編輯打印因素分析標籤
所以它看起來像它的東西,在我呼籲library(reshape)
是打碎的因素的標籤。這不包括在最小的例子中,但現在將被添加。不需要創建該示例,但需要重新創建該問題。我需要圖書館來獲得我的數據,甚至做到factanal()
。任何想法重塑的部分是打破它,以及如何解決它?
原來的問題
我已經運行因素分析了我的數據,並一直有一個間歇性問題與結果的打印方式。
如果我創建一個數據集如下所示:
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)
)
並運行
factanal(mock, factors = 2)
因素分析我得到的輸出爲標籤的使用項目名稱很嬌滴滴打印出來行,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
這種輸出正是我正在尋找的。然而,當我在我自己的數據運行相同類型的分析(我爲長度在這裏表示歉意):
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)
結果是更漂亮,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,而不是有很好的項目名稱作爲加載標籤,我現在得到索引。雖然這對我來說很好,但我會和明天不熟悉R的教授合作,並且可能會因缺乏標籤而感到沮喪。那麼第二種情況下的標籤會發生什麼?我怎樣才能讓他們回來?
我不能重現此。當我運行你的代碼,輸出與標籤和所有內容完全「相當」。 – joran
@joran我進一步麻煩了這個問題並相應地更新了這個問題。 – Ryan