0

这是我在 Rstudio 中运行的代码：

library(caret)
set.seed(1)
demo <-data.frame(x1 = runif(100),x2 <- rnorm(100),y <- rbinom(100,1,0.3))
modelout <- try(train(y~.,data = demo,
                  method = "AMORE"),silent = FALSE)

输出如下所示：

Loading required package: AMORE
Attaching package: ‘AMORE’
The following object is masked from ‘package:caret’:
   train
Error in order(x$nhid) : argument 1 is not a vector
In addition: There were 50 or more warnings (use warnings() to see the first 50)

警告信息与“AMORE”包中的“train”功能有关。如果我输入

>train

在我运行上述代码后的命令窗口中，我得到了这个函数的主体，这从 AMORE 包中很明显。

function (net, P, T, Pval = NULL, Tval = NULL, error.criterium = "LMS", 
   report = TRUE, n.shows, show.step, Stao = NA, prob = NULL, 
   n.threads = 0L) 
{...
}
<environment: namespace:AMORE>

我搜索了我的路径，显然 AMORE 出现在插入符号之前：

>search()

[1] ".GlobalEnv"        "package:AMORE"     "package:caret"     "package:ggplot2"           "package:lattice"  
 [6] "tools:rstudio"     "package:stats"     "package:graphics"  "package:grDevices" "package:utils"    
[11] "package:datasets"  "package:methods"   "Autoloads"         "package:base"

以下是我的会话信息：

R version 3.0.2 (2013-09-25)
Platform: x86_64-w64-mingw32/x64 (64-bit)

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

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

other attached packages:
[1] AMORE_0.2-14    caret_6.0-21    ggplot2_0.9.3.1 lattice_0.20-24

loaded via a namespace (and not attached):
[1] car_2.0-19         codetools_0.2-8    colorspace_1.2-4   compiler_3.0.2     dichromat_2.0-0   
[6] digest_0.6.4       foreach_1.4.1      grid_3.0.2         gtable_0.1.2       iterators_1.0.6   
[11] labeling_0.2       MASS_7.3-29        munsell_0.4.2      nnet_7.3-7         plyr_1.8          
[16] proto_0.3-10       RColorBrewer_1.0-5 reshape2_1.2.2     scales_0.2.3       stringr_0.6.2     
[21] tools_3.0.2

我应该如何解决这个问题，以便 R 将使用“caret”中的“train”函数而不是“AMORE”中的函数？感谢您的宝贵时间，并提前感谢您的帮助。

捷通

4

2 回答 2

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指定命名空间。

caret::train()

library(caret)
set.seed(1)
demo <-data.frame(x1 = runif(100),x2 <- rnorm(100),y <- rbinom(100,1,0.3))
modelout <- try(caret::train(y~.,data = demo,
                  method = "AMORE"),silent = FALSE)
于 2014-01-09T21:08:46.330 回答
0

我通过创建函数的副本来解决这个问题。是的，这是一个命名空间问题。如果我重新定义了一个函数 trainc <- caret::train，并在代码中使用 trainc 而不是 train，问题就解决了。只是好奇为什么不能直接使用 caret::train！不过感谢 SethB 的帮助。

于 2014-01-12T05:03:12.783 回答