Exercise 6.2:

Developing a model to predict permeability (see Sect. 1.4) could save significant resources for a pharmaceutical company, while at the same time more rapidly identifying molecules that have a sufficient permeability to become a drug:

  1. Start R and use these commands to load the data:
library(AppliedPredictiveModeling)
data(permeability)

The matrix fingerprints contains the 1,107 binary molecular predictors for the 165 compounds, while permeability contains permeability response.

  1. The fingerprint predictors indicate the presence or absence of substructures of a molecule and are often sparse meaning that relatively few of the molecules contain each substructure. Filter out the predictors that have low frequencies using the nearZeroVar function from the caret package. How many predictors are left for modeling?
dim(fingerprints)
## [1]  165 1107
fingerprints <- fingerprints[, -nearZeroVar(fingerprints)]

dim(fingerprints)
## [1] 165 388

There are 388 predictors left for modeling.

  1. Split the data into a training and a test set, pre-process the data, and tune a PLS model. How many latent variables are optimal and what is the corresponding resampled estimate of \(R^{2}\)?
set.seed(123)

# index for training
index <- createDataPartition(permeability, p = .8, list = FALSE)

# train 
train_perm <- permeability[index, ]
train_fp <- fingerprints[index, ]

# test
test_perm <- permeability[-index, ]
test_fp <- fingerprints [-index, ]

# 10-fold cross-validation to make reasonable estimates
ctrl <- trainControl(method = "cv", number = 10)

plsTune <- train(train_fp, train_perm, method = "pls", metric = "Rsquared",
             tuneLength = 20, trControl = ctrl, preProc = c("center", "scale"))

# Plotting plsTune
plot(plsTune)

plsTune
## Partial Least Squares 
## 
## 133 samples
## 388 predictors
## 
## Pre-processing: centered (388), scaled (388) 
## Resampling: Cross-Validated (10 fold) 
## Summary of sample sizes: 121, 121, 118, 119, 119, 119, ... 
## Resampling results across tuning parameters:
## 
##   ncomp  RMSE      Rsquared   MAE      
##    1     13.31894  0.3442124  10.254018
##    2     11.78898  0.4830504   8.534741
##    3     11.98818  0.4792649   9.219285
##    4     12.04349  0.4923322   9.448926
##    5     11.79823  0.5193195   9.049121
##    6     11.53275  0.5335956   8.658301
##    7     11.64053  0.5229621   8.878265
##    8     11.86459  0.5144801   9.265252
##    9     11.98385  0.5188205   9.218594
##   10     12.55634  0.4808614   9.610747
##   11     12.69674  0.4758068   9.702325
##   12     13.01534  0.4538906   9.956623
##   13     13.12637  0.4367362   9.878017
##   14     13.44865  0.4140715  10.065088
##   15     13.60135  0.4034269  10.188150
##   16     13.79361  0.3943904  10.247160
##   17     14.00756  0.3845119  10.412776
##   18     14.18113  0.3711378  10.587027
##   19     14.25674  0.3703610  10.575726
##   20     14.33121  0.3723176  10.679764
## 
## Rsquared was used to select the optimal model using the largest value.
## The final value used for the model was ncomp = 6.

The latent variables are 6 and the corresponding resampled estimate of \(R^{2}\) is 0.5335956.

  1. Predict the response for the test set. What is the test set estimate of \(R^{2}\)?
fp_predict <- predict(plsTune, test_fp)

postResample(fp_predict, test_perm)
##       RMSE   Rsquared        MAE 
## 12.3486900  0.3244542  8.2881075

The test set estimate of \(R^{2}\) is 0.3244542.

  1. Try building other models discussed in this chapter. Do any have better predictive performance?
set.seed(123)

# grid of penalties
enetGrid <- expand.grid(.lambda = c(0, 0.01, .1), .fraction = seq(.05, 1, length = 20))

# tuning penalized regression model
enetTune <- train(train_fp, train_perm, method = "enet",
                  tuneGrid = enetGrid, trControl = ctrl, preProc = c("center", "scale"))

plot(enetTune)

enetTune
## Elasticnet 
## 
## 133 samples
## 388 predictors
## 
## Pre-processing: centered (388), scaled (388) 
## Resampling: Cross-Validated (10 fold) 
## Summary of sample sizes: 120, 119, 118, 120, 121, 119, ... 
## Resampling results across tuning parameters:
## 
##   lambda  fraction  RMSE       Rsquared   MAE       
##   0.00    0.05       12.97787  0.4301959    9.877733
##   0.00    0.10       11.80785  0.4931852    9.107986
##   0.00    0.15       11.66393  0.4930871    8.897928
##   0.00    0.20       11.86820  0.4793179    9.030145
##   0.00    0.25       12.10947  0.4728912    9.253825
##   0.00    0.30       12.34485  0.4658690    9.393513
##   0.00    0.35       12.61952  0.4590835    9.612879
##   0.00    0.40       12.75663  0.4601371    9.686790
##   0.00    0.45       12.96171  0.4539654    9.774599
##   0.00    0.50       13.22572  0.4468568    9.900789
##   0.00    0.55       13.44341  0.4427435    9.994948
##   0.00    0.60       13.62322  0.4378770   10.071188
##   0.00    0.65       13.86382  0.4299241   10.153834
##   0.00    0.70       14.18080  0.4200728   10.374171
##   0.00    0.75       14.48298  0.4084205   10.465235
##   0.00    0.80       14.79949  0.3962019   10.559980
##   0.00    0.85       15.11740  0.3847540   10.634954
##   0.00    0.90       15.47863  0.3758087   10.864504
##   0.00    0.95       15.82572  0.3675399   11.039900
##   0.00    1.00       16.20635  0.3601753   11.220501
##   0.01    0.05       19.70369  0.5419408   13.658257
##   0.01    0.10       29.20414  0.5517070   18.690554
##   0.01    0.15       39.34872  0.5464403   24.503744
##   0.01    0.20       49.58867  0.5300114   30.253422
##   0.01    0.25       59.75378  0.5135458   35.825185
##   0.01    0.30       69.86658  0.5025866   41.376372
##   0.01    0.35       80.04013  0.4907761   46.959417
##   0.01    0.40       90.16964  0.4820232   52.431432
##   0.01    0.45      100.22907  0.4769870   57.896985
##   0.01    0.50      110.20374  0.4733163   63.339107
##   0.01    0.55      120.15283  0.4689652   68.748714
##   0.01    0.60      130.10820  0.4634918   74.138916
##   0.01    0.65      140.04834  0.4574078   79.532014
##   0.01    0.70      149.98967  0.4508075   84.943512
##   0.01    0.75      159.92961  0.4435334   90.354687
##   0.01    0.80      169.84077  0.4381988   95.756154
##   0.01    0.85      179.74379  0.4319106  101.148939
##   0.01    0.90      189.66445  0.4256649  106.576223
##   0.01    0.95      199.59443  0.4201282  112.076637
##   0.01    1.00      209.53137  0.4142206  117.588754
##   0.10    0.05       12.48366  0.5107258    9.539035
##   0.10    0.10       11.53534  0.5261893    8.482600
##   0.10    0.15       11.27266  0.5429020    8.204349
##   0.10    0.20       11.27554  0.5488762    8.346509
##   0.10    0.25       11.30648  0.5527622    8.491996
##   0.10    0.30       11.39070  0.5510568    8.624229
##   0.10    0.35       11.39403  0.5533536    8.686883
##   0.10    0.40       11.39420  0.5565505    8.705934
##   0.10    0.45       11.50017  0.5532039    8.805005
##   0.10    0.50       11.62477  0.5502964    8.891474
##   0.10    0.55       11.75005  0.5467109    8.971633
##   0.10    0.60       11.85638  0.5433706    9.039529
##   0.10    0.65       11.92754  0.5414808    9.064147
##   0.10    0.70       11.97002  0.5408816    9.061763
##   0.10    0.75       12.00539  0.5408888    9.064088
##   0.10    0.80       12.02698  0.5416069    9.059826
##   0.10    0.85       12.03704  0.5428213    9.068134
##   0.10    0.90       12.04706  0.5438642    9.086377
##   0.10    0.95       12.05637  0.5446054    9.098177
##   0.10    1.00       12.06264  0.5453869    9.100052
## 
## RMSE was used to select the optimal model using the smallest value.
## The final values used for the model were fraction = 0.15 and lambda = 0.1.
enet_predict <- predict(enetTune, test_fp)

postResample(enet_predict, test_perm)
##       RMSE   Rsquared        MAE 
## 11.5871592  0.3519218  7.2484765

The test set estimate of \(R^{2}\) is 0.3519218, which is higher than PLS model. The Elastic Net model has a lower RMSE than the PLS model, indicating that, on average, its predictions are closer to the observed values. The Elastic Net model has better predictive performance than the Partial Least Squares model across all three evaluated metrics.

  1. Would you recommend any of your models to replace the permeability laboratory experiment?

I would recommend the Elastic Net regression model as it produced better statistics. It had a higher R2 and lower RMSE and MAE.

Exercise 6.3:

A chemical manufacturing process for a pharmaceutical product was discussed in Sect. 1.4. In this problem, the objective is to understand the relationship between biological measurements of the raw materials (predictors), measurements of the manufacturing process (predictors), and the response of product yield. Biological predictors cannot be changed but can be used to assess the quality of the raw material before processing. On the other hand, manufacturing process predictors can be changed in the manufacturing process. Improving product yield by 1% will boost revenue by approximately one hundred thousand dollars per batch:

  1. Start R and use these commands to load the data:
library(AppliedPredictiveModeling)
data(ChemicalManufacturingProcess)

The matrix processPredictors contains the 57 predictors (12 describing the input biological material and 45 describing the process predictors) for the 176 manufacturing runs. yield contains the percent yield for each run.

  1. A small percentage of cells in the predictor set contain missing values. Use an imputation function to fill in these missing values (e.g., see Sect. 3.8).
sum(is.na(ChemicalManufacturingProcess))
## [1] 106
missings <- preProcess(ChemicalManufacturingProcess, method = "bagImpute")
Chemical <- predict(missings, ChemicalManufacturingProcess)

sum(is.na(Chemical))
## [1] 0

There were 106 missing values in ChemicalManufacturingProcess. Bagged trees were used to impute the data. Bagged trees are made using all the other variables.

  1. Split the data into a training and a test set, pre-process the data, and tune a model of your choice from this chapter. What is the optimal value of the performance metric?
# filtering low frequencies
Chemical <- Chemical[, -nearZeroVar(Chemical)]

set.seed(123)

# index for training
index <- createDataPartition(Chemical$Yield, p = .8, list = FALSE)

# train 
train_chem <- Chemical[index, ]

# test
test_chem <- Chemical[-index, ]

Tuning a model of my choice. Here I choose PLS model to process the data.

set.seed(123)

plsTune <- train(Yield ~ ., Chemical , method = "pls", 
             tuneLength = 20, trControl = ctrl, preProc = c("center", "scale"))

plot(plsTune) 

plsTune
## Partial Least Squares 
## 
## 176 samples
##  56 predictor
## 
## Pre-processing: centered (56), scaled (56) 
## Resampling: Cross-Validated (10 fold) 
## Summary of sample sizes: 158, 158, 159, 160, 160, 157, ... 
## Resampling results across tuning parameters:
## 
##   ncomp  RMSE      Rsquared   MAE     
##    1     1.488542  0.4318901  1.179867
##    2     1.907582  0.4397944  1.203172
##    3     1.383700  0.5302843  1.055955
##    4     1.575912  0.5121870  1.119931
##    5     1.803918  0.5225095  1.176135
##    6     1.911910  0.5100661  1.195775
##    7     2.009377  0.5065711  1.224336
##    8     2.020122  0.5086911  1.234196
##    9     2.142552  0.4894267  1.281860
##   10     2.214721  0.4826316  1.301670
##   11     2.255229  0.4735616  1.329375
##   12     2.194734  0.4779426  1.311638
##   13     2.105780  0.4798980  1.292012
##   14     2.089038  0.4773860  1.288809
##   15     2.108072  0.4687723  1.301333
##   16     2.127078  0.4641491  1.303687
##   17     2.132216  0.4579375  1.306510
##   18     2.175989  0.4530179  1.311188
##   19     2.258117  0.4456591  1.332741
##   20     2.290990  0.4381405  1.344932
## 
## RMSE was used to select the optimal model using the smallest value.
## The final value used for the model was ncomp = 3.

The optimal PLS model uses 3 components (ncomp = 3), as it results in the smallest RMSE value among all the tested models. The \(R^2\) is 0.5302843.

  1. Predict the response for the test set.What is the value of the performance metric and how does this compare with the resampled performance metric on the training set?
set.seed(123)

larsTune <- train(Yield ~ ., Chemical , method = "lars", metric = "Rsquared",
                    tuneLength = 20, trControl = ctrl, preProc = c("center", "scale"))

lars_predict <- predict(larsTune, test_chem[ ,-1])

postResample(lars_predict, test_chem[ ,1])
##      RMSE  Rsquared       MAE 
## 1.0867580 0.6596484 0.8315831

Ordinary linear model had the highest \(R^2\), but it comes with consequences. Therefore, the lars method was chosen as it had the highest \(R^2\) The \(R^2\) is 0.6596484, which is higher than the training set.

  1. Which predictors are most important in the model you have trained? Do either the biological or process predictors dominate the list?
varImp(larsTune)
## loess r-squared variable importance
## 
##   only 20 most important variables shown (out of 56)
## 
##                        Overall
## ManufacturingProcess32  100.00
## ManufacturingProcess13   90.02
## BiologicalMaterial06     84.56
## ManufacturingProcess36   75.52
## ManufacturingProcess17   74.88
## BiologicalMaterial03     73.53
## ManufacturingProcess09   70.37
## BiologicalMaterial12     67.97
## BiologicalMaterial02     65.32
## ManufacturingProcess31   60.45
## ManufacturingProcess06   57.36
## ManufacturingProcess33   49.53
## BiologicalMaterial11     48.11
## BiologicalMaterial04     47.12
## BiologicalMaterial08     41.87
## ManufacturingProcess11   41.75
## BiologicalMaterial01     39.13
## ManufacturingProcess12   32.97
## ManufacturingProcess30   32.87
## BiologicalMaterial09     32.41

The 5 most important variables used in the modeling are ManufacturingProcess32, ManufacturingProcess13, BiologicalMaterial06, ManufacturingProcess36, and ManufacturingProcess17. The process predictors dominate the list. The ratio of process to biological predictors is 11:9.

  1. Explore the relationships between each of the top predictors and the response. How could this information be helpful in improving yield in future runs of the manufacturing process?
top10 <- varImp(larsTune)$importance %>%
  arrange(-Overall) %>%
  head(10)


Chemical %>%
  select(c("Yield", row.names(top10))) %>%
  cor() %>%
  corrplot()

ManufacturingProcess32 has a significant positive relationship with Yield, as demonstrated by the correlation plot. Yield and three of the top ten variables have a negative correlation. Given that these were the variables influencing the Yield, this information can be helpful in subsequent manufacturing process runs. They should enhance their biological measures of the raw materials and production process measurements if they hope to increase or optimize their Yield.