我正在尝试探索性因素分析后计算Ω估计值，以估计所找到组件的可靠性。使用omega()包中的psych函数，我得到以下输出：
Alpha: 0.8
G.6: 0.86
Omega Hierarchical: 0.37
Omega H asymptotic: 0.43
Omega Total 0.86
Schmid Leiman Factor loadings greater than
0.2
g F1* F2* F3* h2 u2 p2
EMS1 0.30 0.71 0.59 0.41 0.15
EMS3 -0.21 0.64 0.53 0.47 0.05
EMS4 0.62 0.41 0.59 0.04
EMS7 0.34 0.62 0.50 0.50 0.23
EMS8 0.36 0.42 0.32 0.68 0.40
EMS9 0.57 0.33 0.67 0.00
EMS10 0.39 0.20 0.80 0.11
EMS11 0.72 0.51 0.49 0.02
EMS12 0.68 0.46 0.54 0.02
EMS15 0.54 -0.24 0.41 0.59 0.02
EMS16 0.22 0.77 0.63 0.37 0.08
EMS19 0.65 0.52 0.48 0.01
EMS20 0.27 0.53 0.36 0.64 0.21
EMS21 0.62 0.40 0.60 0.04
EMS23 0.63 0.42 0.58 0.07
EMS24 0.68 0.45 0.55 1.02
EMS25 0.73 0.56 0.44 0.95
EMS27 0.45 0.20 0.25 0.75 0.83
EMS28 0.78 0.59 0.41 1.02
EMS34 0.26 0.31 0.48 0.34 0.66 0.20
With eigenvalues of:
g F1* F2* F3*
2.5 3.4 2.9 0.0
general/max 0.73 max/min = Inf
mean percent general = 0.27 with sd = 0.36 and cv of 1.33
Explained Common Variance of the general factor = 0.28
The degrees of freedom are 133 and the fit is 0.8
The number of observations was 601 with Chi Square = 471.81 with prob < 1.9e-39
The root mean square of the residuals is 0.04
The df corrected root mean square of the residuals is 0.05
RMSEA index = 0.066 and the 10 % confidence intervals are 0.059 0.072
BIC = -379.21
Compare this with the adequacy of just a general factor and no group factors
The degrees of freedom for just the general factor are 170 and the fit is 5.4
The number of observations was 601 with Chi Square = 3195.63 with prob < 0
The root mean square of the residuals is 0.22
The df corrected root mean square of the residuals is 0.24
RMSEA index = 0.173 and the 10 % confidence intervals are 0.167 0.177
BIC = 2107.87
Measures of factor score adequacy
g F1* F2* F3*
Correlation of scores with factors 0.9 0.94 0.93 0
Multiple R square of scores with factors 0.8 0.89 0.86 0
Minimum correlation of factor score estimates 0.6 0.78 0.73 -1
Total, General and Subset omega for each subset
g F1* F2* F3*
Omega total for total scores and subscales 0.86 0.82 0.85 NA
Omega general for total scores and subscales 0.37 0.08 0.34 NA
Omega group for total scores and subscales 0.58 0.75 0.51 NA
Warning messages:
1: In fac(r = r, nfactors = nfactors, n.obs = n.obs, rotate = rotate, :
A loading greater than abs(1) was detected. Examine the loadings carefully.
2: In fac(r = r, nfactors = nfactors, n.obs = n.obs, rotate = rotate, :
An ultra-Heywood case was detected. Examine the results carefully
3: In cov2cor(t(w) %*% r %*% w) :
diag(.) had 0 or NA entries; non-finite result is doubtful
这就是我调用该函数的方式：
omega(df[,items],nfactors=3)
在寻找指导之后，我找不到为什么没有为第三个因素计算欧米茄的原因。我不确定是否与警告消息之一有关的问题：
Warning messages:
1: In fac(r = r, nfactors = nfactors, n.obs = n.obs, rotate = rotate, :
A loading greater than abs(1) was detected. Examine the loadings carefully.
2: In fac(r = r, nfactors = nfactors, n.obs = n.obs, rotate = rotate, :
An ultra-Heywood case was detected. Examine the results carefully
3: In cov2cor(t(w) %*% r %*% w) :
diag(.) had 0 or NA entries; non-finite result is doubtful
答案 0 :(得分：0)
这可能是因为Omega是通过拟合CFA模型计算得出的，并且在您的情况下有3个因子的情况下，出于识别原因使用了因子3。因此，您不会期望为此而计算出欧米茄