one related sets of data, such as a bimodal distribution. This does require you having some knowledge of how many clusters to expect but is fairly efficient and easy to implement.
After you have the means you could then try to find out if any point is far from any of the means. You can define 'far' however you want but I would recommend the suggestions by #Amro as a good starting point.
For a more in-depth discussion of clustering algorithms refer to the wikipedia entry on clustering.
This is an old topic but still it lacks some information.
Evidently, this can be seen as a case of univariate outlier detection. The approaches presented above have several pros and cons. Here are some weak spots:
Detection of outliers with the mean and sigma has the obvious disadvantage of dependence of mean and sigma on the outliers themselves.
The case of the small sample limit (see question for example) is not adequately covered by, 3 sigma, K-Means, IQR etc.
And I could go on... However the statistical literature offers a simple metric: the median absolute deviation. (Medians are insensitive to outliers)
Details can be found here: https://www.sciencedirect.com/book/9780128047330/introduction-to-robust-estimation-and-hypothesis-testing
I think this problem can be solved in a few lines of python code like this:
import numpy as np
import scipy.stats as sts
x = np.array([10, 14, 25, 467, 12]) # your values
np.abs(x - np.median(x))/(sts.median_abs_deviation(x)/0.6745) #MAD criterion
Subsequently you reject values above a certain threshold (97.5 percentile of the distribution of data), in case of an assumed normal distribution the threshold is 2.24. Here it translates to:
array([ 0.6745 , 0. , 1.854875, 76.387125, 0.33725 ])
or the 467 entry being rejected.
Of course, one could argue, that the MAD (as presented) also assumes a normal dist. Therefore, why is it that argument 2 above (small sample) does not apply here? The answer is that MAD has a very high breakdown point. It is easy to choose different threshold points from different distributions and come to the same conclusion: 467 is the outlier.
Both three-sigma rule and IQR test are often used, and there are a couple of simple algorithms to detect anomalies.
The three-sigma rule is correct
mu = mean of the data
std = standard deviation of the data
IF abs(x-mu) > 3*std THEN x is outlier
The IQR test should be:
Q25 = 25th_percentile
Q75 = 75th_percentile
IQR = Q75 - Q25 // inter-quartile range
If x > Q75 + 1.5 * IQR or x < Q25 - 1.5 * IQR THEN x is a mild outlier
If x > Q75 + 3.0 * IQR or x < Q25 – 3.0 * IQR THEN x is a extreme outlier