R中预测函数和零膨胀负二项式模型的误差。

Question

我已经得到了一个 Rdata 文件，其中包含一个回归模型的大量输入和输出。 我已经能够提取模型分析的数据，并重现参数估计。 然而，当我试图使用原始的 predict 语句，我收到一个错误，即使 predict 语句在应用于存储在 Rdata 文件。

我希望有足够的信息呈现在下面，有人能告诉我如何纠正我的预测声明。my.probs尽管我没有提供一个功能上可重复的例子，但我想这是我第一次在这里发表问题而没有提供这样的例子。 我想，这是我第一次在没有提供这样一个例子的情况下在这里发问。 这个数据集包含了100,000个观测值，有点敏感，我不知道如何重现这些数据。Rdata 文件。

library(msm)
library(MASS)
library(pscl)

# model output returned when extracting the model name from the `Rdata` file
original.model
# Call:
# zeroinfl(formula = AA ~ log(BB) + CC + DD + CC:DD | log(BB) + DD, data = original.data, 
#     offset = log(EE), dist = "negbin")
# 
# Count model coefficients (negbin with log link):
#           (Intercept)  log(BB)      CC3      CC4      CC5 DDPrivate CC3:DDPrivate CC4:DDPrivate CC5:DDPrivate  
#              -2.05317  0.31178 -0.41402 -0.71208 -0.92290   0.17878      -0.18476      -0.18674       0.07307  
# Theta = 0.8551 
# 
# Zero-inflation model coefficients (binomial with logit link):
#           (Intercept)  log(BB)        DDPrivate  
#                1.6724 -0.5022            0.9742  
#  
# Warning message:
# In deparse(x$call, width.cutoff = floor(getOption("width") * 0.85)) :
#   invalid 'cutoff' value for 'deparse', using default

# data for new observation for use in the predict statement
new.data
#         DD      EE   CC               BB
# 1  Private       1    4         1118.948

str(new.data)
#'data.frame': 1 obs. of  4 variables:
# $ DD       : Factor w/ 2 levels "Public","Private": 2
# $ EE       : num 1
# $ CC       : Factor w/ 4 levels "2","3","4","5": 3
# $ BB: num 1119

original.probs <- predict(original.model, new.data, type='prob')
original.probs
# truncated probabilities returned by the predict statement.  These sum to one if vector not truncated
c(0.7534319,    0.1552296,    0.05681916,   0.02133936,   0.008116065,  0.003110019,  0.001197667)

# reproduce the original model
my.version <- zeroinfl(formula = AA ~ log(BB) + CC + DD + CC:DD | log(BB) + DD, offset = log(EE), dist = "negbin")

# Error returned by the predict statement
my.probs <- predict(my.version, new.data, type='prob')
my.probs
# Error in exp(X %*% object$coefficients$count + offsetx)[, 1] : 
#   incorrect number of dimensions
# In addition: Warning message:
# In X %*% object$coefficients$count + offsetx :
#   Recycling array of length 1 in array-vector arithmetic is deprecated.
#   Use c() or as.vector() instead.

Answer 1

该 predict 函数在我将输入变量分组为 data.frame 并包括 data 选项中的 zeroinfl 模型声明。

my.data <- data.frame(AA = AA,
                      BB = BB,
                      CC = CC,
                      DD = DD,
                      EE = EE)

my.version <- zeroinfl(formula = AA ~ log(BB) + CC + DD + CC:DD | log(BB) + DD, 
                               offset = log(EE), dist = "negbin", data = my.data)

summary(my.version)

my.probs <- predict(my.version, new.data, type='prob')
my.probs