在给定我拥有的选择数据的情况下，我正在使用该mlogit包来创建医疗保健提供者选择的嵌套多项式 logit 模型。R数据如下所示：
ID RES weight age wealth educ married urban partnerAge totalChildren survivingChildren anyANC
1.0 2468158 FALSE 0.2609153 29 Poor Primary 1 0 31 4 4 1
1.1 2468158 TRUE 0.2609153 29 Poor Primary 1 0 31 4 4 1
1.2 2468158 FALSE 0.2609153 29 Poor Primary 1 0 31 4 4 1
1.3 2468158 FALSE 0.2609153 29 Poor Primary 1 0 31 4 4 1
2.0 14233860 FALSE 0.2754970 19 Poorest Primary 1 0 30 1 1 1
2.1 14233860 TRUE 0.2754970 19 Poorest Primary 1 0 30 1 1 1
2.2 14233860 FALSE 0.2754970 19 Poorest Primary 1 0 30 1 1 1
2.3 14233860 FALSE 0.2754970 19 Poorest Primary 1 0 30 1 1 1
outlier50Km optout alt spa mes dist bobs cobs Q fees chid educSec
1.0 0 -1 0 Home Home 0.000 0.0000000 0.000 0.00 0 1 0
1.1 0 -1 1 Health center Public 13.167 0.4898990 NA 0.64 0 1 0
1.2 0 -1 2 Health center Public 30.596 0.5202020 NA 0.56 0 1 0
1.3 0 -1 3 District hospital Public 41.164 0.7171717 0.825 0.88 0 1 0
2.0 0 -1 0 Home Home 0.000 0.0000000 0.000 0.00 0 2 0
2.1 0 -1 1 Health center Mission 14.756 0.7676768 NA 0.64 1 2 0
2.2 0 -1 2 Health center Public 41.817 0.3787879 NA 0.56 0 2 0
2.3 0 -1 3 District hospital Public 50.419 0.7171717 0.825 0.88 0 2 0
其中spa, mes, dist, bobs, cobs, Q,和fees是提供者的特征以及特定于个人的其余变量。这些数据是长格式的，这意味着每个人有四行，反映了她的四个选择 ( alt = 0:3)，RES作为响应变量。
未嵌套模型的行为适当
f.full <- RES ~ 0 + dist + Q + bobs + fees + spa | 0 + age + wealth + educSec + married + urban + totalChildren + survivingChildren
choice.ml.full <- mlogit(formula = f.full, data = data, weights = weight)
predict(choice.ml.full, data[1:8,])
0 1 2 3
[1,] 0.1124429 0.7739403 0.06893341 0.04468343
[2,] 0.4465272 0.3107375 0.11490317 0.12783210
然而，通过模型拟合的所有度量，嵌套模型比非嵌套模型更好。嵌套模型适当地给了我系数：
ns2 <- mlogit(formula = f.full, nests = list(home = "0", useCare = c("1", "2", "3")), data = data, weight = weight, un.nest.el = TRUE)
summary(ns2)
Call:
mlogit(formula = f.full, data = data, weights = weight, nests = list(home = "0",
useCare = c("1", "2", "3")), un.nest.el = TRUE)
Frequencies of alternatives:
0 1 2 3
0.094378 0.614216 0.194327 0.097079
bfgs method
23 iterations, 0h:0m:13s
g'(-H)^-1g = 9.51E-07
gradient close to zero
Coefficients :
Estimate Std. Error t-value Pr(>|t|)
dist -0.0336233 0.0040136 -8.3773 < 2.2e-16 ***
Q 0.1780058 0.0768181 2.3172 0.0204907 *
bobs -0.0695695 0.0505795 -1.3754 0.1689925
fees -0.8488132 0.1001928 -8.4718 < 2.2e-16 ***
etc...
但是，如果我尝试对单个人进行预测，则会出现以下错误：
predict(ns2, data[1:4,])
Error in apply(Gl, 1, sum) : dim(X) must have a positive length
如果我尝试对多个人进行预测，则会出现不同的错误：
predict(ns2, data[1:8,])
Error in solve.default(crossprod(attr(x, "gradi")[, !fixed])) :
Lapack routine dgesv: system is exactly singular: U[5,5] = 0
任何帮助将不胜感激。