在R中使用Psych库运行Omega

当我在一个构造上运行

alpha

时，我在构造上有五个项目，我得到了以下结果，没有任何错误

 psych::alpha(construct,
         na.rm = TRUE,
         title = 'myscale', 
         n.iter = 1000)

Reliability analysis  myscale  
Call: psych::alpha(x = construct, title = "myscale", na.rm = TRUE, 
n.iter = 1000)

  raw_alpha std.alpha G6(smc) average_r S/N   ase mean   sd
  0.81      0.81    0.78      0.46 4.3 0.013  2.6 0.89

 lower alpha upper     95% confidence boundaries
0.78 0.81 0.84 

 lower median upper bootstrapped confidence intervals
 0.77 0.81 0.84

我一直在阅读《从阿尔法到欧米茄：解决普遍存在的内部一致性估计问题的实用方法》一文

它建议使用下面的代码

MBESS::ci.reliability(construct, interval.type="bca", B=1000, type = "omega") 

$est
[1] 0.8107376

$se
[1] 0.01651936

$ci.lower
[1] 0.7764029

$ci.upper
[1] 0.839944

$conf.level
[1] 0.95

$type
[1] "omega"

$interval.type
[1] "bca bootstrap"

我一直在尝试使用psych软件包在我的样本集上运行omega，以保持分析中的一致性

psych::omega(m = construct, 
      nfactors = 1, fm = "pa", n.iter = 1000, p = 0.05, 
      title = "Omega", plot = FALSE, n.obs = 506)

我收到两条错误消息

在factor.scores中，相关矩阵是奇异的，使用近似值 1因子的ωh没有意义，只有ωt

出现此警告是因为Omega_h的列数较少。上一个关于SO的问题在某种程度上回答了这个问题

我的错误如下

fac中的错误（r=r，nfactors=nfactors，n.obs=n.obs，rotate=rotate，： 很抱歉：相关性矩阵中的缺失值（NAs）不允许我继续。 请删除这些变量，然后重试。 此外：有50个或更多警告（使用warnings（）查看前50个）

没有缺失值，所以我不确定第二个错误

我构造的细节如下所示

    Q1                  Q2          Q3    
 Min.   :0.000   Min.   :0.000   Min.   :0.000  
 1st Qu.:2.000   1st Qu.:2.000   1st Qu.:2.000  
 Median :3.000   Median :2.000   Median :3.000  
 Mean   :2.597   Mean   :2.393   Mean   :3.227  
 3rd Qu.:3.000   3rd Qu.:3.000   3rd Qu.:4.000  
 Max.   :6.000   Max.   :6.000   Max.   :6.000  

Q4              Q5   
 Min.   :0.00   Min.   :0.000  
 1st Qu.:1.00   1st Qu.:2.000  
 Median :2.00   Median :2.000  
 Mean   :2.17   Mean   :2.445  
 3rd Qu.:3.00   3rd Qu.:3.000  
 Max.   :6.00   Max.   :6.000  

编辑 创建了具有相同属性的数据—100个条目（Alpha大约为0.56），但它在omega上生成了相同的错误

structure(list(Q1 = c(4, 5, 3, 5, 4, 5, 3, 5, 5, 5, 6, 
3, 5, 4, 6, 5, 5, 6, 7, 4, 5, 5, 3, 4, 4, 5, 4, 3, 5, 4, 5, 5, 
6, 6, 3, 6, 3, 4, 4, 4, 6, 5, 3, 2, 6, 6, 4, 5, 4, 3, 6, 4, 4, 
5, 6, 2, 4, 3, 4, 6, 4, 6, 4, 5, 5, 6, 4, 6, 5, 5, 4, 5, 6, 6, 
2, 5, 4, 3, 4, 4, 4, 6, 3, 3, 5, 4, 4, 4, 5, 5, 5, 3, 6, 6, 6, 
6, 5, 4, 3, 5), Q2 = c(7, 4, 4, 4, 4, 6, 6, 6, 7, 6, 5, 
6, 5, 4, 5, 6, 6, 6, 7, 5, 4, 4, 6, 6, 4, 4, 6, 2, 6, 5, 4, 6, 
4, 6, 6, 6, 5, 4, 4, 4, 4, 3, 3, 4, 4, 4, 4, 6, 2, 6, 6, 5, 4, 
6, 6, 4, 4, 7, 6, 5, 5, 5, 5, 6, 5, 5, 4, 5, 5, 5, 4, 6, 7, 5, 
5, 5, 6, 5, 6, 5, 6, 7, 2, 6, 5, 7, 3, 5, 5, 3, 3, 3, 7, 4, 5, 
6, 6, 6, 5, 7), Q3 = c(5, 4, 5, 6, 4, 4, 5, 4, 2, 6, 5, 
5, 5, 5, 7, 5, 5, 6, 7, 6, 3, 6, 6, 6, 5, 6, 6, 5, 5, 4, 5, 5, 
6, 6, 5, 6, 5, 5, 4, 4, 6, 4, 4, 4, 4, 4, 4, 5, 5, 4, 5, 5, 4, 
3, 5, 4, 5, 6, 6, 6, 4, 5, 5, 5, 6, 4, 5, 5, 7, 4, 5, 6, 6, 5, 
5, 3, 3, 5, 4, 6, 5, 5, 1, 3, 5, 3, 2, 5, 4, 6, 6, 6, 6, 4, 6, 
3, 6, 6, 6, 5), Q4 = c(6, 6, 4, 7, 4, 6, 7, 6, 7, 6, 6, 
6, 5, 7, 7, 6, 6, 5, 7, 7, 6, 6, 7, 7, 6, 6, 6, 5, 6, 7, 5, 6, 
7, 5, 4, 6, 4, 3, 6, 4, 6, 6, 6, 3, 5, 7, 5, 6, 4, 6, 7, 6, 7, 
4, 6, 3, 5, 7, 5, 4, 6, 6, 4, 6, 5, 5, 5, 5, 7, 7, 7, 6, 6, 6, 
5, 6, 6, 4, 5, 7, 6, 7, 3, 5, 6, 5, 6, 5, 5, 7, 7, 6, 6, 2, 7, 
6, 6, 7, 7, 5)), .Names = c("Q1", "Q2", "Q3", 
"Q4"), row.names = c(NA, 100L), class = "data.frame")

有人能看到我跌倒的地方吗

感谢您抽出时间

因此我尝试了以下方法：

psych::omega(m = construct)

结果是：

Omega 
Call: psych::omega(m = construct)
Alpha:                 0.56 
G.6:                   0.49 
Omega Hierarchical:    0.53 
Omega H asymptotic:    0.89 
Omega Total            0.6 

Schmid Leiman Factor loadings greater than  0.2 
     g   F1*   F2*   F3*   h2   u2   p2
Q1 0.41  0.30             0.26 0.74 0.65
Q2 0.37  0.25             0.20 0.80 0.67
Q3 0.50        0.25       0.31 0.69 0.80
Q4 0.64              0.23 0.46 0.54 0.89

With eigenvalues of:
    g  F1*  F2*  F3* 
 0.95 0.15 0.06 0.05 

 general/max  6.35   max/min =   2.83
mean percent general =  0.75    with sd =  0.11 and cv of  0.15 
Explained Common Variance of the general factor =  0.78 

The degrees of freedom are -3  and the fit is  0 
The number of observations was  100  with Chi Square =  0  with prob <  NA
The root mean square of the residuals is  0 
The df corrected root mean square of the residuals is  NA

Compare this with the adequacy of just a general factor and no group factors
The degrees of freedom for just the general factor are 2  and the fit is  0.01 
The number of observations was  100  with Chi Square =  0.62  with prob <  0.73
The root mean square of the residuals is  0.03 
The df corrected root mean square of the residuals is  0.05 

RMSEA index =  0  and the 90 % confidence intervals are  NA 0.14
BIC =  -8.59 

Measures of factor score adequacy             
                                                 g   F1*   F2*   F3*
Correlation of scores with factors            0.75  0.37  0.27  0.24
Multiple R square of scores with factors      0.57  0.14  0.07  0.06
Minimum correlation of factor score estimates 0.14 -0.72 -0.86 -0.88

 Total, General and Subset omega for each subset
                                                 g  F1*  F2*  F3*
Omega total for total scores and subscales    0.60 0.37 0.31 0.46
Omega general for total scores and subscales  0.53 0.25 0.25 0.41
Omega group for total scores and subscales    0.06 0.12 0.06 0.05

---

使用完整的数据集，您应该能够获得更高的n.iter

，而无需测试数据集，很难知道此错误的原因。无论如何，您是否尝试过在

omega

中更改

fm

参数？此外，尝试为您的构造运行

psych:：fa

。您好@John Smith，请告诉我是否有什么需要修改的内容关于我发给你的邮件，我真的很想帮你嗨@Derek Corcoran，对耽搁表示歉意，本周我们的工作忙得不可开交，所以我没有机会玩弄它。再次感谢你的回答。我今天和明天都会玩弄它，让你知道我是怎么做的。我想我的主要问题是我掌握的数据是desi我想加载一个结构。所以可能只是因为欧米茄不合适，或者正如你所指出的，我需要更多的因素。再次感谢你的帮助。我真的很感激

psych::omega(m = construct, n.iter = 7, p = 0.05, nfactors = 3)