%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Native samples %%%%%%%%%%%%%%%%%%%%%% >> X = [df_aac48, df_apple, df_nero, df_opus, df_vorbis]; >> X = X'; >> % Spearman's rank correlation Y = pdist(X,'spearman'); squareform(Y) Z = linkage(Y,'average'); dendrogram(Z, 'Labels',{'AAC48';'Apple';'Nero';'Opus';'Vorbis'}) Coph = cophenet(Z,Y) trineqtest(Y) ans = 0 0.1143 0.1576 0.2394 0.3274 0.1143 0 0.0788 0.2242 0.2850 0.1576 0.0788 0 0.2158 0.2950 0.2394 0.2242 0.2158 0 0.1172 0.3274 0.2850 0.2950 0.1172 0 Coph = 0.9130 Triangle inequality holds for 60 cases from 60 possible (100.00%). % ----------------------------------------------------------------- Y = pdist(X,@distancecorr); squareform(Y) Z = elinkage(Y); % e-link dendrogram(Z, 'Labels',{'AAC48';'Apple';'Nero';'Opus';'Vorbis'}) Coph = cophenet(Z,Y) trineqtest(Y) ans = 0 0.1184 0.1700 0.2565 0.3443 0.1184 0 0.0936 0.2456 0.3125 0.1700 0.0936 0 0.2450 0.3200 0.2565 0.2456 0.2450 0 0.1294 0.3443 0.3125 0.3200 0.1294 0 Coph = 0.9214 Triangle inequality holds for 60 cases from 60 possible (100.00%). %%%%%%%%%%%%%%%%%%%%%%%%%% The random mix %%%%%%%%%%%%%%%%%%%%%%% >> X = [df_aac48, df_apple, df_nero, df_opus, df_vorbis]; >> X = X'; >> % Spearman's rank correlation Y = pdist(X,'spearman'); squareform(Y) Z = linkage(Y,'average'); dendrogram(Z, 'Labels',{'AAC48';'Apple';'Nero';'Opus';'Vorbis'}) Coph = cophenet(Z,Y) trineqtest(Y) ans = 0 0.2359 0.2619 0.2508 0.2981 0.2359 0 0.1062 0.2079 0.2164 0.2619 0.1062 0 0.1841 0.1956 0.2508 0.2079 0.1841 0 0.1130 0.2981 0.2164 0.1956 0.1130 0 Coph = 0.9591 Triangle inequality holds for 60 cases from 60 possible (100.00%). % ----------------------------------------------------------------- Y = pdist(X,@distancecorr); squareform(Y) Z = elinkage(Y); % e-link dendrogram(Z, 'Labels',{'AAC48';'Apple';'Nero';'Opus';'Vorbis'}) Coph = cophenet(Z,Y) trineqtest(Y) ans = 0 0.2606 0.2945 0.2741 0.3230 0.2606 0 0.1316 0.2341 0.2447 0.2945 0.1316 0 0.2141 0.2261 0.2741 0.2341 0.2141 0 0.1333 0.3230 0.2447 0.2261 0.1333 0 Coph = 0.8380 Triangle inequality holds for 60 cases from 60 possible (100.00%).