无法将xgboost模型导出为R中的pmml格式

Question

我正在尝试使用 library(pmml) 将 XGboost 模型导出为 pmml 格式。这是我使用的简单代码和不断显示的错误。

dev_samp = data.matrix(mtcars[,c(4,6,9)])

set.seed(123)
bst <- xgboost(data=as.matrix(dev_samp[,1:2]), 
                 label=dev_samp[,3],
                 max_depth=2,
                 eta=0.2,  
                 nrounds=2,
                 colsample_bytree = 0.5,
                 lambda = 0.3,
                 objective = "binary:logistic",
                 eval_metric = "error")

# Export to PMML:
pmod <- pmml(bst,input_feature_names = colnames(as.matrix(dev_samp[,1:2])),
             output_label_name = colnames(dev_samp)[3])

Error in pmml.xgb.Booster(bst, input_feature_names = colnames(as.matrix(dev_samp[,  : 
  Input feature names required at present version. Try using colnames function on Matrix, matrix or xgb.DMatrix$data

在此函数之外，代码 colnames(as.matrix(dev_samp[,1:2])) 和 colnames(dev_samp)[3] 有效。

编辑：

感谢您的回答。不幸的是，我被限制从 GitHub.

直接安装

所以我所做的是：

library(xgboost)
library(r2pmml)

#mtcars$wt=as.integer(mtcars$wt*1000)
#mtcars$hp=as.integer(mtcars$hp*2)
dev_samp = data.matrix(mtcars[,c(4,6,9)])

set.seed(123)
bst <- xgboost(data=as.matrix(dev_samp[,1:2]), 
               label=as.integer(dev_samp[,3]),
               max_depth=2,
               eta=0.2,  
               nrounds=2,
               colsample_bytree = 0.5,
               lambda = 0.3,
               objective = "binary:logistic",
               eval_metric = "error")

fmap=data.frame(seq_along(bst$feature_names)-1, bst$feature_names, "q")
fmap[,2] = as.factor(fmap[,2])
fmap[,3] = as.factor(fmap[,3])
fmap[,1] = as.integer(fmap[,1])

pmod <- decorate.xgb.Booster(bst, "mtcars.pmml",
       fmap = fmap,
       response_name = "prediction",
       response_levels = c("0", "1"),
       missing = "")

r2pmml(pmod, "XGBoostModel2.pmml")

但是最后一行r2pmml(pmod, "XGBoostModel2.pmml")returns错误：

Error in .convert(tempfile, file, converter, converter_classpath, verbose) : 
  The JPMML-R conversion application has failed (error code 1). The Java executable should have printed more information about the failure into its standard output and/or standard error streams

Answer 1

或者，如果正确性和转换速度对您很重要，那么您可以使用 R2PMML 包将 R 语言 XGBoost 模型转换为 PMML。

library("r2pmml")

r2pmml(bst, "xgboost.pmml", fmap = as.fmap(dev_samp[,1:2]), response_name = colnames(dev_samp)[3])

务必使用最新的 R2PMML 包版本（直接从 GitHub 存储库安装）。 CRAN版本目前有点过时（不提供as.fmap(..)实用功能）。

无法将xgboost模型导出为R中的pmml格式

Can't export xgboost model to pmml format in R

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pmml

xgboost