1

caret使用vglm()from构建了一个自定义模型VGAM。它适用于简单的效果,但是当我尝试添加交互时,它会失败并显示object 'x1:x2' not found错误消息,其中x1x2是我作为交互输入模型的预测变量。问题与预测有关,除非我弄错了,否则它似乎是因为要么 要么predict.train试图predictvglm用来x1:x2预测类。

我在下面提供了一个工作示例。

# Set up data
set.seed(123)
n <- 100
x1 <- rnorm(n, 175, 7)
x2 <- rnorm(n, 30, 8)
cont <- 0.5 * x1 - 0.3 * x2 + 10 + rnorm(n, 0, 6)
y  <- cut(cont, breaks = quantile(cont), include.lowest = TRUE,
             labels = c("A", "B", "C", "D"), ordered = TRUE)
d <- data.frame(x1, x2, y)

# My custom caret function
vglmTrain <- list(
  label = "VGAM prop odds",
  library = "VGAM",
  loop = NULL,
  type = "Classification",
  parameters = data.frame(parameter = "parameter",
                          class = "character",
                          label = "parameter"),
  grid = function(x, y,
                  len = NULL, search = "grid") data.frame(parameter = "none"),
  fit = function(x, y, wts, param, lev, last, classProbs, ...) {
    dat <- if(is.data.frame(x)) x else as.data.frame(x)
    dat$.outcome <- y
    if(!is.null(wts))
    {
      out <- vglm(.outcome ~ ., propodds, data = dat, weights = wts, ...)
    } else {
      out <- vglm(.outcome ~ ., propodds, data = dat, ...)
    }
    out
  },
  predict = function(modelFit, newdata, preProc = NULL, submodels = NULL) {
    probs <- predict(modelFit, data.frame(newdata), type = "response")

    predClass <- function (x) {
      n <- colnames(x)
      factor(as.vector(apply(x, 1, which.max)),
             levels = 1:length(n),
             labels = n)
    }
    predClass(probs)
  },
  prob = function(modelFit, newdata, preProc = NULL, submodels = NULL)
    predict(modelFit, data.frame(newdata), type = "response"),
  predictors = function(x, ...) names(attributes(terms(x))$dataClasses[-1]),
  levels = function(x) x@misc$ynames,
  sort = function(x) x)

现在,如果我尝试使用该函数,如果我为公式提供交互,它会退出并出现错误。

# Load caret
library(caret)

ctrl <- trainControl(method = "cv", number = 2, verboseIter = T)

# A model with no interactions - works
f1 <- train(y ~ x1 + x2, data = d,
           method = vglmTrain,
           trControl = ctrl)

# A model with interactions - fails
f2 <- train(y ~ x1*x2, data = d,
            method = vglmTrain,
            trControl = ctrl)

Error in train.default(x, y, weights = w, ...) : Stopping
In addition: Warning messages:
1: In eval(expr, envir, enclos) :
  predictions failed for Fold1: parameter=none Error in eval(expr, envir, enclos) : object 'x1:x2' not found

2: In eval(expr, envir, enclos) :
  predictions failed for Fold2: parameter=none Error in eval(expr, envir, enclos) : object 'x1:x2' not found

3: In nominalTrainWorkflow(x = x, y = y, wts = weights, info = trainInfo,  :
  There were missing values in resampled performance measures.

这是我的 sessionInfo():

> sessionInfo()
R version 3.2.4 (2016-03-10)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows >= 8 x64 (build 9200)

locale:
[1] LC_COLLATE=English_United States.1252  LC_CTYPE=English_United States.1252    LC_MONETARY=English_United States.1252 LC_NUMERIC=C                          
[5] LC_TIME=English_United States.1252    

attached base packages:
[1] splines   stats4    stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] VGAM_1.0-0      caret_6.0-64    ggplot2_2.1.0   lattice_0.20-33

loaded via a namespace (and not attached):
 [1] Rcpp_0.12.3        magrittr_1.5       MASS_7.3-45        munsell_0.4.3      colorspace_1.2-6   foreach_1.4.3      minqa_1.2.4        stringr_1.0.0      car_2.1-1         
[10] plyr_1.8.3         tools_3.2.4        nnet_7.3-12        pbkrtest_0.4-6     parallel_3.2.4     grid_3.2.4         gtable_0.2.0       nlme_3.1-125       mgcv_1.8-12       
[19] quantreg_5.21      e1071_1.6-7        class_7.3-14       MatrixModels_0.4-1 iterators_1.0.8    lme4_1.1-11        Matrix_1.2-3       nloptr_1.0.4       reshape2_1.4.1    
[28] codetools_0.2-14   stringi_1.0-1      compiler_3.2.4     scales_0.4.0       SparseM_1.7       

有人知道如何解决这个问题吗?

4

2 回答 2

0

Phiver 的解决方案在本示例中运行良好,但是当我添加虚拟编码变量时,模型再次失败。

我做了更多的挖掘,问题似乎已经发生,因为data.frame更改了要预测的数据集中列的名称。predict在我现在添加的代码中的两个调用中data.frame(newdata, check.names = F),这似乎可以解决问题。

它现在可以使用公式界面

f2 <- train(y ~ x1 * x2, data = d,
            method = vglmTrain,
            trControl = ctrl)

模型矩阵法

f2 <- train(y = y, x = data.frame(model.matrix(y ~ x1*x2 - 1, data = d)),
            method = vglmTrain,
            trControl = ctrl)

这是新代码:

vglmTrain <- list(
  label = "VGAM prop odds",
  library = "VGAM",
  loop = NULL,
  type = "Classification",
  parameters = data.frame(parameter = "parameter",
                          class = "character",
                          label = "parameter"),
  grid = function(x, y,
                  len = NULL, search = "grid") data.frame(parameter = "none"),
  fit = function(x, y, wts, param, lev, last, classProbs, ...) {
    dat <- if(is.data.frame(x)) x else as.data.frame(x)
    dat$.outcome <- y
    if(!is.null(wts))
    {
      out <- vglm(.outcome ~ ., propodds, data = dat, weights = wts, ...)
    } else {
      out <- vglm(.outcome ~ ., propodds, data = dat, ...)
    }
    out
  },
  predict = function(modelFit, newdata, preProc = NULL, submodels = NULL) {
    probs <- predict(modelFit, data.frame(newdata, check.names = F), type = "response")

    predClass <- function (x) {
      n <- colnames(x)
      factor(as.vector(apply(x, 1, which.max)),
             levels = 1:length(n),
             labels = n)
    }
    predClass(probs)
  },
  prob = function(modelFit, newdata, preProc = NULL, submodels = NULL)
   predict(modelFit, data.frame(newdata, check.names = F), type = "response"),
  levels = function(x) x@misc$ynames,
  tags = c("Cumulative Link", "Logistic Regression", "Accepts Case Weights",
           "Probit", "Logit"),
  sort = function(x) x)
于 2016-03-15T08:55:05.470 回答
0

Caret 确实处理交互。但是,我找到了一种解决方法。您可以先调用 model.matrix 来创建具有交互作用的矩阵。您还需要删除拦截。

以您的 f2 为例,我们将数据指定为 x 和 y,而不是公式。x 包含带有交互的 model.matrix 规范,-1 删除截距。这转换为一个 data.frame 并且你准备好了。

f2 <- train(y = y, x = data.frame(model.matrix(y ~ x1*x2 - 1, data = d)),
            method = vglmTrain,
            trControl = ctrl)

编辑:

在调试了 train.default 并检查了您的模型类型规范和其他一些规范之后,我发现在插入符号模型中而不是在您的模型中进行了检查。检查与 predict 和 probs 函数有关。这些都对 Dataframe 进行了检查。如果您将此检查添加到这两个函数,它会按预期工作。

if (!is.data.frame(newdata)) 
  newdata <- as.data.frame(newdata)

整个函数将是:

vglmTrain <- list(
  label = "VGAM prop odds",
  library = "VGAM",
  loop = NULL,
  type = "Classification",
  parameters = data.frame(parameter = "parameter",
                          class = "character",
                          label = "parameter"),
  grid = function(x, y,
                  len = NULL, search = "grid") data.frame(parameter = "none"),
  fit = function(x, y, wts, param, lev, last, classProbs, ...) {
    dat <- if(is.data.frame(x)) x else as.data.frame(x)
    dat$.outcome <- y
    if(!is.null(wts))
    {
      out <- vglm(.outcome ~ ., propodds, data = dat, weights = wts, ...)
    } else {
      out <- vglm(.outcome ~ ., propodds, data = dat, ...)
    }
    out
  },
  predict = function(modelFit, newdata, preProc = NULL, submodels = NULL) {

    if (!is.data.frame(newdata)) 
      newdata <- as.data.frame(newdata)
    probs <- predict(modelFit, newdata, type = "response")

    predClass <- function (x) {
      n <- colnames(x)
      factor(as.vector(apply(x, 1, which.max)),
             levels = 1:length(n),
             labels = n)
    }
    predClass(probs)
  },
  prob = function(modelFit, newdata, preProc = NULL, submodels = NULL) {
    if (!is.data.frame(newdata)) 
      newdata <- as.data.frame(newdata)

    predict(modelFit, newdata, type = "response")
  },

  levels = function(x) x@misc$ynames,

  tags = c("Cumulative Link", "Logistic Regression", "Accepts Case Weights",
           "Probit", "Logit"),

  sort = function(x) x)
于 2016-03-14T20:01:18.393 回答