I want to cluster some curves which contains daily click rate.
The dataset is click rate data in time series.
y1 = [time1:0.10,time2:0.22,time3:0.344,...]
y2 = [time1:0.10,time2:0.22,time3:0.344,...]
I don't know how to measure two curve's similarity using kmeans.
Is there any paper for this purpose or some library?
For similarity, you could use any kind of time series distance. Many of these will perform alignment, also of sequences of different length.
However, k-means will not get you anywhere.
K-means is not meant to be used with arbitrary distances. It actually does not use distance for assignment, but least-sum-of-squares (which happens to be squared euclidean distance) - aka: variance.
The mean must be consistent with this objective. It's not hard to see that the mean also minimizes the sum of squares. This guarantees convergence of k-means: in each single step (both assignment and mean update), the objective is reduced, thus it must converge after a finite number of steps (as there are only a finite number of discrete assignments).
But what is the mean of multiple time series of different length?
Related
I am trying to learn some facial landmark detection model, and notice that many of them use NME(Normalized Mean Error) as performance metric:
The formula is straightforward, it calculate the l2 distance between ground-truth points and model prediction result, then divided it by a normalized factor, which vary from different dataset.
However, when adopting this formula on some landmark detector that some one developed, i have to deal with this non-trivial situation, that is some detector may not able to generate enough number landmarks for some input image(might because of NMS/model inherited problem/image quality etc). Thus some of ground-truth points might not have their corresponding one in the prediction result.
So how to solve this problem, should i just add such missing point result to "failure result set" and use FR to measure the model, and ignore them when doing the NME calculation?
If you have as output of neural network an vector 10x1 as example
that is your points like [x1,y1,x2,y2...x5,y5]. This vector will be fixed length cause of number of neurons in your model.
If you have missing points - this is because (as example you have 4 from 5 points) some points are go beyond the image width and height. Or are with minus (negative) like [-0.1, -0.2, 0.5,0.7 ...] there first 2 points you can not see on image like they are mission but they will be in vector and you can callculate NME.
In some custom neural nets that can be possible, because missing values will be changed to biggest error points.
Given some x data points in an N dimensional space, I am trying to find a fixed length representation that could describe any subset s of those x points? For example the mean of the s subset could describe that subset, but it is not unique for that subset only, that is to say, other points in the space could yield the same mean therefore mean is not a unique identifier. Could anyone tell me of a unique measure that could describe the points without being number of points dependent?
In short - it is impossible (as you would achieve infinite noiseless compression). You have to either have varied length representation (or fixed length with length being proportional to maximum number of points) or dealing with "collisions" (as your mapping will not be injective). In the first scenario you simply can store coordinates of each point. In the second one you approximate your point clouds with more and more complex descriptors to balance collisions and memory usage, some posibilities are:
storing mean and covariance (so basically perofming maximum likelihood estimation over Gaussian families)
performing some fixed-complexity density estimation like Gaussian Mixture Model or training a generative Neural Network
use set of simple geometrical/algebraical properties such as:
number of points
mean, max, min, median distance between each pair of points
etc.
Any subset can be identified by a bit mask of length ceiling(lg(x)), where bit i is 1 if the corresponding element belongs to the subset. There is no fixed-length representation that is not a function of x.
EDIT
I was wrong. PCA is a good way to perform dimensionality reduction for this problem, but it won't work for some sets.
However, you can almost do it. Where "almost" is formally defined by the Johnson-Lindenstrauss Lemma, which states that for a given large dimension N, there exists a much lower dimension n, and a linear transformation that maps each point from N to n, while keeping the Euclidean distance between every pair of points of the set within some error ε from the original. Such linear transformation is called the JL Transform.
In other words, your problem is only solvable for sets of points where each pair of points are separated by at least ε. For this case, the JL Transform gives you one possible solution. Moreover, there exists a relationship between N, n and ε (see the lemma), such that, for example, if N=100, the JL Transform can map each point to a point in 5D (n=5), an uniquely identify each subset, if and only if, the minimum distance between any pair of points in the original set is at least ~2.8 (i.e. the points are sufficiently different).
Note that n depends only on N and the minimum distance between any pair of points in the original set. It does not depend on the number of points x, so it is a solution to your problem, albeit some constraints.
I have a parallel application in which I am computing in each node the variance of each partition of datapoint based on the calculated mean, but how can I compute the global variance (sum of all the variances)?
I thought that it would be a simple sum of the variances and divided by the number of nodes, but it is not giving me a close result...
The global variation is a sum.
You can compute parts of the sum in parallel trivially, and then add them together.
sum(x1...x100) = sum(x1...x50) + sum(x51...x100)
The same way, you can compute the global averages - compute the global sum, compute the sum of the object counts, divide (don't divide by the number of nodes; but by the total number of objects).
mean = sum/count
Once you have the mean, you can compute the sum of squared deviations using the distributed sum formula above (applied to (xi-mean)^2), then divide by count-1 to get the variance.
Do not use E[X^2] - (E[X])^2
While this formula "mean of square minus square of mean" is highly popular, it is numerically unstable when you are using floating point math. It's known as catastrophic cancellation.
Because the two values can be very close, you lose a lot of digits in precision when computing the difference. I've seen people get a negative variance this way...
With "big data", numerical problems gets worse...
Two ways to avoid these problems:
Use two passes. Computing the mean is stable, and gets you rid of the subtraction of the squares.
Use an online algorithm such as the one by Knuth and Welford, then use weighted sums to combine the per-partition means and variances. Details on Wikipedia In my experience, this often is slower; but it may be beneficial on Hadoop due to startup and IO costs.
You need to add the sums and sums of squares of each partition to get the global sum and sum of squares and then use them to calculate the global mean and variance.
UPDATE: E[X2] - E[X]2 and cancellation...
To figure out how important cancellation error is when calculating the standard deviation with
σ = √(E[X2] - E[X]2)
let us assume that we have both E[X2] and E[X]2 accurate to 12 significant decimal figures. This implies that σ2 has an error of order 10-12 × E[X2] or, if there has been significant cancellation, equivalently 10-12 × E[X]2 when σ will have an error of approximate order 10-6 × E[X]; one millionth the mean.
For many, if not most, statistical analyses this is negligable, in the sense that it falls within other sources of error (like measurement error), and so you can in good consciense simply set negative variances to zero before you take the square root.
If you really do care about deviations of this magnitude (and can show that it's a feature of the thing you are measuring and not, for example, an artifact of the method of measurement) then you can start worrying about cancellation. That said, the most likely explanation is that you have used an inappropriate scale for your data, such as measuring daily temperatures in Kelvin rather than Celcius!
my aim is to cluster 126 time-series concerning 26 weeks (so each time-series has 26 observation). I used pam{cluster} = partitioning around medoids to cluster these time-series.
Before clustering I wanted to compare which distance measure is the most appropriate: euclidean, manhattan or dynamic time warping. I used each distance to cluster and compare by silhouette plot. Is there any way I can compare different distance measure?
For example I know that procedure clValid {clValid} to validate cluster results, however I cannot implement dtw to calculate indexes.
So how can I compare different distance metrics (not only by silhouette)?
Additional question: is GAP statistic enough to decide how many clusters choose? Or should I evaluate number of clusters with different methods or compare two or three ways how to do it?
I would be grateful for any suggestions.
I have just read the book "cluster analysis, fifth edition" by Brian S. Everitt, etc. And currently, I adopt the following strategy to select method to calculate distance matrix, clustering and validation:
for distance: using cmdscale{stats} function to calculate multidimentional scaling, and plot the scatterplot of the two scaling dimensions with density information. As expected, if there is distinct clusters or nested clusters, the scatterplot will give some hints.
for clustering: for every clustering method, calculate cophenetic correlation between clustering results and the distance, this can be calculated using cophenetic{stats} function. The best clustering method will give higher correlation. However, this is only working for hierarchical clustering. I haven't idea for other clustering methods, like pam, or kmeans.
for partition evaluation: package {clusterSim} give several function to calculate the index to evaluate the clustering quality. Another package {NbClust} also calculate so many as 30 index to evaluate the combination of "distance", "clustering" and "number of clusters". However, this package partition the hierarchical tree using {cutree}, which is not suitable for nested clustering structure. Another method provided by {dynamicTreeCut} give reasonable results.
for cluster number determination: will added later.
Cluster data for which you have class labels, and use the RAND index to measure cluster quality.
50 such datasets are at the UCR time series archive
This paper does something similar
http://www.cs.ucr.edu/~eamonn/ClusteringTimeSeriesUsingUnsupervised-Shapelets.pdf
I have a matrix of 62 columns and 181408 rows that I am going to be clustering using k-means. What I would ideally like is a method of identifying what the optimum number of clusters should be. I have tried implementing the gap statistic technique using clusGap from the cluster package (reproducible code below), but this produces several error messages relating to the size of the vector (122 GB) and memory.limitproblems in Windows and a "Error in dist(xs) : negative length vectors are not allowed" in OS X. Does anyone has any suggestions on techniques that will work in determining optimum number of clusters with a large dataset? Or, alternatively, how to make my code function (and does not take several days to complete)? Thanks.
library(cluster)
inputdata<-matrix(rexp(11247296, rate=.1), ncol=62)
clustergap <- clusGap(inputdata, FUN=kmeans, K.max=12, B=10)
At 62 dimensions, the result will likely be meaningless due to the curse of dimensionality.
k-means does a minimum SSQ assignment, which technically equals minimizing the squared Euclidean distances. However, Euclidean distance is known to not work well for high dimensional data.
If you don't know the numbers of the clusters k to provide as parameter to k-means so there are three ways to find it automaticaly:
G-means algortithm: it discovers the number of clusters automatically using a statistical test to decide whether to split a k-means center into two. This algorithm takes a hierarchical approach to detect the number of clusters, based on a statistical test for the hypothesis that a subset of data follows a Gaussian distribution (continuous function which approximates the exact binomial distribution of events), and if not it splits the cluster. It starts with a small number of centers, say one cluster only (k=1), then the algorithm splits it into two centers (k=2) and splits each of these two centers again (k=4), having four centers in total. If G-means does not accept these four centers then the answer is the previous step: two centers in this case (k=2). This is the number of clusters your dataset will be divided into. G-means is very useful when you do not have an estimation of the number of clusters you will get after grouping your instances. Notice that an inconvenient choice for the "k" parameter might give you wrong results. The parallel version of g-means is called p-means. G-means sources:
source 1
source 2
source 3
x-means: a new algorithm that efficiently, searches the space of cluster locations and number of clusters to optimize the Bayesian Information Criterion (BIC) or the Akaike Information Criterion (AIC) measure. This version of k-means finds the number k and also accelerates k-means.
Online k-means or Streaming k-means: it permits to execute k-means by scanning the whole data once and it finds automaticaly the optimal number of k. Spark implements it.
This is from RBloggers.
https://www.r-bloggers.com/k-means-clustering-from-r-in-action/
You could do the following:
data(wine, package="rattle")
head(wine)
df <- scale(wine[-1])
wssplot <- function(data, nc=15, seed=1234){
wss <- (nrow(data)-1)*sum(apply(data,2,var))
for (i in 2:nc){
set.seed(seed)
wss[i] <- sum(kmeans(data, centers=i)$withinss)}
plot(1:nc, wss, type="b", xlab="Number of Clusters",
ylab="Within groups sum of squares")}
wssplot(df)
this will create a plot like this.
From this you can choose the value of k to be either 3 or 4. i.e
there is a clear fall in 'within groups sum of squares' when moving from 1 to 3 clusters. After three clusters, this decrease drops off, suggesting that a 3-cluster solution may be a good fit to the data.
But like Anony-Mouse pointed out, the curse of dimensionality affects due to the fact that euclidean distance being used in k means.
I hope this answer helps you to a certain extent.