Exploratory Factor Analysis: Share of variance explained by retained factors using ipf

[Rebecca Mason]

I am conducting an exploratory factor analysis using the iterated principal factor method (ipf). After determining the number of factors to retain (using a combination of screeplot, Velicer's MAP, and PA), it would be helpful to calculate the % of total variance accounted for by retained factors.

have found information indicating that, for principal component analysis (pca), the % of total variance explained = sum of eigenvalues / sum of all positive eigenvalues. Is this possible to calculate the share of total variance explained by retained factors using ipf? If so, what is the Stata command?

When I specify the number of factors to retain, it seems like stata only reports the % explained variance explained by each of the retained factors (which is always 100% cumulatively). See code and output below.

factor varlist, ipf factors(1) blanks(0.3)

(obs=612)
Factor analysis/correlation Number of obs = 612
Method: iterated principal factors Retained factors = 1
Rotation: (unrotated) Number of params = 13
-----------	-------------	-------------	------------------	----------------
Factor	Eigenvalue	Difference	Proportion	Cumulative
-----------	-------------	-------------	------------------	----------------
Factor1	6.92805	5.65735	1	1
Factor2	1.27069	0.90124	0.1834	1.1834
Factor3	0.36945	0.19711	0.0533	1.2367
Factor4	0.17234	0.12943	0.0249	1.2616
Factor5	0.04291	0.03149	0.0062	1.2678
Factor6	0.01142	0.08806	0.0016	1.2695
Factor7	-0.07664	0.02108	-0.0111	1.2584
Factor8	-0.09773	0.05138	-0.0141	1.2443
Factor9	-0.14911	0.04597	-0.0215	1.2228
Factor10	-0.19508	0.09969	-0.0282	1.1946
Factor11	-0.29476	0.15309	-0.0425	1.1521
Factor12	-0.44785	0.15784	-0.0646	1.0874
Factor13	-0.60569	.	-0.0874	1
--------------------------------------------------------------------------
LR test: independent vs. saturated: chi2(78) = 1.1e+04 Prob>chi2 = 0.0000

[Oded Mcdossi]

In order to get the proportion and cumulative share of each factor as a percentage of all positive eigenvalues, just omit the number of factors to retain in the option part. Then, the values of the eigenvalues for each factor will be the same, but the proportion will change, the total will be the sum of the positive eigenvalues.

[Rebecca Mason]

I am concerned about using this proportion/cumulative share because the Eigenvalues change depending on whether # factors retained is specified or not.

factor varlist, ipf

Factor analysis/correlation Number of obs = 612
Method: iterated principal factors Retained factors = 12
Rotation: (unrotated) Number of params = 78
Beware: solution is a Heywood case
(i.e., invalid or boundary values of uniqueness)
------------	--------------	---------------	---------------	---------------
Factor	Eigenvalue	Difference	Proportion	Cumulative
------------	--------------	---------------	---------------	---------------
Factor1	7.1392	5.37029	0.6852	0.6852
Factor2	1.76891	1.12573	0.1698	0.855
Factor3	0.64318	0.25871	0.0617	0.9167
Factor4	0.38447	0.1743	0.0369	0.9536
Factor5	0.21017	0.06742	0.0202	0.9738
Factor6	0.14275	0.08187	0.0137	0.9875
Factor7	0.06088	0.03152	0.0058	0.9934
Factor8	0.02936	0.00487	0.0028	0.9962
Factor9	0.0245	0.01333	0.0024	0.9985
Factor10	0.01116	0.00779	0.0011	0.9996
Factor11	0.00337	0.00221	0.0003	0.9999
Factor12	0.00116	0.00147	0.0001	1
Factor13	-0.00031	.	0	1
------------	--------------	---------------	---------------	---------------
LR test: independent vs. saturated: chi2(78) = 1.1e+04 Prob>chi2 = 0.0000

[Oded Mcdossi]

You are correct, mistake on my side. I haven't noticed the type of factor method.

[Oded Mcdossi]

It seems like this is impossible since with ipf method the number of iterations depends on the number of extracted factors unless you fixed the communality reestimation iterations to 0 with the citerate() option as explained by the help file. See also examples 4-5 in the PDF Documentation.