Haishan Liu.  to identify correspondences (“mappings”) between ERP patterns derived from:  different data decomposition methods (temporal PCA and spatial.

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Presentation transcript:

Haishan Liu

 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

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

 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.

 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.

 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))

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

 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

 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 = Sim(C, C’) = V c ·V c’ = 90 = c' 1 c' 2 c1c c2c2 3345

 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 = Sim(C, C’) = V c ·V c’ = 90 = ADCO(C, C’) = 110 / max(|V c |,|V c’ |) = 110 / max(152, 120)=0.724

MFNN1N3P1P3 MFN N N P P SG01_tPCA_m1 SG02_tPCA_m1 ADCO: Can be solved by the Hungarian Method. Time complexity O(n 3 ).

SG01-sICA-m1SG01-tPCA-m1SG01-sICA-m2SG01-tPCA-m2SG02-sICA-m1SG02-tPCA-m1SG02-sICA-m2SG02-tPCA-m2 SG01-sICA-m SG01-tPCA-m SG01-sICA-m SG01-tPCA-m SG02-sICA-m SG02-tPCA-m SG02-sICA-m210.8 SG02-tPCA-m21

SG01-sICA-m1SG01-tPCA-m1SG01-sICA-m2SG01-tPCA-m2SG02-sICA-m1SG02-tPCA-m1SG02-sICA-m2SG02-tPCA-m2 SG01-sICA-m SG01-tPCA-m SG01-sICA-m SG01-tPCA-m SG02-sICA-m SG02-tPCA-m SG02-sICA-m210.8 SG02-tPCA-m21 SG01-tPCA-m1SG02-tPCA-m1SG01-tPCA-m2SG02-tPCA-m2 SG01-sICA-m SG02-sICA-m SG01-sICA-m SG02-sICA-m 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