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factor score from PCA
2 messages · ya, Sarah Goslee
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 <xinxi813 at 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