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我已经开发了一个graphlearner包mlr3，我想在Rplumber服务中发布它。但是，当我收到数据进行预测（JSON 格式的数据）时，graphlearner无法识别数据，因为 的fromJSON函数jsonlite无法推断出正确的类型（在其上学习了图形）。您对此有解决方案吗？在预测阶段是否有一种机制来管理 mlr3 中的 JSON 数据？

学习步骤

library(mlr3)
imp_missind = po("missind")
imp_fct     = po("imputenewlvl", param_vals =list(affect_columns = selector_type("factor")))
imp_num     = po("imputehist", param_vals =list(affect_columns = selector_type("numeric")))
learner = lrn('regr.ranger')
graph = po("copy", 2) %>>% 
  gunion(list(imp_missind, imp_num %>>% imp_fct)) %>>%
  po("featureunion") %>>%
  po(learner)
t1 = tsk("boston_housing")
g1 = GraphLearner$new(graph)
g1$train(t1)
saveRDS(g1,'my-model')

预测步骤：有效 （模拟数据进行预测，删除目标 col）

data=t1$data()[1:1,-1]
model = readRDS('my-model')
model$predict_newdata(newdata=data)

预测步骤：它不起作用 （模拟 JSON 数据进行预测）

model = readRDS('my-model')
data = t1$data()[1:1,-1]
json = fromJSON(toJSON(data, na="string"))
model$predict_newdata(newdata=json)

和错误：

错误：无法 rbind 任务：类型与列不匹配：cmedv（数字！=整数）

更新 可重现的例子

library(mlr3learners)
library(mlr3)
library(mlr3pipelines)
library(jsonlite)



imp_missind = po("missind")

imp_fct     = po("imputenewlvl", param_vals =list(affect_columns = selector_type("factor")))

imp_num     = po("imputehist", param_vals =list(affect_columns = selector_type("numeric")))

learner = lrn('regr.ranger')

graph = po("copy", 2) %>>% 
  gunion(list(imp_missind, imp_num %>>% imp_fct)) %>>%
  po("featureunion") %>>%
  po(learner)


task = tsk("boston_housing")


graphlearner = GraphLearner$new(graph)

#train model 
graphlearner$train(task)

# create data to predict  (juste one observation)

data= task$data()
data[1:1, chas := NA]
data = data[1:1,-1]




# look the the types of columns
str(data)

# predictin, this works fine 
predict(graphlearner, data)


# simulate the case when json data is received

json_data = toJSON(data, na="string")

print(json_data)

# get R data from json formatted data
data_from_json = fromJSON(json_data)

# look the types of columns, some are different numeric != integer, factor != char
str(data_from_json)

# try to predict, this does not work, get erro  :    cmedv (numeric != integer)
predict(graphlearner,data_from_json)
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1 回答 1

4

我想我们可能想在某个时候解决这个问题，但是在修复之前我建议通过修复你保存的架构来解决这个问题task$feature_types：

library(mlr3misc)
repair_schema = function(data, feature_types) {
   imap_dtc(data, function(v, k) {
    ft_type = feature_types[id == k,][["type"]]
    if (typeof(v) != ft_type) {
      fn = switch(ft_type,
        "character" = as.character,
        "factor" = as.factor,
        "numeric" = as.numeric,
        "integer" = as.integer
      )
      v = fn(v)
    }
    return(v)
  })
}
data_from_json2 = repair_schema(data_from_json, task$feature_types)
predict(graphlearner,data_from_json2)

这种方法还可以为您提供更大的灵活性，因为您可能会遇到一系列无法​​始终预料到的编码问题。

于 2020-03-03T14:36:53.457 回答