RE: st: RE: principal component analysis-creating linear combinations

[Nick Cox <n.j.cox@durham.ac.uk>]

Not so: There is an explicit example for exactly your need in the help. 

Individual scores for the components are obtained via predict
        . predict f1
        . predict f1 f2

That is, for 2, 3, ... components, specify as many names as you need. 

I am looking at Stata 11 documentation; if you are using an earlier version, you should state that as requested in the Statalist FAQ. 

Nick 
n.j.cox@durham.ac.uk 

James Wu [mailto:jwu736@gmail.com] 

Nick, thank you very much.

But how can I obtain the second component scores (that would
correspond to Y2 that I called earlier) by using predict?
I read the manual on pca postestimation, but there is no indication on
it (only the first component scores).

On Thu, Mar 10, 2011 at 9:56 AM, Nick Cox <n.j.cox@durham.ac.uk> wrote:

> The easiest and best way to create the principal components themselves is use -predict- after -pca-. There is no need for you to do the calculation by typing out coefficients in a linear equation. That is even at best problematic in terms of keeping precision.
>
> The default of -pca- is to use the correlation matrix; that is entirely equivalent to using standardised variables, so that there is absolutely no need to standardise yourself, except possibly as an exercise.
>
> I wouldn't call the eigenvectors the PCs myself, although there are varying habits on this.

James Wu

> Suppose we ran pca on four variables, x1, x2, x3, x4 as follows:

> . pca  x1 x2 x3 x4, components (3)
>
> Principal components/correlation                  Number of obs    =       659
> Number of comp.  =         3
> Trace            =         4
> Rotation: (unrotated = principal)             Rho              =    0.9550
> --------------------------------------------------------------------------
> Component    Eigenvalue   Difference         Proportion   Cumulative
> -------------+------------------------------------------------------------
> Comp1       2.42894      1.67142             0.6072       0.6072
> Comp2       .757515      .124084             0.1894       0.7966
> Comp3       .633431      .453314             0.1584       0.9550
> Comp4       .180117            .             0.0450       1.0000
> --------------------------------------------------------------------------
> Principal components (eigenvectors)
> ----------------------------------------------------------
> Variable     Comp1     Comp2     Comp3  Unexplained
> -------------+------------------------------+-------------
> x1    0.3894    0.8726   -0.2945    .00004265
> x2    0.4517    0.0966    0.8858     .0003491
> x3    0.5733   -0.3179   -0.2218       .09384
> x4    0.5619   -0.3580   -0.2817       .08588
> ----------------------------------------------------------
>
>
> Now, suppose that you decide to retain the firs two principal
> components, and then you want to create two variables that are linear
> combinations of the original four variables.
>
> Question1:  Would it be simply to create by multiply the Principal
> Components (eigenvectors, columns)  with the orginal variables, say,
> Y1=0.3894*x1+0.4517*x2+0.5733*x3+0.5619*x4 and
> Y2=0.8726*x1+0.0966*x2-0.3179*x3-0.3580*x4?
>
> Question 2: Assuming that I am correct in creating new variables by
> simply multiplying the Principal components (eigenvectors) with the
> orginal variables (Question 1),
> if these four original variables are in different units of
> measurement, then should we standardize the original four variables
> (so that each of standardized original variable has mean 0 and std of
> 1) before computing the multiproducts as in my Question 1?
>
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