R误差中联合分析的ChoiceModelR包：估计Beta

我进行了一项联合研究，现在尝试使用R软件包来估算效用。我已经阅读了软件包文档，并查看了有关StackOverflow（和）的类似答案/问题，但得到了下面的错误消息

我的数据文件（您可以从中下载的.csv具有所需的格式（UnitID Set Alt X_1…X_natts y）是离散的。在每个选项集中，我有2个可选项和“无”选项。我已从数据文件中排除了带有“无”选项的行。对于“无”选项集，选项号设置为“可选项数+1”，即3

据我所知，该数据是必需的格式。但是，我无法估计实用程序并得到以下错误：

Error: invalid values of y present in data - values must be 1 to 3

我的输入如下所示：

id    set_number card_number att1 att2 att3 att4 dv
1 932          1           1    1    1    1    3  2
2 932          1           2    2    2    4    4  0
4 932          2           1    3    3    3    1  2
5 932          2           2    4    2    2    4  0
7 932          3           1    5    4    1    3  2
8 932          3           2    6    3    3    2  0
...

我的代码如下所示

library(ChoiceModelR)

c_data <- read.csv("c_data.csv")

# Set parameter for calculation
# R is the total number of iterations of the Markov Chain Monte Carlo to be performed
# use is the number of iterations to be used in parameter estimation 

mcmc = list(R = 4000, use = 2000)

# Set parameter of datainput
# xcoding specifies the way in which each attribute will be coded (0 = categorical, 1 = nominal)
# none = TRUE to estimate a none parameter, and the data does not include a row for “none” (i.e., no choice)
# keep is the thinning parameter that defines the number of random draws to save

xcoding = c(0, 0, 0, 0)
options = list(none = TRUE, keep = 5)

# Putting it together to get betas
choicemodelr(data = c_data, xcoding = xcoding, mcmc = mcmc, options = options)

20个带dput的顶行：

structure(list(id = c(932L, 932L, 932L, 932L, 932L, 932L, 932L, 
932L, 932L, 932L, 932L, 932L, 932L, 932L, 932L, 932L, 933L, 933L, 
933L, 933L), set_number = c(1L, 1L, 2L, 2L, 3L, 3L, 4L, 4L, 5L, 
5L, 6L, 6L, 7L, 7L, 8L, 8L, 1L, 1L, 2L, 2L), card_number = c(1L, 
2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 
2L, 1L, 2L), att1 = c(1L, 2L, 3L, 4L, 5L, 6L, 3L, 4L, 5L, 2L, 
3L, 4L, 2L, 6L, 6L, 4L, 3L, 5L, 6L, 2L), att2 = c(1L, 2L, 3L, 
2L, 4L, 3L, 1L, 3L, 1L, 2L, 1L, 4L, 4L, 3L, 4L, 3L, 3L, 1L, 2L, 
4L), att3 = c(1L, 4L, 3L, 2L, 1L, 3L, 2L, 4L, 1L, 2L, 3L, 4L, 
1L, 4L, 2L, 3L, 1L, 4L, 3L, 4L), att4 = c(3L, 4L, 1L, 4L, 3L, 
2L, 2L, 1L, 3L, 4L, 2L, 4L, 2L, 3L, 2L, 3L, 4L, 2L, 3L, 1L), 
    dv = c(2, 0, 2, 0, 2, 0, 2, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 
    0, 1, 0)), row.names = c(NA, -20L), class = "data.frame")

如果您能给我提些建议，我将不胜感激

structure(list(id = c(932L, 932L, 932L, 932L, 932L, 932L, 932L, 
932L, 932L, 932L, 932L, 932L, 932L, 932L, 932L, 932L, 933L, 933L, 
933L, 933L), set_number = c(1L, 1L, 2L, 2L, 3L, 3L, 4L, 4L, 5L, 
5L, 6L, 6L, 7L, 7L, 8L, 8L, 1L, 1L, 2L, 2L), card_number = c(1L, 
2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 
2L, 1L, 2L), att1 = c(1L, 2L, 3L, 4L, 5L, 6L, 3L, 4L, 5L, 2L, 
3L, 4L, 2L, 6L, 6L, 4L, 3L, 5L, 6L, 2L), att2 = c(1L, 2L, 3L, 
2L, 4L, 3L, 1L, 3L, 1L, 2L, 1L, 4L, 4L, 3L, 4L, 3L, 3L, 1L, 2L, 
4L), att3 = c(1L, 4L, 3L, 2L, 1L, 3L, 2L, 4L, 1L, 2L, 3L, 4L, 
1L, 4L, 2L, 3L, 1L, 4L, 3L, 4L), att4 = c(3L, 4L, 1L, 4L, 3L, 
2L, 2L, 1L, 3L, 4L, 2L, 4L, 2L, 3L, 2L, 3L, 4L, 2L, 3L, 1L), 
    dv = c(2, 0, 2, 0, 2, 0, 2, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 
    0, 1, 0)), row.names = c(NA, -20L), class = "data.frame")