如何获得 R 中探索性因素分析的初始共同性？

Question

我想获得 R 中探索性因素分析的初始公性（即，由分析中包含的其他项目预测时每个项目的 R 平方）。

有没有办法用 jmv::efa 或 psych::fa 做到这一点？

我只看到唯一性，它告诉我因子提取后的共同性（1-唯一性）...

感谢您的考虑:)

Answer 1

数据集

这是我正在使用的数据的输入，以下称为hwk:

hwk <- structure(list(V1 = structure(c(4, 4, 2, 2, 2, 2, 2, 2, 4, 4, 
2, 3, 2, 3, 4, 2, 2, 2, 3, 3, 2, 3, 1, 3, 3, 3, 3, 4, 1, 2, 4, 
1, 2, 3, 2, 3, 1, 1, 2, 2, 4, 3, 2, 1, 2, 3, 3, 4, 3, 3, 2, 3, 
1, 4, 3, 2, 3, 4, 1, 3, 3, 3, 2, 2, 1, 2, 3, 4, 4, 2, 4, 3, 2, 
3, 3, 3, 3, 2, 4, 3, 3, 3, 2, 2, 3, 4, 2, 4, 4, 2, 2, 3, 3), format.spss = "F8.0"), 
    V2 = structure(c(4, 4, 3, 4, 3, 4, 3, 2, 4, 1, 3, 3, 3, 4, 
    3, 3, 2, 3, 4, 3, 1, 4, 2, 3, 4, 2, 4, 3, 3, 2, 3, 2, 3, 
    3, 4, 3, 3, 3, 3, 3, 3, 2, 4, 2, 2, 2, 4, 3, 4, 4, 2, 4, 
    2, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 4, 3, 3, 4, 4, 4, 4, 4, 
    3, 4, 3, 3, 3, 4, 2, 4, 3, 4, 3, 3, 2, 3, 3, 4, 3, 4, 3, 
    4, 4, 3), format.spss = "F8.0"), V3 = structure(c(4, 4, 4, 
    4, 4, 4, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 
    3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 
    3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 
    3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 
    4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4), format.spss = "F8.0"), 
    V4 = structure(c(4, 4, 3, 4, 3, 4, 2, 1, 3, 2, 3, 1, 4, 4, 
    2, 3, 2, 2, 2, 4, 1, 2, 2, 2, 3, 2, 3, 2, 2, 1, 3, 1, 1, 
    2, 4, 1, 1, 2, 3, 2, 2, 1, 1, 1, 3, 2, 4, 3, 3, 3, 3, 3, 
    3, 4, 3, 1, 4, 3, 4, 3, 2, 3, 2, 1, 4, 1, 4, 1, 2, 4, 4, 
    4, 3, 3, 3, 2, 2, 1, 4, 3, 2, 3, 2, 1, 3, 4, 1, 2, 4, 3, 
    4, 2, 2), format.spss = "F8.0"), V5 = structure(c(3, 3, 3, 
    4, 3, 4, 3, 1, 1, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 2, 
    2, 2, 2, 4, 2, 3, 2, 3, 4, 1, 4, 2, 3, 3, 2, 2, 3, 2, 2, 
    3, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 3, 3, 2, 2, 3, 3, 2, 3, 
    2, 2, 3, 3, 3, 2, 3, 3, 3, 4, 3, 2, 3, 3, 3, 3, 3, 3, 4, 
    3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 4, 3, 3), format.spss = "F8.0"), 
    V6 = structure(c(4, 4, 3, 4, 3, 4, 4, 1, 3, 3, 3, 3, 2, 3, 
    4, 2, 4, 3, 3, 3, 3, 4, 4, 3, 3, 3, 4, 4, 4, 3, 4, 4, 3, 
    3, 3, 4, 2, 2, 3, 3, 3, 4, 2, 4, 3, 4, 4, 4, 3, 4, 2, 4, 
    3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 3, 1, 4, 4, 4, 4, 4, 4, 
    4, 3, 4, 4, 4, 4, 2, 4, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 3, 
    4, 4, 4), format.spss = "F8.0"), V7 = structure(c(4, 4, 2, 
    4, 2, 4, 4, 3, 3, 3, 2, 2, 4, 4, 3, 3, 1, 4, 3, 3, 1, 2, 
    4, 3, 4, 2, 4, 4, 3, 3, 2, 2, 3, 2, 4, 3, 3, 3, 3, 3, 3, 
    1, 4, 3, 2, 2, 4, 3, 4, 4, 2, 4, 2, 3, 4, 3, 3, 3, 4, 3, 
    4, 4, 3, 4, 4, 3, 4, 4, 4, 4, 3, 4, 4, 4, 3, 3, 4, 3, 4, 
    3, 3, 3, 3, 2, 2, 4, 4, 4, 4, 2, 4, 4, 3), format.spss = "F8.0"), 
    V8 = structure(c(4, 4, 2, 1, 2, 1, 1, 1, 3, 3, 2, 3, 2, 3, 
    4, 2, 2, 2, 3, 3, 2, 3, 1, 3, 3, 3, 3, 4, 1, 2, 4, 1, 2, 
    3, 2, 3, 1, 1, 2, 2, 3, 1, 1, 1, 2, 3, 3, 4, 3, 3, 2, 3, 
    1, 3, 4, 2, 3, 4, 1, 3, 3, 3, 2, 2, 1, 2, 3, 4, 4, 2, 4, 
    3, 4, 4, 4, 4, 3, 2, 4, 3, 3, 3, 2, 2, 3, 4, 2, 4, 4, 2, 
    1, 3, 4), format.spss = "F8.0"), V9 = structure(c(4, 4, 4, 
    4, 4, 4, 4, 4, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 2, 3, 4, 4, 
    4, 4, 3, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 4, 3, 2, 4, 3, 4, 
    4, 4, 4, 4, 4, 3, 3, 3, 3, 4, 4, 4, 4, 4, 3, 4, 3, 2, 4, 
    3, 3, 4, 4, 4, 3, 4, 4, 4, 4, 4, 3, 4, 3, 4, 3, 4, 4, 4, 
    4, 3, 4, 4, 4, 4, 4, 3, 2, 4, 4, 4, 4, 4), format.spss = "F8.0"), 
    V10 = structure(c(4, 4, 2, 4, 2, 4, 3, 2, 3, 3, 3, 2, 4, 
    4, 2, 2, 1, 3, 4, 4, 1, 4, 2, 3, 3, 2, 4, 3, 2, 3, 3, 1, 
    3, 2, 4, 3, 2, 3, 3, 3, 3, 1, 2, 4, 2, 3, 4, 4, 3, 3, 2, 
    4, 2, 4, 3, 3, 4, 3, 4, 3, 4, 4, 4, 1, 4, 3, 3, 4, 3, 4, 
    4, 3, 3, 3, 3, 3, 4, 1, 4, 3, 3, 3, 3, 2, 3, 4, 4, 2, 4, 
    2, 4, 4, 3), format.spss = "F8.0"), V11 = structure(c(3, 
    3, 1, 4, 1, 4, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 2, 2, 1, 
    2, 3, 1, 2, 3, 3, 2, 1, 2, 2, 2, 3, 2, 2, 3, 2, 1, 2, 2, 
    1, 1, 4, 3, 1, 3, 2, 3, 1, 2, 1, 2, 1, 2, 2, 1, 2, 2, 3, 
    2, 2, 2, 2, 2, 2, 1, 1, 1, 3, 3, 4, 2, 1, 2, 2, 3, 3, 3, 
    3, 4, 3, 2, 3, 3, 2, 2, 2, 2, 1, 3, 1, 4, 1, 3), format.spss = "F8.0"), 
    V12 = structure(c(4, 4, 3, 2, 3, 2, 3, 1, 3, 3, 3, 3, 2, 
    3, 3, 2, 4, 3, 3, 4, 4, 3, 3, 4, 4, 3, 3, 3, 4, 3, 4, 4, 
    3, 3, 3, 4, 2, 2, 3, 3, 3, 4, 2, 4, 3, 4, 4, 4, 3, 4, 2, 
    4, 3, 3, 3, 3, 4, 3, 3, 2, 2, 1, 1, 3, 1, 4, 4, 4, 4, 4, 
    4, 4, 3, 3, 2, 2, 2, 2, 4, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 
    3, 2, 3, 4), format.spss = "F8.0")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -93L))

全民教育

我在初步回答后做了一些研究，似乎有一个名为 EFA 工具的软件包。有一个名为 EFA 的函数允许您指定您想要初始社区。首先，运行图书馆和下面的 EFA 本身：

# Load EFA Tools library:
library(EFAtools)

# Run EFA:
hwkfa <- EFA(hwk,
    n_factors = 3,
    start_method = "psych",
    method = "PAF",
    rotation = "promax",
    init_comm = "smc", # selected initial communalities
    type = "SPSS")

获得初始共同点：

然后您可以使用以下代码简单地 select 初始社区：

hwkfa$h2_init

这会为您提供以下输出向量：

       V1        V2        V3        V4        V5        V6        V7 
0.8034001 0.5583605 0.5487691 0.3255253 0.5685402 0.4643686 0.5227481 
       V8        V9       V10       V11       V12 
0.8050573 0.3474202 0.5564858 0.3496354 0.3783390

我运行在 SPSS 中做了同样的事情并得到了匹配值：

Answer 2

正如您所注意到的，因子分析中的初始公性是每个变量与其余变量的平方多重相关 (SMC)。以 built-in attitude 数据集为例，无需额外的软件包即可轻松计算它们：

1 - 1 / diag(solve(cor(attitude)))

    rating complaints privileges   learning     raises   critical    advance 
 0.7326020  0.7700868  0.3831176  0.6194561  0.6770498  0.1881465  0.5186447

为了方便起见，psych 软件包包含 smc() 函数：

psych::smc(attitude)

    rating complaints privileges   learning     raises   critical    advance 
 0.7326020  0.7700868  0.3831176  0.6194561  0.6770498  0.1881465  0.5186447

如何获得 R 中探索性因素分析的初始共同性？

How can I get the initial communalities for an exploratory factor analysis in R?

r

factor-analysis

数据集

全民教育

获得初始共同点：