difference between pca and pcaFromStatToolbox
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It might sound stupid, but I actually am confused with the results of the pca and pcaFromStatToolbox. I noticed the different output just now and I am wondering why there are different:
[coeff1,score1]=pcaFromStatToolbox(ran)
[coeff2,score2]=pca(x)
so lets have an example:
>> pcaData=rand(4,5)
pcaData =
0.4638 0.7937 0.6250 0.1400 0.4149
0.7046 0.5080 0.3831 0.8778 0.0977
0.0153 0.8616 0.8466 0.7827 0.0962
0.5929 0.9365 0.0800 0.0978 0.8779
----
>> [n,nn]=pcaFromStatToolbox(pcaData)
n =
-0.2633 0.6508 -0.4266
-0.1495 -0.3995 0.3599
0.4276 -0.4884 -0.4736
0.6094 0.4031 0.5820
-0.5951 -0.1256 0.3542
nn =
-0.1772 -0.2040 -0.2480
0.3372 0.5222 -0.0219
0.6069 -0.3321 0.1240
-0.7668 0.0139 0.1458
------
>> [m,mm]=pca(pcaData)
netlab pca: using eig
netlab pca: sorting evec
m =
0.3671
0.1416
0.0329
0.0000
0.0000
mm =
0.2633 -0.6508 -0.4266 -0.1072 0.5601
0.1495 0.3995 0.3599 0.3753 0.7400
-0.4276 0.4884 -0.4736 -0.5079 0.3106
-0.6094 -0.4031 0.5820 -0.2916 0.2056
0.5951 0.1256 0.3542 -0.7104 0
3 Comments
the cyclist
on 23 Feb 2016
Edited: the cyclist
on 23 Feb 2016
I just did a quick search KPMstats and MATLAB. I found this annotation:
"KPMstats is a directory of miscellaneous statistics functions written by Kevin Patrick Murphy and various other people (see individual file headers)."
Personally, I would need to dig in to get more confidence in Murphy et al. (who are surely fine fellows). I have a fair amount of experience with the MATLAB pca, and I am very confident in its output.
Answers (1)
the cyclist
on 23 Feb 2016
Edited: the cyclist
on 23 Feb 2016
Even without knowing the source of the other function, I can make a guess.
Notice that for the input, you have 4 observations (4 rows) of 5 variables. So, you can fully explain 100% of the variation with just 4 principal components. Furthermore, because MATLAB centers the variables, you can do it with 3 principal components.
Notice that MATLAB outputs 3 principal component coefficients, where your other software outputs 5 vectors. That other software It is clearly making a different assumption in the case where you only actually need 3 to fully span the space. My guess is that the 4th and 5th vectors (the ones that are different from MATLAB) are linear combinations of the first 3.
1 Comment
the cyclist
on 23 Feb 2016
My speculation that the other two vector outputs are linear combinations of the other three seems not to be true. I'm not sure what's going on there.
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