The number of PCs from ordination of DEICODE distance matrix

[11132 (yun du)]

Hi dear friends,
I’ve a question about the number of PCs from ordination of DEICODE distance matrix.
My DEICODE matrix had 482 samples. The ordination showed 4503 features. But there was only 3 PCs and the proportion of variation was summed to 100%.
Is it right? And how come?
Because for traditional PCA, the number of PCs is either the number of samples or the number of features, whichever the smaller is the upper bound.
Thanks and look forward to our discussion.

Best,
Yun

[SoilRotifer (Mike Robeson)]

Hi @11132,

The following threads may help:

Help understanding DEICODE

Hi Cameron, Sorry, one other question. When I use the qiime plugin I now I understand that the --p-rank parameter defines the output rank of the matrix which means the total variance of your axes in PCA will be defined by that number (rank # = number of PCA vectors to sum to 100% variance). When I analyze my own dataset I find that if I use the default rank setting of 3 that the axes explain 47%,27%,26% of the variance respectively. When I rerun this analysis with --p-rank set to 5 the axes exp…

when things look good, I wonder where I've gone wrong...

Hi @devonorourke, You are correct in lowering the minimum count sum perameters. In this case lowering them to 1 or zero may even be warranted. The DIECODE percent variance explained is a relative percentage across each axis (in this case only three). For example, if the rank is two there will only be two axis and they will sum to 100% of the variance. This can make the % variance increase dramatically over other methods with many PCs. So I would not compare it between methods (like PCoA) and o…

-Mike

[11132 (yun du)]

Hi Mike,

Thanks for your reply. I used the default, 3 ranks. But it seems award. I’m confused how it comes.

best,
Yun