Data 624 - Homework 8
Ohannes (Hovig) Ohannessian
4/12/2019
7.2. Friedman (1991) introduced several benchmark data sets create by simulation. One of these simulations used the following nonlinear equation to create data: \(y = 10sin(πx_1x_2)+20(x3 −0.5)^2 +10x_4 +5x_5 +N(0,σ^2)\) where the x values are random variables uniformly distributed between [0, 1] (there are also 5 other non-informative variables also created in the simulation). The package mlbench contains a function called mlbench.friedman1 that simulates these data:
trainingData <- mlbench.friedman1(200, sd = 1)
## We convert the ' x ' data from a matrix to a data frame
## One reason is that this will give the columns names.
trainingData$x <- data.frame(trainingData$x)
## Look at the data using
#featurePlot(trainingData$x, trainingData$y)
## or other methods.
## This creates a list with a vector ' y ' and a matrix
## of predictors ' x ' . Also simulate a large test set to
## estimate the true error rate with good precision:
testData <- mlbench.friedman1(5000, sd = 1)
testData$x <- data.frame(testData$x)Tune several models on these data. For example:
knnModel <- train(x = trainingData$x, y = trainingData$y, method = "knn", preProc = c("center", "scale"), tuneLength = 10)
knnModel## k-Nearest Neighbors
##
## 200 samples
## 10 predictor
##
## Pre-processing: centered (10), scaled (10)
## Resampling: Bootstrapped (25 reps)
## Summary of sample sizes: 200, 200, 200, 200, 200, 200, ...
## Resampling results across tuning parameters:
##
## k RMSE Rsquared MAE
## 5 3.426070 0.5384105 2.735971
## 7 3.358310 0.5554782 2.653725
## 9 3.286518 0.5821513 2.580048
## 11 3.255192 0.6032095 2.558699
## 13 3.259073 0.6133974 2.574079
## 15 3.274781 0.6194655 2.601801
## 17 3.304555 0.6213184 2.634957
## 19 3.325879 0.6282695 2.669575
## 21 3.342864 0.6314736 2.691948
## 23 3.362049 0.6378854 2.714370
##
## RMSE was used to select the optimal model using the smallest value.
## The final value used for the model was k = 11.RMSE was used to select the optimal model using the smallest value. The final value used for the model was k = 19.
knnPred <- predict(knnModel, newdata = testData$x)
## The function 'postResample' can be used to get the test set
## perforamnce values
postResample(pred = knnPred, obs = testData$y)## RMSE Rsquared MAE
## 3.1172384 0.6595098 2.4896405Which models appear to give the best performance? Does MARS select the informative predictors (those named X1–X5)?
KNN:
Given in the question, the result of KNN RMSE is:
- KNN_RMSE = 3.117
MARS:
Next we will take a look at MARS model
set.seed(101)
marsGrid <- expand.grid(degree =1:2, nprune=seq(2,14,by=2))
marsModel <- train(x = trainingData$x, y = trainingData$y, method='earth', tuneGrid = marsGrid, trControl = trainControl(method = "cv"))
marsModel## Multivariate Adaptive Regression Spline
##
## 200 samples
## 10 predictor
##
## No pre-processing
## Resampling: Cross-Validated (10 fold)
## Summary of sample sizes: 180, 180, 180, 180, 180, 180, ...
## Resampling results across tuning parameters:
##
## degree nprune RMSE Rsquared MAE
## 1 2 4.282319 0.3209299 3.647872
## 1 4 3.252892 0.6177937 2.596347
## 1 6 2.422492 0.7949329 1.951867
## 1 8 1.818816 0.8843736 1.443801
## 1 10 1.710585 0.8942351 1.320219
## 1 12 1.713885 0.8938793 1.318051
## 1 14 1.719069 0.8935452 1.327885
## 2 2 4.282319 0.3209299 3.647872
## 2 4 3.235833 0.6141738 2.589953
## 2 6 2.612814 0.7594481 2.101367
## 2 8 2.018077 0.8545566 1.578292
## 2 10 1.606696 0.9113584 1.281839
## 2 12 1.286895 0.9434766 1.071257
## 2 14 1.253988 0.9469622 1.038891
##
## RMSE was used to select the optimal model using the smallest value.
## The final values used for the model were nprune = 14 and degree = 2.#Variable Importance
varImp(marsModel)## earth variable importance
##
## Overall
## X4 100.00
## X2 83.05
## X1 67.99
## X5 55.45
## X3 43.90
## X10 0.00
## X6 0.00
## X9 0.00
## X7 0.00
## X8 0.00marsPred <- predict (marsModel, testData$x)
postResample(pred = marsPred, obs = testData$y)## RMSE Rsquared MAE
## 1.3021741 0.9319866 1.0256858- It appears MARS is only using x1-x5 to make a prediction.
- MARS_RMSE = 1.3021741, better improvement than KNN
Neural Network:
nnetPred <- predict(nnetModel, newdata = testData$x)
pr <- postResample(pred = nnetPred, obs = testData$y)- NNET_RMSE = 14.3192529
SVM:
set.seed(102)
svmModel <- train(x = trainingData$x, y = trainingData$y, method='svmRadial', tuneLength = 14, trControl = trainControl(method = "cv"))
svmModel$finalModel## Support Vector Machine object of class "ksvm"
##
## SV type: eps-svr (regression)
## parameter : epsilon = 0.1 cost C = 8
##
## Gaussian Radial Basis kernel function.
## Hyperparameter : sigma = 0.0592686035661848
##
## Number of Support Vectors : 165
##
## Objective Function Value : -92.0455
## Training error : 0.010848svmPred <- predict(svmModel, newdata = testData$x)
postResample(pred = svmPred, obs = testData$y)## RMSE Rsquared MAE
## 1.9501784 0.8481132 1.5150040SVM_RMSE = 1.9501784
Conclusion:
MARS has the most accurate result among the 4 models here with best RMSE value of 1.3021741
7.5. Exercise 6.3 describes data for a chemical manufacturing process. Use the same data imputation, data splitting, and pre-processing steps as before and train several nonlinear regression models.
Which nonlinear regression model gives the optimal resampling and test set performance?
Which predictors are most important in the optimal nonlinear regression model? Do either the biological or process variables dominate the list? How do the top ten important predictors compare to the top ten predictors from the optimal linear model?
Explore the relationships between the top predictors and the response for the predictors that are unique to the optimal nonlinear regression model. Do these plots reveal intuition about the biological or process predictors and their relationship with yield?
set.seed(1234)
registerDoParallel(cores=3)
chem_imput <- missForest(ChemicalManufacturingProcess, ntree = 100)$ximp## missForest iteration 1 in progress...done!
## missForest iteration 2 in progress...done!
## missForest iteration 3 in progress...done!
## missForest iteration 4 in progress...done!ds <- list(org.data.set = chem_imput)
trans_ds <- ds$org.data.set %>%
mutate(BiologicalMaterial04.boxcoxtrans = BoxCox(BiologicalMaterial04, -0.99992424816297),
ManufacturingProcess01.boxcoxtrans = BoxCox(ManufacturingProcess01, 1.99995900720725),
ManufacturingProcess02.boxcoxtrans = BoxCox(ManufacturingProcess02, 1.99995900720725),
ManufacturingProcess03.boxcoxtrans = BoxCox(ManufacturingProcess03, -0.99992424816297),
ManufacturingProcess06.boxcoxtrans = BoxCox(ManufacturingProcess06, -0.99992424816297),
ManufacturingProcess07.boxcoxtrans = BoxCox(ManufacturingProcess07,-0.99992424816297),
ManufacturingProcess08.boxcoxtrans = BoxCox(ManufacturingProcess08, 1.99992424816297),
ManufacturingProcess12.boxcoxtrans = BoxCox(ManufacturingProcess12, 0.0000410225926326142),
ManufacturingProcess16.boxcoxtrans = BoxCox(ManufacturingProcess16, 1.99995900720725),
ManufacturingProcess18.boxcoxtrans = BoxCox(ManufacturingProcess18, 1.99995900720725))
data.set <- c(ds, list(trans.data.set = trans_ds)) %>%
c(., list(train.test.split = floor(0.75 * nrow(chem_imput)))) %>%
c(., train.index = list(sample(seq_len(nrow(chem_imput)), size = .$train.test.split))) %>%
c(., train = list(as.data.frame(trans_ds[.$train.index,])), test = list(as.data.frame(trans_ds[-.$train.index,])))a.
- Neural Network:
nnetFit <- nnet(data.set$train %>% dplyr::select(-Yield), data.set$train$Yield, size = 5, decay = .01, maxit = 500, trace = FALSE)
nnetPred <- predict(nnetFit, data.set$test %>% dplyr::select(-Yield)) %>%
cbind(.,data.set$test$Yield) %>%
as.data.frame()
colnames(nnetPred) <- c("predicted", "actual")
rmse(nnetPred$actual, nnetPred$predicted)## [1] 39.27299- MARS:
marsFit <- earth(data.set$train %>% dplyr::select(-Yield), data.set$train$Yield)
marsPred <- predict(marsFit, data.set$test %>% dplyr::select(-Yield)) %>%
cbind(.,data.set$test$Yield) %>%
as.data.frame()
colnames(marsPred) <- c("predicted", "actual")
rmse(marsPred$actual, marsPred$predicted)## [1] 1.20368- MARS model has better performance and with RMSE value of 26.3391164.
b.
dt <- setDT(varImp(marsFit), keep.rownames = TRUE)[]
options(scipen=99); dt[order(-Overall)]## rn Overall
## 1: ManufacturingProcess32 100.000000
## 2: ManufacturingProcess09 59.796136
## 3: ManufacturingProcess13 33.003558
## 4: BiologicalMaterial03 15.563278
## 5: ManufacturingProcess33 14.981677
## 6: ManufacturingProcess28 12.625159
## 7: ManufacturingProcess45 11.930599
## 8: BiologicalMaterial02 9.980304
## 9: BiologicalMaterial01 0.000000
## 10: BiologicalMaterial04 0.000000
## 11: BiologicalMaterial05 0.000000
## 12: BiologicalMaterial06 0.000000
## 13: BiologicalMaterial07 0.000000
## 14: BiologicalMaterial08 0.000000
## 15: BiologicalMaterial09 0.000000
## 16: BiologicalMaterial10 0.000000
## 17: BiologicalMaterial11 0.000000
## 18: BiologicalMaterial12 0.000000
## 19: ManufacturingProcess01 0.000000
## 20: ManufacturingProcess02 0.000000
## 21: ManufacturingProcess03 0.000000
## 22: ManufacturingProcess04 0.000000
## 23: ManufacturingProcess05 0.000000
## 24: ManufacturingProcess06 0.000000
## 25: ManufacturingProcess07 0.000000
## 26: ManufacturingProcess08 0.000000
## 27: ManufacturingProcess10 0.000000
## 28: ManufacturingProcess11 0.000000
## 29: ManufacturingProcess12 0.000000
## 30: ManufacturingProcess14 0.000000
## 31: ManufacturingProcess15 0.000000
## 32: ManufacturingProcess16 0.000000
## 33: ManufacturingProcess17 0.000000
## 34: ManufacturingProcess18 0.000000
## 35: ManufacturingProcess19 0.000000
## 36: ManufacturingProcess20 0.000000
## 37: ManufacturingProcess21 0.000000
## 38: ManufacturingProcess22 0.000000
## 39: ManufacturingProcess23 0.000000
## 40: ManufacturingProcess24 0.000000
## 41: ManufacturingProcess25 0.000000
## 42: ManufacturingProcess26 0.000000
## 43: ManufacturingProcess27 0.000000
## 44: ManufacturingProcess29 0.000000
## 45: ManufacturingProcess30 0.000000
## 46: ManufacturingProcess31 0.000000
## 47: ManufacturingProcess34 0.000000
## 48: ManufacturingProcess35 0.000000
## 49: ManufacturingProcess36 0.000000
## 50: ManufacturingProcess37 0.000000
## 51: ManufacturingProcess38 0.000000
## 52: ManufacturingProcess39 0.000000
## 53: ManufacturingProcess40 0.000000
## 54: ManufacturingProcess41 0.000000
## 55: ManufacturingProcess42 0.000000
## 56: ManufacturingProcess43 0.000000
## 57: ManufacturingProcess44 0.000000
## 58: BiologicalMaterial04.boxcoxtrans 0.000000
## 59: ManufacturingProcess01.boxcoxtrans 0.000000
## 60: ManufacturingProcess02.boxcoxtrans 0.000000
## 61: ManufacturingProcess03.boxcoxtrans 0.000000
## 62: ManufacturingProcess06.boxcoxtrans 0.000000
## 63: ManufacturingProcess07.boxcoxtrans 0.000000
## 64: ManufacturingProcess08.boxcoxtrans 0.000000
## 65: ManufacturingProcess12.boxcoxtrans 0.000000
## 66: ManufacturingProcess16.boxcoxtrans 0.000000
## 67: ManufacturingProcess18.boxcoxtrans 0.000000
## rn Overall- Manufacturing Processes shows up to be important with few important biological materials
- Manufacturing processes contain a lot of information on the variance of yield.
c.
subset.data.set <- data.set$trans.data.set %>%
dplyr::select(ManufacturingProcess32, ManufacturingProcess09, ManufacturingProcess13,
ManufacturingProcess33, BiologicalMaterial02, BiologicalMaterial03,
ManufacturingProcess28, ManufacturingProcess43, ManufacturingProcess04,
BiologicalMaterial05, Yield)
correlations <- cor(subset.data.set)
corrplot::corrplot(correlations, order = "FPC", diag = TRUE)
- The correlation plot illustrates some insight to these important variables.
- Yield has a mostly positive correlation with many of the processes but with only
- Manufacturing processes have Yield with the strongest negative correlation seen in top variables.
- Some very high correlations is noticed which it may impicate for possible issues of multi-collinearity.