ML.NET无法提取训练模型参数的问题求助

问题场景 #

我参照微软官网代码尝试重新训练模型，执行提取训练模型参数的代码时触发类型转换错误。

提取参数的代码 #

// Extract trained model parameters
var originalModelParameters =
    ((ISingleFeaturePredictionTransformer<object>)trainedModel).Model as PoissonRegressionModelParameters;

报错信息 #

System.InvalidCastException: 'Unable to cast object of type 'Microsoft.ML.Data.TransformerChain`1[Microsoft.ML.ITransformer]' to type 'Microsoft.ML.ISingleFeaturePredictionTransformer`1[System.Object]'.'

相关代码 #

保存数据预处理和模型的代码 #

var csvmodel = csvpipeline.Fit(csvTrainingDataView);

// Define data preparation estimator
IEstimator<ITransformer> dataPrepEstimator =
    mlContext.Transforms.Concatenate("Features", new string[] { "t" })
        .Append(mlContext.Transforms.NormalizeMinMax("Features"));

// Create data preparation transformer
ITransformer dataPrepTransformer = dataPrepEstimator.Fit(csvTrainingDataView);

// Define StochasticDualCoordinateAscent regression algorithm estimator
var Estimator = mlContext.Regression.Trainers.LbfgsPoissonRegression();

// Pre-process data using data prep operations
IDataView transformedData = dataPrepTransformer.Transform(csvTrainingDataView);

//savemodel
//csv savemodel
savemodel(csvmlContext, csvmodel, transformedData);
/*
var modelPath = string.Format("{0}/MODEL/{1}-{2}.zip", Environment.CurrentDirectory,"Model","Testing");
using (var fileStream = new FileStream(modelPath, FileMode.Create, FileAccess.Write, FileShare.Write))
    csvmlContext.Model.Save(csvmodel, csvTrainingDataView.Schema, fileStream);*/

// Save Data Prep transformer
savedatprep(csvmlContext, dataPrepTransformer, csvTrainingDataView);

保存方法代码 #

static void savemodel(MLContext mlcontext,ITransformer model,IDataView Data)
{
    var modelPath = string.Format("{0}/MODEL/{1}-{2}.zip", Environment.CurrentDirectory, "Model", "Testing");
    using (var fileStream = new FileStream(modelPath, FileMode.Create, FileAccess.Write, FileShare.Write))
        mlcontext.Model.Save(model, Data.Schema, fileStream);
}
static void savedatprep(MLContext mlcontext, ITransformer model, IDataView Data)
{
    var modelDataPrepPath = string.Format("{0}/MODEL/{1}-{2}.zip", Environment.CurrentDirectory, "Model", "Testing_DataPrep");
    using (var fileStream = new FileStream(modelDataPrepPath, FileMode.Create, FileAccess.Write, FileShare.Write))
        mlcontext.Model.Save(model, Data.Schema, fileStream);
}

数据文件结构（表头及第一行） #

t,o,h,l,c,s,z,q,n,v,x,y,Ema,Rsi,Macd,MacdSign,MacdHistN3,MacdHistN2,MacdHistN1,MacdHistN0,FuturePrice
16636653,1.283,1.283,1.271,1.278,133642.8,1663666199999,170622.876,751,67266.6,85853.53,0,0,45.80812565453984,69.7112186336032,69.38782605382958,3.3440424757016984,2.8439819612832054,1.640807326496386,0.3233925797736106,0

已尝试操作 #

• 调整保存流程
• 查阅同类问题，确认参数类型与保存模型一致

解决建议 #

1. 从TransformerChain中定位训练模型节点
   你保存的csvmodel是由多个转换器组成的TransformerChain，而非单独的Poisson回归模型。需要先从链中筛选出对应的训练模型节点：

// 遍历链找到训练模型节点
var trainedModelNode = ((TransformerChain<ITransformer>)trainedModel)
    .LastOrDefault(t => t is ISingleFeaturePredictionTransformer<object>);
if (trainedModelNode != null)
{
    var originalModelParameters = ((ISingleFeaturePredictionTransformer<object>)trainedModelNode).Model as PoissonRegressionModelParameters;
    // 后续参数操作
}

2. 合并预处理与训练流程后保存完整模型
   将数据预处理和训练估算器合并为一个完整流程，拟合后保存整个链，后续加载后可直接使用，也便于定位参数：

// 合并预处理与训练流程
var fullPipeline = dataPrepEstimator.Append(Estimator);
// 拟合完整模型
var fullModel = fullPipeline.Fit(csvTrainingDataView);
// 保存完整模型
savemodel(csvmlContext, fullModel, csvTrainingDataView);

提取参数时可借助TryGetScoreTransformer辅助获取：

if (mlContext.Model.TryGetScoreTransformer(fullModel, out var scoreTransformer))
{
    var modelParams = scoreTransformer.Model as PoissonRegressionModelParameters;
    // 使用模型参数
}

3. 单独保存训练模型
   如果仅需提取训练模型参数，可在训练完成后单独保存训练模型，而非整个预处理链：

// 单独训练模型
var trainedModel = Estimator.Fit(transformedData);
// 保存训练模型
savemodel(csvmlContext, trainedModel, transformedData);

此时加载后可直接转换为ISingleFeaturePredictionTransformer<object>获取参数。

内容的提问来源于stack exchange，提问作者remington.