r - 如何在没有模型对象的情况下根据 ns 样条参数进行预测

Question

我有 R 中拟合的 glm 的系数，我想预测一组新数据的预期值。如果我有模型对象，这会很简单，使用predict()。但是，我现在不在现场，出于数据保密原因，我不再拥有模型对象。我只有使用summary(model) 生成的摘要对象，其中包含模型系数。

使用系数来预测简单模型的期望值非常容易。但是，我想知道当模型包含三次样条 ns() 时如何执行此操作。当模型还包含分类变量时，任何快捷方式也将受到赞赏。

这是一个简单的例子。

library(splines)
dat <- data.frame(x=1:500, z=runif(500), k=as.factor(sample(c("a","b"), size=500, replace=TRUE)))
kvals <- data.frame(kn=c("a","b"),kv=c(20,30))
dat$y = dat$x + (40*dat$z)^2 + kvals$kv[match(dat$k,kvals$kn)] + rnorm(500,0,30)
# Fit model
library(splines)
mod <- glm(y ~ x + ns(z,df=2) + k,data=dat)
# Create new dataset
dat.new <- expand.grid(x=1:3,z=seq(0.2,0.4,0.1),k="b")
# Predict expected values in the usual way
predict(mod,newdata=dat.new)
summ <- summary(mod)
rm(mod)
# Now, how do I predict using just the summary object and dat.new?

Answer 1

可能有一种更有效的方法来解决这个问题，但这里是一个起点，可以帮助您实现罗兰简要建议的策略。 summ 对象具有定义样条函数所需的信息，但它是隐藏的:

    names(summ)
     [1] "call"           "terms"          "family"         "deviance"       "aic"           
     [6] "contrasts"      "df.residual"    "null.deviance"  "df.null"        "iter"          
    [11] "deviance.resid" "coefficients"   "aliased"        "dispersion"     "df"            
    [16] "cov.unscaled"   "cov.scaled"    

查看 terms 叶的结构，我们看到样条细节埋藏在 predvars 子叶内更深处:

str(summ$terms)
Classes 'terms', 'formula'  language y ~ x + ns(z, df = 2) + k
  ..- attr(*, "variables")= language list(y, x, ns(z, df = 2), k)
  ..- attr(*, "factors")= int [1:4, 1:3] 0 1 0 0 0 0 1 0 0 0 ...
  .. ..- attr(*, "dimnames")=List of 2
  .. .. ..$ : chr [1:4] "y" "x" "ns(z, df = 2)" "k"
  .. .. ..$ : chr [1:3] "x" "ns(z, df = 2)" "k"
  ..- attr(*, "term.labels")= chr [1:3] "x" "ns(z, df = 2)" "k"
  ..- attr(*, "order")= int [1:3] 1 1 1
  ..- attr(*, "intercept")= int 1
  ..- attr(*, "response")= int 1
  ..- attr(*, ".Environment")=<environment: R_GlobalEnv> 
  ..- attr(*, "predvars")= language list(y, x, ns(z, knots = structure(0.514993450604379, .Names = "50%"), Boundary.knots = c(0.00118412892334163,  0.99828373757191), intercept = FALSE), k)
  ..- attr(*, "dataClasses")= Named chr [1:4] "numeric" "numeric" "nmatrix.2" "factor"
  .. ..- attr(*, "names")= chr [1:4] "y" "x" "ns(z, df = 2)" "k"

所以拉出属性:

str(attributes(summ$terms)$predvars)
 language list(y, x, ns(z, knots = structure(0.514993450604379, .Names = "50%"),
               Boundary.knots = c(0.00118412892334163,  0.99828373757191), intercept = FALSE), k)

您可以看到，如果提供所需的 x、y、z 和 k 值，则可以恢复样条线:

with(dat, ns(z, knots = 0.514993450604379, Boundary.knots = c(0.00118412892334163, 
 0.99828373757191), intercept = FALSE) )
#---
                 1             2
  [1,] 5.760419e-01 -1.752762e-01
  [2,] 2.467001e-01 -1.598936e-01
  [3,] 4.392684e-01  4.799757e-01
snipping ....
[498,] 4.965628e-01 -2.576437e-01
[499,] 5.627389e-01  1.738909e-02
[500,] 2.393920e-02 -1.611872e-02
attr(,"degree")
[1] 3
attr(,"knots")
[1] 0.5149935
attr(,"Boundary.knots")
[1] 0.001184129 0.998283738
attr(,"intercept")
[1] FALSE
attr(,"class")
[1] "ns"     "basis"  "matrix"

如果您知道数据的极端情况，您可以构建替代的dat。请参阅 ?ns 及其链接到的其他帮助页面。