1. Coordinated weighted sampling for estimating aggregates over multiple weight assignments Edith Cohen, AT&T Research Haim Kaplan, Tel Aviv University Shubho Sen, AT&T Research

2. Data model Universe U of items: i1i1 i2i2 i3i3 i4i4 ………… w1(i1)w1(i1) w1(i2)w1(i2) w1(i3)w1(i3) w1(i4)w1(i4) w2(i1)w2(i1) w2(i2)w2(i2) w2(i3)w2(i3) w2(i4)w2(i4) w3(i1)w3(i1) w3(i2)w3(i2) w3(i3)w3(i3) w3(i4)w3(i4) Multiple weight assignments defined over the items IP addresses traffic day 1 traffic day 2 traffic day 3

3. Data model (example) Universe U of items: i1i1 i2i2 i3i3 i4i4 w1(i1)w1(i1) w1(i2)w1(i2) w1(i3)w1(i3) w1(i4)w1(i4) w2(i1)w2(i1) w2(i2)w2(i2) w2(i3)w2(i3) w2(i4)w2(i4) w3(i1)w3(i1) w3(i2)w3(i2) w3(i3)w3(i3) w3(i4)w3(i4) facebook users male friends female friends online per day …………

4. Data model (example) Universe U of items: i1i1 i2i2 i3i3 i4i4 ………… w1(i1)w1(i1) w1(i2)w1(i2) w1(i3)w1(i3) w1(i4)w1(i4) w2(i1)w2(i1) w2(i2)w2(i2) w2(i3)w2(i3) w2(i4)w2(i4) w3(i1)w3(i1) w3(i2)w3(i2) w3(i3)w3(i3) w3(i4)w3(i4) customers wait time item 1 wait time item 2 wait time item 3

5. Data model (example) Universe U of items: i1i1 i2i2 i3i3 i4i4 ………… w1(i1)w1(i1) w1(i2)w1(i2) w1(i3)w1(i3) w1(i4)w1(i4) w2(i1)w2(i1) w2(i2)w2(i2) w2(i3)w2(i3) w2(i4)w2(i4) w3(i1)w3(i1) w3(i2)w3(i2) w3(i3)w3(i3) w3(i4)w3(i4) license plates GW bridge day 1 GW bridge day 2 GW bridge day 3

6. Data model (example) Universe U of items: i1i1 i2i2 i3i3 i4i4 ………… w1(i1)w1(i1) w1(i2)w1(i2) w1(i3)w1(i3) w1(i4)w1(i4) w2(i1)w2(i1) w2(i2)w2(i2) w2(i3)w2(i3) w2(i4)w2(i4) w3(i1)w3(i1) w3(i2)w3(i2) w3(i3)w3(i3) w3(i4)w3(i4) web pages bytes out links in links

7. Aggregation queries Selection predicate d(i) over items in U compute the sum of the weights of all items for which the predicate holds: Examples: Total number of links out of pages of some web site Total number of flows into a particular network

8. This work:Aggregates depending on more than one weight function Selection predicate d(i) over items in U Want to compute the sum of f(i) for all items for which the predicate holds: f(i) depends on some subset of the weights given to i Simple examples: f(i) = max b w b (i), f(i) = min b w b (i), f(i)=max b w b (i)-min b w b (i)

9. Data model (example) Universe U of items: i1i1 i2i2 i3i3 i4i4 ………… w1(i1)w1(i1) w1(i2)w1(i2) w1(i3)w1(i3) w1(i4)w1(i4) w2(i1)w2(i1) w2(i2)w2(i2) w2(i3)w2(i3) w2(i4)w2(i4) w3(i1)w3(i1) w3(i2)w3(i2) w3(i3)w3(i3) w3(i4)w3(i4) customers wait time item 1 wait time item 2 wait time item 3 f(i) = max b w b (i) aggregated over a subset gives the total waiting time of these customers

10. Data model Universe U of items: i1i1 i2i2 i3i3 i4i4 ………… w1(i1)w1(i1) w1(i2)w1(i2) w1(i3)w1(i3) w1(i4)w1(i4) w2(i1)w2(i1) w2(i2)w2(i2) w2(i3)w2(i3) w2(i4)w2(i4) w3(i1)w3(i1) w3(i2)w3(i2) w3(i3)w3(i3) w3(i4)w3(i4) IP addresses traffic day 1 traffic day 2 traffic day 3 f(i) = max b w b (i)-min b w b (i) aggregated over a network gives sum of maximum changes in # of flows

11. Massive data volumes, too large to store or to transport elsewhere w 1 may not be available when we process w 2, etc, collected at different place/time Many different types of queries (d(i),f(i)), not known in advance Challenges  Exact aggregation can be very resource consuming or simply impossible Challenge: which summary to keep and how to estimate the aggregate ??

12. Desirable Properties of Sampling Scheme: Scalability: efficient on data streams and distributed data, decouple the summarization of different sets Applicable to a wide range of d(i), f(i) Good estimators: unbiased, small variance Keep a sample Solution: Use Sampling

13. Independent (Bernoulli) with probability kw(i)/W Poisson, PPS sampling k times without replacement Order (bottom-k) sampling (includes PPSWOR, priority sampling (Rosen 72, Rosen 97, CK07, DLT 07) Methods can be implemented efficiently in a stream/reservoir/distributed setting Order (bottom-k) sampling dominates other methods (fixed sample size, with unbiased estimators [CK07, DLT07] ) Sampling a single weight function

14. Rank-based definition of Order/bottom-k sampling For each item draw a random rank value from some distribution f w(i) Order/bottom-k: Pick the k smallest-ranked items to your sample r(i)=u(i)/w(i) : priority order samples r(i)=-ln(u(i))/w(i) : PPSWOR order samples Rank distributions: (The seed)

15. Independent sampling: Get a sample from each weight function independently Coordinated Sampling: If w 2 (i) ≥ w 1 (i) then r 2 (i) ≤ r 1 (i). For example by using the same seed u(i) for all w i : r 1 (i) = u(i)/w 1 (i) r 2 (i) = u(i)/w 2 (i) Relations between samples by different weight function Can be achieved efficiently via hashing

16. Coordination is critical for tight estimates We develop estimators for coordinated samples and analyze them Coordination is critical A lot of previous work on coordination in the context of survey sampling and when weights are uniform..

17. Horvitz Thompson Estimators [HT52]  a(Q) is an unbiased estimator of w(Q) HT estimator: Give each item i an adjusted weight a(i) Let p(i) be the prob. that item i is sampled If i is in sample, a(i)=w(i)/p(i) (otherwise a(i) = 0 )

18. Using HT for order samples Cannot compute p(i) from an order sample Therefore cannot compute HT adjusted weights a(i)=w(i)/p(i) p(i)? Problem For item i, if we can find a partition of the sample space such that we can compute the conditioned p(i) for item i for the cell that the sampled set belongs to Then apply HT using that conditioned p(i) p 1 (i) p 2 (i) p 4 (i) p 3 (i) p 5 (i) p 6 (i)  Obtain unbiased adjusted weights for each item Solution: Apply HT on partitions

19. Estimators for aggregates over multiple weights Suppose we want to estimate: We have a sample for each weight function using independent or coordinated ranks

20. Estimator for Identify the items i for which d(i)=true and you know min b w b (i) (=Items contained in all samples) Compute p(i), the probability of such item to be sampled for all weight functions (conditioned in a subspace that depends on the other ranks) Set a(i) = min b w b (i)/p(i) Sum up these adjusted weights

21. Independent vs Coordinated Variance is small when p(i) is large Coordinated sketches have larger p(i)

22. Estimator for Identify the items i for which d(i)=true and you know max b w b (i) If you see all weights then you know max b w b (i) But you never see them all if min b w b (i) = 0

23. Estimator for If the largest weight you see has rank smaller than min b {k th smallest rank of an item  I\{i}} then you know it’s the maximum. Use the consistency of the ranks:

24. Estimator for Compute p(i), the probability of this to happen Set a(i) = min b w b (i)/p(i) Sum up these adjusted weights If the largest weight you see has rank smaller than min b {k th smallest rank of an item  I\{i}} then you know it’s the maximum.

25. More estimators Now we can get an unbiased estimator for the L1 difference: Can prove that this estimator is nonnegative

26. Empirical Evaluation Data sets: IP packet traces: Items: (src,dest,src-port,dest-port). w b : total # of bytes in hour b Netflix data sets: Items: movies. w b : total # of ratings in month b Stock data: Items: ticker symbols w b : high value on day b

27. Netflix:

28. Netflix: for various f

29. Summary Coordinated sketches improves accuracy and sometimes estimating is not feasible without it It can be achieved with little overhead using state of the art sampling techniques

30. Thank you!

31. Application: Sensor nodes recording daily vehicle traffic in different locations in a city Items: vehicles (license plate numbers) Sets: All vehicles observed at a particular location/date Example queries: Number of distinct vehicles in Manhattan on election day (size of the union of all Manhattan locations on election day) Number of trucks with PA license plates that crossed both the Hudson and the East River on June 18, 2009

32. Application: Items are IP addresses. Sets h1,…,h24 correspond to destination IP’s observed in different 1- hour time periods. Example Queries: number of distinct IP dests In the first 12 hours (size of union of h1,…,h12) In at most one/at least 4 out of h1,…,h24 In the intersection of h9,…,h17 (all business hours) Same queries restricted to: blacklisted IP’s, servers in a specific country etc. Uses: Traffic Analysis, Anomaly detection

33. Application: Text/HyperText corpus: items are features/terms, sets are documents (or vice versa) Example Queries: Number of distinct hyperlinks to financial news websites from websites in the.edu domain Fraction of pages citing A that also cite B Similarity (Jaccard coefficient) of two documents Uses: research, duplicate elimination, similarity search or clustering

34. Application: Market basket data set. Items are consumers/baskets, subsets are goods (or vice versa) Example queries: Likelihood of purchasing beer if diapers were purchased Number of baskets from zip code 07932 containing shampoo Number of consumers purchasing baby food and not diapers Uses: Marketing, placement, pricing, choosing products

35. Application: Online digital library, items are viewers, each set corresponds to viewers for a specific movie Example Queries: total number of distinct viewers of any or all James Bond titles (union, intersection) any independent title from 2008 same restricted to selected viewers: females, certain geographic location, under age 30. Uses: marketing research, content placement,….. Many other applications: makrket basket analysis, text/hypertext analysis etc.

36. Rank-based definition of sampling For each item draw a random rank value from some distribution f w(i) Poisson: Pick all items with rank value below Tau(k) Order/bottom-k: Pick the k smallest-ranked items to your sample r(i)=u/w(i) : Poisson PPS, priority order samples r(i)=-ln(u)/w(i) : PPSWOR order samples Rank distributions:

37. Horvitz Thompson Estimators [HT52] Let p(i) be the prob. that item i is sampled If i is in the sample a(i)=w(i)/p(i) (otherwise a(i) = 0 )  a(i) is an unbiased estimator of w(i) Give each item i an adjusted weight a(i):

38. Rank Conditioning Estimator for single bottom-k sample set [CK07, DLT07] A 0.10 0.38 0.04 0.15 0.87 0.79 0.62 0.51 0.42 0.94 4-sample of A < 0.42 Use the principal of applying HT on a partition For item i in sample, partition is based on the k th smallest rank value among all other items (excluding i) in the sample being some fixed value. In this specific case, k=4 and the fixed value = 0.42. Compute p(i) : the probability that the rank value of i is smaller than 0.42. For priority order samples, r(i)=u/w(i); p(i)=prob(r(i)<0.42)= min{1,0.42w(i)}; a(i)=w(i)/p(i)=max{w(i),100/42}