>From ?princomp
Arguments:
scores: a logical value indicating whether the score on each
principal component should be calculated.
Value:
scores: if ‘scores = TRUE’, the scores of the supplied data on the
principal components. These are non-null only if ‘x’ was
supplied, and if ‘covmat’ was also supplied if it was a
covariance list. For the formula method, ‘napredict()’ is
applied to handle the treatment of values omitted by the
‘na.action’.
## NA-handling
USArrests[1, 2] <- NA
pc.cr <- princomp(~ Murder + Assault + UrbanPop,
data = USArrests, na.action=na.exclude, cor = TRUE)
pc.cr$scores[1:5, ]
>From the help for psych:::principal:
Arguments:
scores: If TRUE, find component scores
Value:
scores: If scores=TRUE, then estimates of the factor scores are
reported
So the scores are returned as pc$scores in your example, and princomp
will also calculate them.
Sarah
On Fri, Oct 19, 2012 at 11:30 AM, ya <xinxi...@126.com> wrote:
>
>
>
> Hi everyone,
>
> I am trying to get the factor score for each individual case from a principal
> component analysis, as I understand, both princomp() and prcomp() can not
> produce this factor score, the principal() in psych package has this option:
> scores=T, but after running the code, I could not figure out how to show the
> factor score results. Here is my code, could anyone give me some advice
> please? Thank you very much.
>
>> pc <- principal(a,rotate="varimax",scores=TRUE)
>> pc
> Principal Components Analysis
> Call: principal(r = a, rotate = "varimax", scores = TRUE)
> Standardized loadings (pattern matrix) based upon correlation matrix
> PC1 h2 u2
> V1 0.80 6.4e-01 0.36
> V2 0.03 7.9e-04 1.00
> V3 -0.92 8.4e-01 0.16
> V4 0.00 2.0e-06 1.00
> V5 0.54 2.9e-01 0.71
>
> PC1
> SS loadings 1.77
> Proportion Var 0.35
>
> Test of the hypothesis that 1 component is sufficient.
>
> The degrees of freedom for the null model are 10 and the objective function
> was 0.87
> The degrees of freedom for the model are 5 and the objective function was
> 0.39
> The number of observations was 20 with Chi Square = 6.14 with prob < 0.29
>
> Fit based upon off diagonal values = 0.61
>
>> summary(pc)
>
> Factor analysis with Call: principal(r = a, rotate = "varimax", scores = TRUE)
>
> Test of the hypothesis that 1 factor is sufficient.
> The degrees of freedom for the model is 5 and the objective function was
> 0.39
> The number of observations was 20 with Chi Square = 6.14 with prob < 0.29
>
>
>
--
Sarah Goslee
http://www.functionaldiversity.org
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