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Published byParker Stripe Modified about 1 year ago

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Haishan Liu

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to identify correspondences (“mappings”) between ERP patterns derived from: different data decomposition methods (temporal PCA and spatial ICA) different spatial and temporal metrics that are used to summarize data different subject groups

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tPCAsICA Metric Set 1SG01 SDv2_SG01_tPCA_10ICm1.xlsSDv2_SG01_sICA_10ICm1.xls Metric Set 1SG02 SDv2_SG02_tPCA_10ICm1.xlsSDv2_SG02_sICA_10ICm1.xls Metric Set 2SG01 SDv2_SG01_tPCA_10ICm2.xlsSDv2_SG01_sICA_10ICm2.xls Metric Set 2SG02 SDv2_SG02_tPCA_10ICm2.xlsSDv2_SG02_sICA_10ICm2.xls SG01/SG02 refer to subject groups #1 and #2, respectively. tPCA and sICA are distinct methods for transforming data into discrete, rank-1, spatiotemporal patterns Metric Sets 1-2 are alternative sets of spatial/temporal attributes to summarize the simulated ERP patterns

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Detecting structural dissimilarity To incorporate distribution information of data points along each attribute. Comparing non-overlapping clusterings To compare clusterings which do not share any common data points.

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density of an attribute-bin region for cluster c k in clustering C dens C (k, i, j): the number of points in the region (i, j), which belongs to the cluster c k of clustering C.

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density profile of clustering C V C = (dens C (1, 1, 1), dens C (1, 1, 2),.., dens C (1, 1,Q), dens C (1, 2, 1),.., dens C (1,R,Q), dens C (2, 1, 1),.., dens C (K,R,Q))

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V C = (8, 0, 5, 3, 0, 6, 3, 3) V C ′ = (5, 2, 2, 5, 3, 4, 6, 1)

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V C = (8, 0, 5, 3, 0, 6, 3, 3) V C ′ = (5, 2, 2, 5, 3, 4, 6, 1) c1c1 c2c2 c’ 1 c’ 2 Sim(C, C’) = V c ·V c’ = 110

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V C = (8, 0, 5, 3, 0, 6, 3, 3) V C ′ = (5, 2, 2, 5, 3, 4, 6, 1) V C = (8, 0, 5, 3, 0, 6, 3, 3) V C ′ = (3, 4, 6, 1, 5, 2, 2, 5) c1c1 c2c2 c’ 1 c’ 2 c1c1 c2c2 c’ 1 Sim(C, C’) = V c ·V c’ = 90 Sim(C, C’) = V c ·V c’ = 110

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V C = (8, 0, 5, 3, 0, 6, 3, 3) V C ′ = (5, 2, 2, 5, 3, 4, 6, 1) V C = (8, 0, 5, 3, 0, 6, 3, 3) V C ′ = (3, 4, 6, 1, 5, 2, 2, 5) c1c1 c2c2 c’ 1 c’ 2 c1c1 c2c2 c’ 1 Sim(C, C’) = V c ·V c’ = 110 = 65 + 45 Sim(C, C’) = V c ·V c’ = 90 = 57 + 33 c' 1 c' 2 c1c1 6557 c2c2 3345

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V C = (8, 0, 5, 3, 0, 6, 3, 3) V C ′ = (5, 2, 2, 5, 3, 4, 6, 1) V C = (8, 0, 5, 3, 0, 6, 3, 3) V C ′ = (3, 4, 6, 1, 5, 2, 2, 5) c1c1 c2c2 c’ 1 c’ 2 c1c1 c2c2 c’ 1 Sim(C, C’) = V c ·V c’ = 110 = 65 + 45 Sim(C, C’) = V c ·V c’ = 90 = 57 + 33 ADCO(C, C’) = 110 / max(|V c |,|V c’ |) = 110 / max(152, 120)=0.724

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MFNN1N3P1P3 MFN157471673781891769313266 N1656811401354181945258 N350773981282846843350 P19792109904788126208524 P394861010346171183311478 SG01_tPCA_m1 SG02_tPCA_m1 ADCO: 0.667 Can be solved by the Hungarian Method. Time complexity O(n 3 ).

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SG01-sICA-m1SG01-tPCA-m1SG01-sICA-m2SG01-tPCA-m2SG02-sICA-m1SG02-tPCA-m1SG02-sICA-m2SG02-tPCA-m2 SG01-sICA-m110.7890.9060.8680.8540.770.8730.884 SG01-tPCA-m110.8280.7290.8770.6670.7090.845 SG01-sICA-m210.8240.8790.7140.8670.912 SG01-tPCA-m210.7840.8620.8110.808 SG02-sICA-m110.740.7870.917 SG02-tPCA-m110.7480.735 SG02-sICA-m210.8 SG02-tPCA-m21

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SG01-sICA-m1SG01-tPCA-m1SG01-sICA-m2SG01-tPCA-m2SG02-sICA-m1SG02-tPCA-m1SG02-sICA-m2SG02-tPCA-m2 SG01-sICA-m110.7890.9060.8680.8540.770.8730.884 SG01-tPCA-m110.8280.7290.8770.6670.7090.845 SG01-sICA-m210.8240.8790.7140.8670.912 SG01-tPCA-m210.7840.8620.8110.808 SG02-sICA-m110.740.7870.917 SG02-tPCA-m110.7480.735 SG02-sICA-m210.8 SG02-tPCA-m21 SG01-tPCA-m1SG02-tPCA-m1SG01-tPCA-m2SG02-tPCA-m2 SG01-sICA-m10.7890.770.8680.884 SG02-sICA-m10.8770.740.7840.917 SG01-sICA-m20.8280.7140.8240.912 SG02-sICA-m20.7090.7480.8110.8 Difference in only DM (decomposition method) Difference DM and SG (subject group) Difference DM and MS (metric set) Difference in DM and SG and MS

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