SensoMineR面板错误在对比中，对比只能适用于2级或2级以上的因素
我试图使用SensoMineR包中的panelperf函数运行以下代码：
panelperf<-panelperf(data,formul="~Product+Panelist+rep+Product:Panelist+Product:rep+Panelist:rep", firstvar=4)
我的数据框架包括:第1栏:小组成员，充满了小组成员的名字，第2栏:产品:填充了产品的名称，第3栏: rep，产品的复制，第4栏，直到被测量的变量(轻水果，黑果，酒精，酸味等)--我所有的变量都是dbl。
但是，在运行该函数时会出现以下错误：Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) : contrasts can be applied only to factors with 2 or more levels In addition: Warning messages: 1: In xtfrm.data.frame(x) : cannot xtfrm data frames 2: In xtfrm.data.frame(x) : cannot xtfrm data frames 3: In xtfrm.data.frame(x) : cannot xtfrm data frames
sapply(lapply(data, unique), length)
Product Panelist rep Lightfruit Darkfruit Applepear Citrus DryFruit Nutty Vegetables Earthy Floral
13 12 2 87 83 72 67 76 69 76 67 66
Woddy hgt Chemical Chocolate Honey Cheesy Alcohol Overallaroma Astringent Sour Hot Viscocity
62 64 80 57 65 69 86 85 88 85 85 83
Sweet Bitter
87 86
So none of the variables has only 2 levels as the error suggests
I have been reading answers about this error, but either the answers dont apply to my case or, as a non very experienced R user, I do not follow what they suggest.
I would appreciate your help a lot!
Thank you!
And let me know if you need more information!回答 1
Stack Overflow用户
发布于 2022-10-08 18:01:33
检查因子级别的代码可以是
i1 <- sapply(data, \(x) is.factor(x) && nlevels(x) > 1)
i2 <- !sapply(data, is.factor)
data1 <- data[i1|i2]下面是一个例子，说明了这个问题？？当数据的nlevel的因子大于1时，它可以工作。
library(SensoMineR)
data(chocolates)
res <- panelperf(sensochoc, firstvar = 5, formul = "~Product+Panelist+
Session+Product:Panelist+Session:Product+Panelist:Session")
> str(res)
List of 4
$ p.value : num [1:14, 1:6] 8.85e-14 6.44e-08 1.75e-28 3.74e-40 1.18e-22 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : chr [1:14] "CocoaA" "MilkA" "CocoaF" "MilkF" ...
.. ..$ : chr [1:6] "Product " "Panelist " "Session " "Product:Panelist" ...
$ variability: num [1:14, 1:6] 0.139 0.103 0.397 0.532 0.267 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : chr [1:14] "CocoaA" "MilkA" "CocoaF" "MilkF" ...
.. ..$ : chr [1:6] "Product " "Panelist " "Session " "Product:Panelist" ...
$ res : num [1:14, 1] 1.87 1.89 1.41 1.47 1.63 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : chr [1:14] "CocoaA" "MilkA" "CocoaF" "MilkF" ...
.. ..$ : chr "stdev residual"
$ r2 : num [1:14, 1] 0.673 0.761 0.846 0.882 0.862 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : chr [1:14] "CocoaA" "MilkA" "CocoaF" "MilkF" ...
.. ..$ : chr "r2"
> str(sensochoc)
'data.frame': 348 obs. of 18 variables:
$ Panelist : Factor w/ 29 levels "1","2","3","4",..: 1 1 1 1 1 1 2 2 2 2 ...
$ Session : Factor w/ 2 levels "1","2": 1 1 1 1 1 1 1 1 1 1 ...
$ Rank : Factor w/ 6 levels "1","2","3","4",..: 1 6 3 5 2 4 1 4 3 5 ...
$ Product : Factor w/ 6 levels "choc1","choc2",..: 6 3 2 1 4 5 4 3 6 2 ...
$ CocoaA : int 7 6 8 7 8 7 6 4 5 5 ...
$ MilkA : int 6 7 6 8 5 5 1 2 1 2 ...
$ CocoaF : int 6 2 5 8 4 3 8 3 8 8 ...
$ MilkF : int 5 7 4 3 4 5 1 4 1 1 ...
$ Caramel : int 5 8 7 3 4 6 0 0 0 0 ...
$ Vanilla : int 3 4 4 2 4 2 0 0 0 0 ...
$ Sweetness : int 7 7 5 4 5 5 1 5 1 0 ...
$ Acidity : int 2 2 5 7 6 4 0 0 0 0 ...
$ Bitterness : int 4 2 6 8 6 7 3 0 3 6 ...
$ Astringency: int 5 3 6 6 4 4 0 0 0 0 ...
$ Crunchy : int 8 3 7 3 6 6 8 4 6 8 ...
$ Melting : int 3 8 5 2 3 6 5 8 2 2 ...
$ Sticky : int 4 6 4 3 7 4 0 3 1 4 ...
$ Granular : int 3 5 3 5 3 7 0 1 1 0 ...如果我们将“会话”级别2更改为NA (只有2个级别)，则会显示错误。
levels(sensochoc$Session)[2] <- NA
res1 <- panelperf(sensochoc, firstvar = 5, formul = "~Product+Panelist+
Session+Product:Panelist+Session:Product+Panelist:Session")
Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) :
contrasts can be applied only to factors with 2 or more levels 使用OP的代码，它仍然显示数据具有大于1的级别，因为如果存在，unique也会返回NA，因此length将包含NA元素，这里总共是2
> sapply(lapply(sensochoc, unique), length)
Panelist Session Rank Product CocoaA MilkA CocoaF MilkF Caramel Vanilla Sweetness Acidity
29 2 6 6 11 11 11 11 11 10 11 11
Bitterness Astringency Crunchy Melting Sticky Granular
11 11 11 11 11 11 与本文中的特定代码一样，nlevels只删除非NA级别的计数返回NA。
i1 <- sapply(sensochoc, \(x) is.factor(x) && nlevels(x) > 1)
i2 <- !sapply(sensochoc, is.factor)
names(sensochoc)[i1]
[1] "Panelist" "Rank" "Product"
names(sensochoc)[sapply(sensochoc, is.factor)]
[1] "Panelist" "Session" "Rank" "Product" 省略了Session。我们可能需要修改公式，以省略包含Session的条件
https://stackoverflow.com/questions/73999342
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