编辑
因此，在我的呼吁中，这似乎library(reshape)打破了因素的标签。这未包含在最小示例中，但现在将添加。不需要创建示例，但需要重新创建问题。我需要图书馆来整理我的数据，甚至可以做到factanal()。任何想法重塑的哪一部分正在破坏它以及如何修复它？
原始问题
我一直在对我的数据进行因子分析，并且在打印结果的方式上遇到了间歇性问题。
如果我创建如下数据集：
library(reshape)
mock <- data.frame(
sample_name1 = sample(1:100),
sample_name2 = sample(1:100),
sample_name3 = sample(1:100),
s_amplename_4 = sample(1:100),
samplename5 = sample(1:100),
sa_mplen_a_me_6 = sample(1:100),
samplename7 = sample(1:100),
samplename8 = sample(1:100)
)
并使用
factanal(mock, factors = 2)
我得到输出以非常漂亮的方式打印出来，其中项目名称作为行的标签，例如：
# Snip snip
Loadings:
Factor1 Factor2
sample_name1 -0.126 -0.105
sample_name2 -0.414
sample_name3 0.665
s_amplename_4 -0.314
samplename5 0.850
sa_mplen_a_me_6 -0.117
samplename7 0.442
samplename8 -0.139
这种输出正是我想要的。但是，当我对自己的数据进行相同类型的分析时（我为这里的长度道歉）：
miniset <- structure(list(`clarity1` = c(2, 2, 2, 3, 4.5, 1.5, 1.5, 3.5,
2, 6, 2.5, 4, 1, 1.5, 6, 2, 5.5, 2, 2, 3, 1.5, 5, 3.5, 2, 1.5,
2.5, 3, 3, 2, 1),
`clarity2` = c(1.5, 2, 2, 2, 3.5, 5, 3, 5,
2, 4, 2, 2.5, 1, 1.5, 2, 4, 5, 2, 2, 3.5, 6, 1, 2, 1.5, 1, 2,
2, 3, 6.5, 1),
`clarity3` = c(3, 3.5, 2, 3.5, 5.5, 4, 6, 5.5,
2, 3, 3, 3.5, 1, 2.5, 2, 5, 5, 5, 2, 6.5, 5.5, 5, 5.5, 6, 3,
2, 2, 5, 4.5, 5.5),
`detail1` = c(3, 4, 2, 6, 5, 6.5, 5.5,
4, 3, 6, 2.5, 4, 1, 4, 2, 4.5, 7, 6.5, 2, 6.5, 6, 2, 6, 5, 2.5,
5.5, 4, 5.5, 6, 1.5),
`detail2` = c(3.5, 4, 4, 6.5, 4.5, 6,
4, 4.5, 2, 6, 2.5, 5, 2, 4, 3, 6, 7, 7, 2, 6.5, 6, 3, 6, 6, 2.5,
6, 3, 5, 6.5, 2.5),
`detail3` = c(2.5, 4, 2, 6, 5, 6, 6, 4,
2, 6, 2, 5, 2, 3, 3, 5, 6.5, 6, 2, 6.5, 7, 7, 5.5, 5, 3.5, 2,
3, 5, 6, 2),
`complete1` = c(2, 2.5, 2, 3, 3.5, 5.5, 2.5, 2.5,
2, 3, 3, 3.5, 2, 4, 3, 3, 7, 4, 2, 3, 6, 3, 5.5, 2, 3, 2, 2,
3, 6, 3),
`complete2` = c(3, 4.5, 2, 3, 4.5, 6, 6, 4.5, 3,
3, 3.5, 4, 2, 5, 3, 4, 7, 4, 2, 6, 7, 5, 5, 6, 3, 3, 5, 5, 6,
2),
`complete3` = c(3, 4.5, 2, 2.5, 4.5, 6.5, 5, 5, 2, 6.5,
3.5, 3.5, 1, 3, 3, 2.5, 7, 4, 2, 6, 1.5, 7, 5.5, 6.5, 3.5, 5.5,
3, 3, 2.5, 1),
`truthful1` = c(2.5, 2, 2, 3, 3.5, 2, 2, 2.5,
2, 3, 3, 2.5, 2, 3, 2, 2, 3.5, 3, 2, 3.5, 1.5, 1, 3.5, 2.5, 3,
2, 2, 3, 1.5, 1.5),
`truthful2` = c(2.5, 1.5, 2, 2, 3, 1.5,
2, 1, 1, 5.5, 3, 3.5, 1, 4.5, 2, 2, 5, 2, 2, 1.5, 4.5, 1, 3.5,
2, 3.5, 2.5, 2, 2, 4.5, 1),
`truthful3` = c(2, 1.5, 2, 3.5,
2.5, 2, 2, 2.5, 2, 2, 3.5, 2.5, 1, 1.5, 3, 2, 5, 3, 3, 2, 3.5,
1, 2, 1, 3.5, 2, 2, 2.5, 4.5, 1),
`relevant1` = c(1.5, 1.5,
2, 5, 2.5, 1.5, 2, 3.5, 2, 4.5, 2.5, 3.5, 1, 3.5, 3, 1.5, 5.5,
3.5, 2, 2, 6, 3, 3.5, 3, 1.5, 2, 3, 3, 6, 1),
`relevant2` = c(1.5,
3, 2, 2, 3.5, 1.5, 2.5, 5.5, 1, 2, 3.5, 2, 1, 1.5, 2, 4, 5.5,
2, 3, 5.5, 5.5, 1, 4, 5, 1.5, 2, 3, 2.5, 3, 1),
`relevant3` = c(1.5,
2, 2, 3, 2, 1, 2, 2, 1, 2, 1.5, 2.5, 1, 1.5, 2, 1.5, 5.5, 5,
2, 1, 7, 1, 1, 2, 1, 2, 3, 3, 2.5, 1)),
.Names = c("clarity1",
"clarity2", "clarity3", "detail1", "detail2", "detail3",
"complete1", "complete2", "complete3", "truthful1", "truthful2",
"truthful3", "relevant1", "relevant2", "relevant3"),
row.names = c(NA, 30L), class = c("cast_df", "data.frame"))
factanal(miniset, factors = 3)
结果不那么漂亮，例如：
Loadings:
Factor1 Factor2 Factor3
[1,] 0.222 0.664
[2,] 0.559 0.524
[3,] 0.824
[4,] 0.740 0.361 0.282
[5,] 0.698 0.374 0.251
[6,] 0.783 0.278 0.265
[7,] 0.498 0.598 0.140
[8,] 0.796 0.227 0.204
[9,] 0.490 -0.240 0.835
[10,] 0.147 0.156 0.348
[11,] 0.697 0.324
[12,] 0.756
[13,] 0.319 0.811 0.204
[14,] 0.567 0.252 0.108
[15,] 0.320 0.690
因此，我现在没有将漂亮的项目名称作为加载的标签，而是得到索引。虽然这对我来说很好，但我明天将与一位不太熟悉 R 并且可能会因缺乏标签而感到沮丧的教授一起工作。那么在第二种情况下标签会发生什么？我怎样才能让他们回来？