2

我正在使用gam模型caret.train（caret使用gam来自包mgcv）：

> fit <- train(P~II+TH+DR+TT,data=training,method="gam",trControl=ctrl,metric="Rsquared",preProc=c("center","scale"))
> names(fit)
 [1] "method"       "modelType"    "results"      "pred"         "bestTune"     "call"        
 [7] "dots"         "metric"       "control"      "finalModel"   "preProcess"   "trainingData"
 [13] "resample"     "resampledCM"  "perfNames"    "maximize"     "yLimits"      "times"       
 [19] "terms"        "coefnames"    "xlevels"

我没有看到fitted.values上面，但一个gam对象应该返回fitted.values- http://hosho.ees.hokudai.ac.jp/~kubo/Rdoc/library/gam/html/gam.html 结果，我无法绘制拟合与残差，以及其他一些功能也失败了。一种解决方法是gam直接使用而不是caret，但我也计划使用其他模型，并且想要一个一致的界面。

请指教。

编辑：

  1. 数据快照 -dput(head(training))输出：

    structure(list(TT = c(1.810376, 0.089206, 0.623906, 0.676775, 
0.206524, 1.014849), P = c(682L, 674L, 681L, 679L, 655L, 682L
), II = c(846000000L, 4790000L, 38600000L, 40600000L, 1379632L, 
7526080L), WSM = c(5272L, 144L, 576L, 576L, 2336L, 18696L), TSM = c(168704L, 
4608L, 18432L, 18432L, 74752L, 598272L), L2M = c(1.49e+09, 12600000, 
85700000, 1.24e+08, 4214560, 33560200), DR = c(2.52e+09, 18400000, 
1.3e+08, 1.8e+08, 5559030, 44681000), DW = c(11600000L, 5440000L, 
39600000L, 46400000L, 4920550L, 36812430L), TH = c(32.032843125, 
0.1880727305, 0.2003506939, 0.1983195715, 0.558498625, 0.495952125
)), .Names = c("TT", "P", "II", "WSM", "TSM", "L2M", "DR", "DW", 
"TH"), row.names = c(3L, 5L, 7L, 8L, 9L, 10L), class = "data.frame")

  2. str(fit)按照@nograpes 的建议显示fitted.values在里面。finalModel

    $ finalModel  :List of 50
..$ coefficients     : Named num [1:37] 761 -1839 -377 745 -473 ...
.. ..- attr(*, "names")= chr [1:37] "(Intercept)" "s(II).1" "s(II).2" "s(II).3" ...
..$ residuals        : num [1:44] -8.229 0.402 -11.41 -26.357 -8.202 ...
..$ fitted.values    : Named num [1:44] 690 674 683 707 687 ...
4

2 回答 2

4

如果您提供了一些示例数据，这将使这个问题更容易回答。

检查names对象的 不是找出其中内容的好方法。尝试跑步str(fit)，你会看到你缺少什么。

fitted.values对象嵌套更深一层。

library(mgcv)
library(caret)

dat <- gamSim(1,n=400,dist="normal",scale=2)
test<-train(y~x1, data=dat, method='gam') # Next time, provide any data like this.

test$finalModel$fitted.values # If that isn't what you want, try str(test)
于 2013-04-03T23:24:09.370 回答
1

见 http://caret.r-forge.r-project.org/modelList.html

使用 method = "gam" 可以从 mgcv 包中获取 gam() 并使用 "gamLoess" 和 "gamSpline" 使用 gam 包中的 gam()。

> library(gam)
> data(kyphosis)
> mod <- train(Kyphosis ~ Age + Number, data = kyphosis, method = "gamSpline")
> mod$finalModel
Call:
gam:::gam(formula = smootherFormula(data[, !(names(data) %in% 
    ".outcome"), drop = FALSE], smoother = "s", df = tuneValue$.df), 
    family = if (type == "Regression") gaussian() else binomial(), 
    data = data)

Degrees of Freedom: 80 total; 76.99992 Residual
Residual Deviance: 64.9097 
> head(mod$finalModel$fitted.values)
         1          2          3          4          5          6 
0.15217276 0.10961985 0.27168469 0.07017040 0.04072463 0.01414652

最大限度

于 2013-04-04T18:43:25.340 回答