从 mlr3 中拟合的 regr.nnet 对象中提取权重

0. （我自己添加）如果您使用 mlr3 创建教材并想分享它，您可以在 mlr-org/mlr3website 存储库中创建问题或 PR。在 mlr-org 网站上，我们有一个资源选项卡，我们可以在其中链接类似的东西:)https://mlr-org.com/resources.html

1. 是否可以很容易地提取其他措施，例如 从基准对象？msr("regr.rmse")

该方法采用一系列度量值（例如，可由 构造）。（见下面的代码）$aggregate()msrs()

2. 我对我的代码行.....不满意，但还没有找到更好的方法。pr <- as.data.table()

您可以不进行转换。best$predictions()[[1]]$response

3. 最后，我想访问拟合 nnet 的内部数据结构，以提取原始权重，以便在 R 之外“离线”使用神经网络。

我们不干涉拟合物体的内部结构。它们可以通过训练有素的学习者的插槽访问（请参阅下面的代码）。$model

library(mlr3)
library(mlr3learners)

learner = lrns(c("classif.rpart", "classif.nnet"))
task = tsk("iris")
resampling = rsmp("holdout")

design = benchmark_grid(
  tasks = task,
  learners = learner,
  resamplings = resampling
)

bmr = benchmark(design, store_models = TRUE)
#> INFO  [07:13:52.802] [mlr3] Running benchmark with 2 resampling iterations
#> INFO  [07:13:52.889] [mlr3] Applying learner 'classif.rpart' on task 'iris' (iter 1/1)
#> INFO  [07:13:52.921] [mlr3] Applying learner 'classif.nnet' on task 'iris' (iter 1/1)
#> # weights:  27
#> initial  value 118.756408 
#> iter  10 value 58.639749
#> iter  20 value 45.676852
#> iter  30 value 21.336083
#> iter  40 value 8.646964
#> iter  50 value 6.041813
#> iter  60 value 5.906140
#> iter  70 value 5.902865
#> iter  80 value 5.898339
#> final  value 5.898161 
#> converged
#> INFO  [07:13:52.946] [mlr3] Finished benchmark

bmr$aggregate(msrs(c("classif.acc", "time_train")))
#>    nr      resample_result task_id    learner_id resampling_id iters
#> 1:  1 <ResampleResult[21]>    iris classif.rpart       holdout     1
#> 2:  2 <ResampleResult[21]>    iris  classif.nnet       holdout     1
#>    classif.acc time_train
#> 1:        0.98      0.007
#> 2:        1.00      0.005

# get the first resample result
rr1 = bmr$resample_result(1)

# Get the model from the first resampling iteration of this ResampleResult
rr1$learners[[1]]$model
#> n= 100 
#> 
#> node), split, n, loss, yval, (yprob)
#>       * denotes terminal node
#> 
#> 1) root 100 66 versicolor (0.32000000 0.34000000 0.34000000)  
#>   2) Petal.Length< 2.45 32  0 setosa (1.00000000 0.00000000 0.00000000) *
#>   3) Petal.Length>=2.45 68 34 versicolor (0.00000000 0.50000000 0.50000000)  
#>     6) Petal.Width< 1.75 37  4 versicolor (0.00000000 0.89189189 0.10810811) *
#>     7) Petal.Width>=1.75 31  1 virginica (0.00000000 0.03225806 0.96774194) *

^{创建于 2023-03-08 with reprex v2.0.2}

根据 @Sebastian 的回答，我已经能够解决一些剩余的小细节，尤其是如何将其与上一个线程相结合。我采用了与 @Sebastian 的答案类似的示例，但将其修改为作为输入和目标的回归任务。然后，我集成了“蛮力”方法，以使用相同的设置训练多个网络，并将相同的细分应用于训练和测试数据。irisSpeciesPetal.width

library(mlr3learners)
#> Loading required package: mlr3
set.seed(123)

data(iris) # to show that we use this data set

## half of indices for training and other for test subset
id_train  <- sample(1:nrow(iris), nrow(iris) %/% 2)
id_test   <- which(!((1:nrow(iris)) %in% id_train))

## create task, learner and custom resampling strategy
task <- as_task_regr(iris, target = "Petal.Width")
learner <- lrn("regr.nnet", size = 5, trace = FALSE)
resampling <- rsmp("custom")
resampling$instantiate(task, train = list(id_train), test = list(id_test))

## replicate the learners, create and run a benchmark design
learners <- replicate(10, learner$clone())
design <- benchmark_grid(
  tasks = task,
  learners = learners,
  resamplings = resampling
)

bmr <- benchmark(design, store_models = TRUE)

## summary of results and "best" result according to mse
bmr$aggregate(msrs(c("regr.mse", "regr.rmse", "regr.rsq", "time_train")))

(i_best <- which.min(bmr$aggregate()$regr.mse))
#> [1] 3
best <- bmr$resample_result(i_best)

## prediction for the test set
best$predictions()[[1]]$response

## or for both subsets
best$learners[[1]]$predict(task, row_ids = id_train)$response
best$learners[[1]]$predict(task, row_ids = id_test)$response

## extract raw model and the weights with coef() method from package "nnet"
best$learners[[1]]$model
#> a 5-5-1 network with 36 weights
#> inputs: Petal.Length Sepal.Length Sepal.Width Speciesversicolor Speciesvirginica 
#> output(s): Petal.Width 
#> options were - linear output units

coef(best$learners[[1]]$model)
#>        b->h1       i1->h1       i2->h1       i3->h1       i4->h1       i5->h1 
#>   0.33696598   5.87087353   5.86170711   2.01521854  -0.40507201   0.59342370 
#>        b->h2       i1->h2       i2->h2       i3->h2       i4->h2       i5->h2 
#>  -0.28521282  -0.66194804  -2.53390441  -1.83519953  -0.36281388   0.08545735 
#>        b->h3       i1->h3       i2->h3       i3->h3       i4->h3       i5->h3 
#>  -0.73454761  -0.57450360   0.15459231  -0.92357099  -1.44787867   3.87019807 
#>        b->h4       i1->h4       i2->h4       i3->h4       i4->h4       i5->h4 
#>   4.15247439  -0.31732640  -0.07613404  -0.33243948  -1.59663476  -5.46270442 
#>        b->h5       i1->h5       i2->h5       i3->h5       i4->h5       i5->h5 
#>   1.31088094  -0.51583287   3.79240044   3.43102418  -0.56412793  -0.46786541 
#>         b->o        h1->o        h2->o        h3->o        h4->o        h5->o 
#>   0.30750849  12.55643667   3.18489708  -2.57517235  -2.36851810 -10.44863295

^{创建于 2023-03-08 with reprex v2.0.2}