dendrogram from precomputed distance matrix
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I have looked around for an answer for this by have not been able to find one so I have come here. I have a precomputed distance matrix from an all vs all comparison of the root mean squared deviation (RMSD) of different protein structures.
0 0.5770 0.4910 1.6840 0.6660 0.8970 1.0920 0.5380 0.6390 0.8100
0.5770 0 0.6540 2.2020 0.2370 0.8790 0.9440 0.6120 0.4030 1.2530
0.4910 0.6540 0 1.7360 0.6300 0.7520 1.4530 0.3460 0.4630 1.1190
1.6840 2.2020 1.7360 0 2.2740 2.0790 2.5400 1.8480 2.1110 1.3380
0.6660 0.2370 0.6300 2.2740 0 0.8560 1.0880 0.5900 0.3360 1.3490
0.8970 0.8790 0.7520 2.0790 0.8560 0 1.4620 0.6840 0.7260 1.4480
1.0920 0.9440 1.4530 2.5400 1.0880 1.4620 0 1.3330 1.2400 1.3950
0.5380 0.6120 0.3460 1.8480 0.5900 0.6840 1.3330 0 0.3260 1.0900
0.6390 0.4030 0.4630 2.1110 0.3360 0.7260 1.2400 0.3260 0 1.2800
0.8100 1.2530 1.1190 1.3380 1.3490 1.4480 1.3950 1.0900 1.2800 0
I want to draw a dendrogram from this matrix:
tree=linkage(matrix,'average')
dengrogram(tree,0)
but if I do this the y-axis distance is all wrong because it's calculating the euclidean distances in the linkage. Is there a way to draw dendrograms directly from a precomputed distance matrix?
답변 (2개)
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Note that a much simpler and faster way to obtain the same result would be:
v = A(tril(true(size(A)), -1))';
crow white
2017년 9월 23일
0 개 추천
I would like to do the same: I have a precomputed distance matrix (calculated on data using the bray-curtis index) and I want to draw a dendogram of the results. I don't see how your solution draws the dendrogram. Thanks for your help
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