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I have df.core.new dataframe with 60 items.
I use PCA to explore the factors:

   psych::principal(df.core.new,rotate ='promax')
Principal Components Analysis
Call: psych::principal(r = df.core.new, rotate = "promax")
Standardized loadings (pattern matrix) based upon correlation matrix
      PC1      h2   u2 com
q01  0.06 3.5e-03 1.00   1
q02  0.12 1.6e-02 0.98   1
q04 -0.26 7.0e-02 0.93   1
q05 -0.06 3.3e-03 1.00   1
q06  0.09 7.6e-03 0.99   1
q07  0.09 7.6e-03 0.99   1
q08  0.01 2.7e-05 1.00   1
q09  0.12 1.3e-02 0.99   1
q10  0.12 1.5e-02 0.99   1
q11  0.09 8.0e-03 0.99   1
q12  0.06 3.6e-03 1.00   1
q13 -0.03 1.2e-03 1.00   1
q14  0.02 6.1e-04 1.00   1
q15  0.11 1.1e-02 0.99   1
q16  0.32 9.9e-02 0.90   1
q17  0.12 1.4e-02 0.99   1
q18  0.19 3.7e-02 0.96   1
q20  0.20 4.2e-02 0.96   1
q21  0.28 7.9e-02 0.92   1
q22  0.13 1.8e-02 0.98   1
q23  0.40 1.6e-01 0.84   1
q24  0.38 1.4e-01 0.86   1
q25  0.34 1.2e-01 0.88   1
q26  0.34 1.2e-01 0.88   1
q27  0.35 1.2e-01 0.88   1
q28  0.50 2.5e-01 0.75   1
q29  0.49 2.4e-01 0.76   1
q30  0.33 1.1e-01 0.89   1
q31  0.34 1.2e-01 0.88   1
q32  0.23 5.5e-02 0.95   1
q33  0.26 6.7e-02 0.93   1
q34  0.13 1.6e-02 0.98   1
q36  0.45 2.0e-01 0.80   1
q38  0.42 1.7e-01 0.83   1
q39  0.46 2.1e-01 0.79   1
q40  0.26 6.7e-02 0.93   1
q41  0.41 1.7e-01 0.83   1
q42  0.26 6.7e-02 0.93   1
q44  0.29 8.2e-02 0.92   1
q46  0.62 3.9e-01 0.61   1
q47  0.35 1.2e-01 0.88   1
q48  0.70 4.8e-01 0.52   1
q49  0.69 4.8e-01 0.52   1
q51  0.60 3.6e-01 0.64   1
q52  0.71 5.0e-01 0.50   1
q53  0.49 2.4e-01 0.76   1
q54  0.52 2.7e-01 0.73   1
q55  0.71 5.0e-01 0.50   1
q56  0.65 4.2e-01 0.58   1
q57  0.59 3.4e-01 0.66   1
q58  0.42 1.8e-01 0.82   1
q59  0.70 4.9e-01 0.51   1
q60  0.64 4.1e-01 0.59   1

                PC1
SS loadings    8.11
Proportion Var 0.15

Mean item complexity =  1
Test of the hypothesis that 1 component is sufficient.

The root mean square of the residuals (RMSR) is  0.12 
 with the empirical chi square  62848.22  with prob <  0 

Fit based upon off diagonal values = 0.55     

In PCA, suggest 1 factor, then I explore the factors number:

    > psych::fa.parallel(df.core.new,fa="both",n.iter=100)
Parallel analysis suggests that the number of factors =  8  and the number of components =  7   

1.How many factors should I use in EFA?
2.If 7 factors,what the usage of PCA result?I really don't know when should I use PCA. PCA seemed not worthy to consider.

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  • $\begingroup$ This question is too broad. Besides, it is potentially too tied with a specific program/software. Have you read any texts (books, articles) about how to choose the number of factors? (Some of your questions about FA on this site makes one wish you first to read something substantial on the topic before asking questions.) $\endgroup$
    – ttnphns
    Jul 29, 2017 at 8:38
  • $\begingroup$ I read textbook, and then have this question.This is not a general queston,then I post script in R. $\endgroup$
    – WhiteGirl
    Jul 29, 2017 at 8:53
  • $\begingroup$ I vote to leave this open; there is an answerable statistics question here. $\endgroup$
    – Peter Flom
    Jul 29, 2017 at 12:01
  • $\begingroup$ One of the historic limitations of all dimension reducing tools such as PCA, EFA, clustering, etc., are the many subjective decisions leading to a solution that are required from the analyst. There is no cookbook answer to any of them, only rules of thumb and widely agreed upon conventions. One fact, not assumption, is that PCA is deterministic (i.e., it does not have an error term and therefore is not a "model") while the latter does have an error term and can be considered a "model." $\endgroup$ Jul 29, 2017 at 12:54

1 Answer 1

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principal in the psych package extracts one component by default. It did not tell you there was one component, you told it to extract one component.

Determining the number of factors is a complex task; there are several statistical guidelines (e.g. eigenvalue over 1, scree test, proportion of variance) and you can read about them in any book on factor analysis. But really, the way to do it is to look at various solutions and see which one makes the most substantive sense. You have to use your subject matter knowledge to figure this out. Of course, sometimes, the answer is obvious. But usually not.

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  • $\begingroup$ Then I always would like to use EFA instead of PCA.I don't see any advantage of PCA $\endgroup$
    – WhiteGirl
    Jul 29, 2017 at 12:14
  • 2
    $\begingroup$ You are just using the function incorrectly. PCA and EFA do different things. PCA is a dimension reduction method, EFA is used to find latent variables. $\endgroup$
    – Peter Flom
    Jul 29, 2017 at 12:17
  • $\begingroup$ I knew that but I did not meet any case of using PCA, and there's no specific example in textbook $\endgroup$
    – WhiteGirl
    Jul 29, 2017 at 12:42

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