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BU CS 565 - Time-series data analysis

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Time-series data analysisWhy deal with sequential data?• Because all data is sequential • All data items arrive in the data store in some order• Examples– transaction data– documents and words• In some (or many) cases the order does not matter• In many cases the order is of interestTime-series data Financial time series, process monitoring…Questions• What is the structure of sequential data?• Can we represent this structure compactlyand accurately?Sequence segmentation• Gives an accurate representation of the structure of sequential data• How?– By trying to find homogeneous segments • Segmentation question:• Can a sequence T={t1,t2,…,tn} be described as a concatenation of subsequences S1,S2,…,Sksuch that each Siis in some sense homogeneous?• The corresponding notion of segmentation in unordered data is clusteringDynamic-programming algorithm• Sequence T, length n, k segments, cost function E(), table M• For i=1 to n– Set M[1,i]=E(T[1…i]) //Everything in one cluster• For j=1 to k– Set M[j,j] = 0 //each point in its own cluster• For j=2 to k– For i=j+1 to n• Set M[j,i] = mini’<i{M[j-1,i]+E(T[i’+1…i])}• To recover the actual segmentation (not just the optimal cost) store also the minimizing values i’• Takes time O(n2k), space O(kn)ExampletRtRBasic definitions• Sequence T = {t1,t2,…,tn}: an ordered set of n d-dimensional real points tiЄRd• A k-segmentation S: a partition of T into k contiguous segments {s1,s2,…,sk}– Each segment sЄS is represented by a single value μsЄRd(the representative of the segment)• Error Ep(S): The error of replacing individual points with representativespSs stpsptSE1||)( The k-segmentation problem• Common cases for the error functionEp: p = 1 and p = 2.– When p = 1, the best μscorresponds the median of the points in segment s.– When p = 2, the best μscorresponds to the mean of the points in segment s. Given a sequence T of length n and a value k, find a k-segmentation S = {s1, s2, …,sk} of T such that the Eperror is minimized.Optimal solution for the k-segmentation problem• Running time O(n2k)– Too expensive for large datasets! [Bellman’61] The k-segmentation problem can be solved optimally using a standard dynamic-programming algorithmHeuristics• Bottom-up greedy (BU): O(nlogn)– *Keogh and Smyth’97, Keogh and Pazzani’98+• Top-down greedy (TD): O(nlogn)– *Douglas and Peucker’73, Shatkay and Zdonik’96, Lavrenko et. al’00+• Global Iterative Replacement (GiR): O(nI)– *Himberg et. al ’01+• Local Iterative Replacement (LiR): O(nI)– *Himberg et. al ’01+Approximation algorithm [Theorem] The segmentation problem can be approximated within a constant factor of 3 for both E1and E2error measures. That is,2,1 )(3)(  pSESEOPTpDnSp The running time of the approximation algorithm is:)(3/53/4knODivide ’n Segment (DnS) algorithm• Main idea– Split the sequence arbitrarily into subsequences– Solve the k-segmentation problem in each subsequence– Combine the results• Advantages– Extremely simple– High quality results– Can be applied to other segmentation problems*Gionis’03, Haiminen’04,Bingham’06+DnS algorithm - DetailsInput: Sequence T, integer kOutput: a k-segmentation of T1. Partition sequence T arbitrarily into m disjoint intervals T1,T2,…,Tm2. For each interval Tisolve optimally the k- segmentation problem using DP algorithm3. Let T’ be the concatenation of mk representatives produced in Step 2. Each representative is weighted with the length of the segment it represents4. Solve optimally the k-segmentation problem for T’ using the DP algorithm and output this segmentation as the final segmentationThe DnS algorithmInput sequence T consisting of n=20 points (k=2)102030405060708090100The DnS algorithm – Step 1Partition the sequence into m=3 disjoint intervals102030405060708090100T1T2T3The DnS algorithm – Step 2102030405060708090100T1T2T3Solve optimally the k-segmentation problem into each partition (k=2)The DnS algorithm – Step 2102030405060708090100T1T2T3Solve optimally the k-segmentation problem into each partition (k=2)The DnS algorithm – Step 3102030405060708090100Sequence T’ consisting of mk=6 representantivesw=8w=4w=4w=1w=1w=2The DnS algorithm – Step 4102030405060708090100Solve k-segmentation on T’ (k=2)w=8w=4w=4w=1w=1w=2Running time• In the case of equipartition in Step 1, the running time of the algorithm as a function of m is: Running time3/53/402)( knmR kmkkmnmmR22)()(  The function R(m) is minimized for320knmThe segmentation error [Theorem] The segmentation error of the DnS algorithms is at most three times the error of the optimal (DP) algorithm for both E1and E2error measures.2,1 )(3)(  pSESEOPTpDnSpProof for E1– λt: the representative of point t in the optimal segmentation– τ: the representative of point t in the segmentation of Step 2102030405060708090100  Tt Ttttdtd ),(),(11Lemma :ProofTttDnStdSE ),()(11 Tttdtd ),(),(11 Tttdtd ),(),(11 Ttttdtdtd ),(),(),(111TttTtttdtd ),(),(211)(3OPTSE(triangle inequality)(optimality of DP)(triangle inequality)(Lemma)(triangle inequality)(optimality of DP)(triangle inequality)(Lemma)  Tt Ttttdtd ),(),(11Lemma : λt: the representative of point t in the optimal segmentation τ: the representative of point t in the segmentation of Step 2 μt: the representative of point t in the final segmentation in Step 4Trading speed for accuracy• Recursively divide (into m pieces) and segment• If χ=(ni)1/2, where nithe length of the sequence in the i-th recursive level (n1=n) then– running time of the algorithm is O(nloglogn)– the segmentation error is at most O(logn) worse than the optimal• If χ =const, the running time of the algorithm is O(n), but there are no guarantees for the segmentation errorReal datasets – DnS algorithmReal datasets – DnS algorithmSpeed vs. accuracy in


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