Question

让我们从简单的线性回归输出（copied from here）开始，

Call:
lm(formula = a1 ~ ., data = clean.algae[, 1:12])

Residuals:
  Min      1Q  Median      3Q     Max 
  -37.679 -11.893  -2.567   7.410  62.190 

  Coefficients:
                Estimate Std. Error t value Pr(>|t|)   
  (Intercept)  42.942055  24.010879   1.788  0.07537 . 
  seasonspring  3.726978   4.137741   0.901  0.36892   
  seasonsummer  0.747597   4.020711   0.186  0.85270   
  seasonwinter  3.692955   3.865391   0.955  0.34065   
  sizemedium    3.263728   3.802051   0.858  0.39179   
  sizesmall     9.682140   4.179971   2.316  0.02166 * 
  speedlow      3.922084   4.706315   0.833  0.40573   
  speedmedium   0.246764   3.241874   0.076  0.93941   
  mxPH         -3.589118   2.703528  -1.328  0.18598   
  mnO2          1.052636   0.705018   1.493  0.13715   
  Cl           -0.040172   0.033661  -1.193  0.23426   
  NO3          -1.511235   0.551339  -2.741  0.00674 **
  NH4           0.001634   0.001003   1.628  0.10516   
  oPO4         -0.005435   0.039884  -0.136  0.89177   
  PO4          -0.052241   0.030755  -1.699  0.09109 . 
  Chla         -0.088022   0.079998  -1.100  0.27265   
  ---
  Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

  Residual standard error: 17.65 on 182 degrees of freedom
  Multiple R-squared:  0.3731,    Adjusted R-squared:  0.3215 
  F-statistic: 7.223 on 15 and 182 DF,  p-value: 2.444e-12

从此输出中，我们可以看到模型适用性以及哪些变量显着影响目标变量。此外，我们可以通过查看系数的符号来查看变量是正影响还是负影响。

现在来看这个示例from the mlr package manual，

### Select features
sfeats = selectFeatures(learner = "surv.coxph", task = wpbc.task, resampling = rdesc,
  control = ctrl, show.info = FALSE)
sfeats
## FeatSel result:
## Features (14): mean_radius, mean_compactness, mean_concavepoints, mean_symmetry, mean_fractaldim, SE_perimeter, SE_area, SE_concavity, SE_fractaldim, worst_radius, worst_perimeter, worst_concavity, worst_concavepoints, tsize
## cindex.test.mean=0.6718346

从上面的输出中，我可以看到重要的功能列表。我的问题是，如何看待方向（正向或负向）特征（独立变量）正在影响目标变量？有人在这个问题上有帮助吗？推荐的阅读材料将不胜感激。

添加

我正在尝试在示例多标签分类yeast.task example中实现您的建议，

library(mlr)
library(mmpf)

yeast <-  getTaskData(yeast.task)
labels <- colnames(yeast)[1:14]
yeast.task <- makeMultilabelTask(id = "multi", data = yeast, target = labels)

lrn.br <- makeLearner("classif.rpart", predict.type = "prob")
lrn.br <- makeMultilabelBinaryRelevanceWrapper(lrn.br)

mod <- mlr::train(lrn.br, yeast.task, subset = 1:1500, weights = rep(1/1500, 1500))
pred <- predict(mod, newdata = yeast[1501:1600,])

performance(pred, measures = list(multilabel.subset01, multilabel.hamloss, multilabel.acc,
                                   multilabel.f1, timepredict))

rdesc <- makeResampleDesc(method = "CV", stratify = FALSE, iters = 3)
r <- resample(learner = lrn.br, task = yeast.task, resampling = rdesc, show.info = FALSE)

getMultilabelBinaryPerformances(pred, measures = list(acc, mmce, auc))
getMultilabelBinaryPerformances(r$pred, measures = list(acc, mmce))

getLearnerModel(mod)

pd <- generatePartialDependenceData(mod, yeast.task)
plotPartialDependence(pd)

最后三行三行给了我以下输出。我不确定这些是否有用。知道我做错了什么吗？

> getLearnerModel(mod)
$label1
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label1; obs = 1500; features = 103
Hyperparameters: xval=0

$label2
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label2; obs = 1500; features = 103
Hyperparameters: xval=0

$label3
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label3; obs = 1500; features = 103
Hyperparameters: xval=0

$label4
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label4; obs = 1500; features = 103
Hyperparameters: xval=0

$label5
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label5; obs = 1500; features = 103
Hyperparameters: xval=0

$label6
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label6; obs = 1500; features = 103
Hyperparameters: xval=0

$label7
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label7; obs = 1500; features = 103
Hyperparameters: xval=0

$label8
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label8; obs = 1500; features = 103
Hyperparameters: xval=0

$label9
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label9; obs = 1500; features = 103
Hyperparameters: xval=0

$label10
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label10; obs = 1500; features = 103
Hyperparameters: xval=0

$label11
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label11; obs = 1500; features = 103
Hyperparameters: xval=0

$label12
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label12; obs = 1500; features = 103
Hyperparameters: xval=0

$label13
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label13; obs = 1500; features = 103
Hyperparameters: xval=0

$label14
Model for learner.id=classif.rpart; learner.class=classif.rpart
Trained on: task.id = label14; obs = 1500; features = 103
Hyperparameters: xval=0

> 
> pd <- generatePartialDependenceData(mod, yeast.task)
Error in data.table(preds, design[, vars, drop = FALSE], key = vars) : 
  column or argument 1 is NULL
> plotPartialDependence(pd)
Error in checkClass(x, classes, ordered, null.ok) : object 'pd' not found

Answer 1

如果要提取基础学习者模型的系数，则必须在mlr中使用getLearnerModel()：

library(mlr)
mod = train(learner = "surv.coxph", task = lung.task)
getLearnerModel(mod)

输出：

Call:
survival::coxph(formula = f, data = data)

               coef exp(coef)  se(coef)     z       p
inst      -3.04e-02  9.70e-01  1.31e-02 -2.31 0.02062
age        1.28e-02  1.01e+00  1.19e-02  1.07 0.28340
sex       -5.67e-01  5.67e-01  2.01e-01 -2.81 0.00489
ph.ecog    9.07e-01  2.48e+00  2.39e-01  3.80 0.00014
ph.karno   2.66e-02  1.03e+00  1.16e-02  2.29 0.02223
pat.karno -1.09e-02  9.89e-01  8.14e-03 -1.34 0.18016
meal.cal   2.60e-06  1.00e+00  2.68e-04  0.01 0.99224
wt.loss   -1.67e-02  9.83e-01  7.91e-03 -2.11 0.03465

Likelihood ratio test=33.7  on 8 df, p=5e-05
n= 167, number of events= 120

如果您对不依赖于 learner 的可插拔性感兴趣，则可以查看部分依赖图。对于coxph，它们不是令人惊讶的线性：

pd = generatePartialDependenceData(mod, lung.task)
plotPartialDependence(pd)

但是您也可以使用随机森林获取生存数据：

mod2 = train(learner = "surv.randomForestSRC", task = lung.task)
pd2 = generatePartialDependenceData(mod2, lung.task)
plotPartialDependence(pd2)

但是，部分依赖图也必须谨慎解释。所以你应该阅读它here。您也可以看看ICE图。

Answer 2

您可以使用结果的$opt.path成员来获得有关在特征选择过程中发生的事情的更多信息，但是一般来说，为单个特征给出正相关和负相关/效果是没有意义的。对于大多数模型，您无法像线性回归一样定义方向相关性，因为它对特定模型没有意义。功能选择功能与模型无关。

即使评估某个特定功能是提高模型的性能（正效应）还是降低模型的性能（负效应），对于这种类型的选择也没有意义，因为它考虑了要素相互作用。对于特定功能，您可能会对其他功能的特定子集产生积极影响，而对其他功能却产生负面影响。

最后，将要素与输出关联不仅取决于模型，还取决于要素-仅适用于数字要素。

mlr-如何查看影响目标变量的方向（正向或负向）特征

2 个答案: