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:
Start R and use these commands to load the data:
library(AppliedPredictiveModeling)
data(permeability)
summary(permeability)## permeability
## Min. : 0.06
## 1st Qu.: 1.55
## Median : 4.91
## Mean :12.24
## 3rd Qu.:15.47
## Max. :55.60The matrix fingerprints contains the 1,107 binary molecular predictors for the 165 compounds, while permeability contains permeability response.
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 1107fingerprints2 <- fingerprints[, -nearZeroVar(fingerprints)]
dim(fingerprints2)## [1] 165 388Filtering out those predictors is reducing our matrix from 165x1107 to 65x388 which would be better for building a model.
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 R2?
set.seed(300)
trainingRows <- createDataPartition(permeability, p = 0.8, list = FALSE)
trainPredictors <- fingerprints2[trainingRows, ]
trainClasses <- permeability[trainingRows]
testPredictors <- fingerprints2[-trainingRows, ]
testClasses <- permeability[-trainingRows]set.seed(300)
plsTune <- train(trainPredictors, trainClasses,
method = "pls",
tuneLength = 10,
trControl = trainControl(method = "repeatedcv"),
preProc = c("center", "scale"))
plsTune$results## ncomp RMSE Rsquared MAE RMSESD RsquaredSD MAESD
## 1 1 13.02635 0.3221317 9.912382 3.028684 0.2478879 2.608384
## 2 2 11.85397 0.4194745 8.273539 2.372950 0.2925410 1.634590
## 3 3 11.54114 0.4707489 8.636745 1.847878 0.2458678 1.625189
## 4 4 11.43967 0.4886680 8.615876 1.492251 0.2296421 1.522032
## 5 5 11.27615 0.5052312 8.171202 1.873386 0.2319873 1.440496
## 6 6 11.30242 0.4940627 8.276156 1.881103 0.2303962 1.345800
## 7 7 11.32204 0.4815905 8.579845 1.793302 0.2138101 1.407066
## 8 8 11.38629 0.4749483 8.776290 1.843288 0.2122199 1.409170
## 9 9 11.49191 0.4736676 8.697322 1.843241 0.1973815 1.485186
## 10 10 11.73679 0.4631460 8.759930 1.656584 0.1934454 1.523329The best R^2 value is with ncomp = 5.
Predict the response for the test set. What is the test set estimate of R2?
output <- predict(plsTune, testPredictors, ncomp = 5)
postResample(pred = output, obs = testClasses)## RMSE Rsquared MAE
## 13.6757568 0.3000775 9.8477466R^2 of the test set is 0.3.
Try building other models discussed in this chapter. Do any have better predictive performance?
I will try several models discussed in the reading and use RMSE to identify the best performance.
df_train <- as.data.frame(cbind(trainPredictors,trainClasses))
df_test <- as.data.frame(cbind(testPredictors,testClasses))
set.seed(300)
lmFitAllPredictors <- lm(trainClasses ~ ., df_train)
predict.lm <- (predict(lmFitAllPredictors, df_test))## Warning in predict.lm(lmFitAllPredictors, df_test): prediction from a rank-
## deficient fit may be misleadingpostResample(pred = predict.lm, obs = df_test$testClasses)## RMSE Rsquared MAE
## 35.74087092 0.02768807 20.39171249Linear model is not performing as well as the PLS model - the RMSE is significantly higher.
set.seed(300)
suppressWarnings(
rlmpred <- train(trainPredictors, trainClasses,
method = "rlm",
trControl = trainControl(method = "repeatedcv"),
preProc = "pca")
)
rlmpred## Robust Linear Model
##
## 133 samples
## 388 predictors
##
## Pre-processing: principal component signal extraction (388),
## centered (388), scaled (388)
## Resampling: Cross-Validated (10 fold, repeated 1 times)
## Summary of sample sizes: 119, 119, 120, 120, 120, 121, ...
## Resampling results across tuning parameters:
##
## intercept psi RMSE Rsquared MAE
## FALSE psi.huber 16.72655 0.5061180 13.169903
## FALSE psi.hampel 16.67739 0.5067777 13.418069
## FALSE psi.bisquare 16.98977 0.4451603 13.293771
## TRUE psi.huber 11.32632 0.5126722 7.783743
## TRUE psi.hampel 12.21488 0.4391843 8.190154
## TRUE psi.bisquare 13.04293 0.4176581 7.883718
##
## RMSE was used to select the optimal model using the smallest value.
## The final values used for the model were intercept = TRUE and psi
## = psi.huber.predict.rlm <- (predict(rlmpred, testPredictors))
postResample(pred = predict.rlm, obs = df_test$testClasses)## RMSE Rsquared MAE
## 12.5415135 0.3427535 9.0255228Robust Linear model is performing much better than regular Lineal Model and is slightly better than PLS model - the RMSE is slightly lower.
enetpred <- train(trainPredictors, trainClasses,
method='enet',
metric='RMSE',
trControl = trainControl(method = "repeatedcv"),
preProcess=c('center','scale'))
predict.enet <- (predict(enetpred, testPredictors))
postResample(pred = predict.enet, obs = df_test$testClasses)## RMSE Rsquared MAE
## 11.5742116 0.4267167 7.9885963ENET model is performing much better than regular Lineal Model and is slightly better than the Robust Linear Model and PLS model - the RMSE is slightly lower.
Overall PLS, Robust Linear Model and ENET models have a good performance overall and ENET model is the best.
Would you recommend any of your models to replace the permeability laboratory experiment?
ENET Model is performing well so I would recommend using it to replace permeability labaratory experiment or at least be used together with the exeriment to make it more efficient and cost effective.
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), 6.5 Computing 139 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:
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.
df1<-ChemicalManufacturingProcess
dim(df1)## [1] 176 58head(df1)## Yield BiologicalMaterial01 BiologicalMaterial02 BiologicalMaterial03
## 1 38.00 6.25 49.58 56.97
## 2 42.44 8.01 60.97 67.48
## 3 42.03 8.01 60.97 67.48
## 4 41.42 8.01 60.97 67.48
## 5 42.49 7.47 63.33 72.25
## 6 43.57 6.12 58.36 65.31
## BiologicalMaterial04 BiologicalMaterial05 BiologicalMaterial06
## 1 12.74 19.51 43.73
## 2 14.65 19.36 53.14
## 3 14.65 19.36 53.14
## 4 14.65 19.36 53.14
## 5 14.02 17.91 54.66
## 6 15.17 21.79 51.23
## BiologicalMaterial07 BiologicalMaterial08 BiologicalMaterial09
## 1 100 16.66 11.44
## 2 100 19.04 12.55
## 3 100 19.04 12.55
## 4 100 19.04 12.55
## 5 100 18.22 12.80
## 6 100 18.30 12.13
## BiologicalMaterial10 BiologicalMaterial11 BiologicalMaterial12
## 1 3.46 138.09 18.83
## 2 3.46 153.67 21.05
## 3 3.46 153.67 21.05
## 4 3.46 153.67 21.05
## 5 3.05 147.61 21.05
## 6 3.78 151.88 20.76
## ManufacturingProcess01 ManufacturingProcess02 ManufacturingProcess03
## 1 NA NA NA
## 2 0.0 0 NA
## 3 0.0 0 NA
## 4 0.0 0 NA
## 5 10.7 0 NA
## 6 12.0 0 NA
## ManufacturingProcess04 ManufacturingProcess05 ManufacturingProcess06
## 1 NA NA NA
## 2 917 1032.2 210.0
## 3 912 1003.6 207.1
## 4 911 1014.6 213.3
## 5 918 1027.5 205.7
## 6 924 1016.8 208.9
## ManufacturingProcess07 ManufacturingProcess08 ManufacturingProcess09
## 1 NA NA 43.00
## 2 177 178 46.57
## 3 178 178 45.07
## 4 177 177 44.92
## 5 178 178 44.96
## 6 178 178 45.32
## ManufacturingProcess10 ManufacturingProcess11 ManufacturingProcess12
## 1 NA NA NA
## 2 NA NA 0
## 3 NA NA 0
## 4 NA NA 0
## 5 NA NA 0
## 6 NA NA 0
## ManufacturingProcess13 ManufacturingProcess14 ManufacturingProcess15
## 1 35.5 4898 6108
## 2 34.0 4869 6095
## 3 34.8 4878 6087
## 4 34.8 4897 6102
## 5 34.6 4992 6233
## 6 34.0 4985 6222
## ManufacturingProcess16 ManufacturingProcess17 ManufacturingProcess18
## 1 4682 35.5 4865
## 2 4617 34.0 4867
## 3 4617 34.8 4877
## 4 4635 34.8 4872
## 5 4733 33.9 4886
## 6 4786 33.4 4862
## ManufacturingProcess19 ManufacturingProcess20 ManufacturingProcess21
## 1 6049 4665 0.0
## 2 6097 4621 0.0
## 3 6078 4621 0.0
## 4 6073 4611 0.0
## 5 6102 4659 -0.7
## 6 6115 4696 -0.6
## ManufacturingProcess22 ManufacturingProcess23 ManufacturingProcess24
## 1 NA NA NA
## 2 3 0 3
## 3 4 1 4
## 4 5 2 5
## 5 8 4 18
## 6 9 1 1
## ManufacturingProcess25 ManufacturingProcess26 ManufacturingProcess27
## 1 4873 6074 4685
## 2 4869 6107 4630
## 3 4897 6116 4637
## 4 4892 6111 4630
## 5 4930 6151 4684
## 6 4871 6128 4687
## ManufacturingProcess28 ManufacturingProcess29 ManufacturingProcess30
## 1 10.7 21.0 9.9
## 2 11.2 21.4 9.9
## 3 11.1 21.3 9.4
## 4 11.1 21.3 9.4
## 5 11.3 21.6 9.0
## 6 11.4 21.7 10.1
## ManufacturingProcess31 ManufacturingProcess32 ManufacturingProcess33
## 1 69.1 156 66
## 2 68.7 169 66
## 3 69.3 173 66
## 4 69.3 171 68
## 5 69.4 171 70
## 6 68.2 173 70
## ManufacturingProcess34 ManufacturingProcess35 ManufacturingProcess36
## 1 2.4 486 0.019
## 2 2.6 508 0.019
## 3 2.6 509 0.018
## 4 2.5 496 0.018
## 5 2.5 468 0.017
## 6 2.5 490 0.018
## ManufacturingProcess37 ManufacturingProcess38 ManufacturingProcess39
## 1 0.5 3 7.2
## 2 2.0 2 7.2
## 3 0.7 2 7.2
## 4 1.2 2 7.2
## 5 0.2 2 7.3
## 6 0.4 2 7.2
## ManufacturingProcess40 ManufacturingProcess41 ManufacturingProcess42
## 1 NA NA 11.6
## 2 0.1 0.15 11.1
## 3 0.0 0.00 12.0
## 4 0.0 0.00 10.6
## 5 0.0 0.00 11.0
## 6 0.0 0.00 11.5
## ManufacturingProcess43 ManufacturingProcess44 ManufacturingProcess45
## 1 3.0 1.8 2.4
## 2 0.9 1.9 2.2
## 3 1.0 1.8 2.3
## 4 1.1 1.8 2.1
## 5 1.1 1.7 2.1
## 6 2.2 1.8 2.0A 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).
I will use KNN method to impute the missing values - it will not be too computationally intense since the dataset is not that large.
df1pp <- preProcess(df1[,2:ncol(df1)], method=c('knnImpute'))
df1 <- cbind(df1$Yield,predict(df1pp, df1[,2:ncol(df1)]))
colnames(df1)[1] <- "Yield"Split the data into a training and a test set, pre-process the data, and tune a model of your choice from this chapter. What is the optimal value of the performance metric?
Splitting the data into test and train is the first step - I will use 80%/20% split.
set.seed(333)
trainingRows <- createDataPartition(df1$Yield, p = 0.8, list = FALSE)
df1_train <- df1[trainingRows, ]
df1_test <- df1[-trainingRows, ]I will tune an ENET model for this data:
df1_enet <- train(df1_train[,2:ncol(df1_train)], df1_train$Yield,
method='enet',
metric='RMSE',
trControl = trainControl(method = "repeatedcv"),
preProcess=c('center','scale'))
predict.df1_enet <- (predict(df1_enet, df1_train[,2:ncol(df1_train)]))
postResample(pred = predict.df1_enet, obs = df1_train$Yield)## RMSE Rsquared MAE
## 1.2910514 0.6032864 1.0545831Training set RMSE is 1.291
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?
Now let’s predict the test set responses.
predict.df1_enet <- (predict(df1_enet, df1_test[,2:ncol(df1_test)]))
postResample(pred = predict.df1_enet, obs = df1_test$Yield)## RMSE Rsquared MAE
## 1.3641651 0.6637226 1.0725533The RMSE is slighlty higher than train set RMSE and R^2 is a bit lower - but overall looks pretty good.
Which predictors are most important in the model you have trained? Do either the biological or process predictors dominate the list?
Here are the most important predictors for my model:
library(elasticnet)
enetCoef<- predict(df1_enet$finalModel, newx = as.matrix(df1_test[,2:ncol(df1_test)]),
s = .1, mode = "fraction",
type = "coefficients")
head(sort(enetCoef$coefficients),5)## ManufacturingProcess13 ManufacturingProcess37 ManufacturingProcess17
## -0.25153249 -0.11488070 -0.06413573
## ManufacturingProcess36 ManufacturingProcess24
## -0.05316984 -0.02116651tail(sort(enetCoef$coefficients),5)## ManufacturingProcess06 ManufacturingProcess15 ManufacturingProcess34
## 0.1034313 0.1581922 0.1798552
## ManufacturingProcess09 ManufacturingProcess32
## 0.5102941 0.7201123All the most important coefficients are Process predictors, there is not a single biological predictors on the list. Process 32, 9, 34, 15 and 13 are the stongest predictors.
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?
lm <- lm(Yield ~ ManufacturingProcess32+ManufacturingProcess09+ManufacturingProcess34+ManufacturingProcess13+ManufacturingProcess15, data=df1_train)
summary(lm)##
## Call:
## lm(formula = Yield ~ ManufacturingProcess32 + ManufacturingProcess09 +
## ManufacturingProcess34 + ManufacturingProcess13 + ManufacturingProcess15,
## data = df1_train)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.6296 -0.7670 0.0493 0.7040 3.2429
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 40.14021 0.09213 435.676 < 2e-16 ***
## ManufacturingProcess32 0.87274 0.10091 8.649 1.19e-14 ***
## ManufacturingProcess09 0.45736 0.15924 2.872 0.004721 **
## ManufacturingProcess34 0.30862 0.09032 3.417 0.000832 ***
## ManufacturingProcess13 -0.50055 0.17206 -2.909 0.004225 **
## ManufacturingProcess15 0.34418 0.10932 3.148 0.002012 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.105 on 138 degrees of freedom
## Multiple R-squared: 0.6521, Adjusted R-squared: 0.6395
## F-statistic: 51.73 on 5 and 138 DF, p-value: < 2.2e-16Manufacturing process 32, 09 and 34, 13 and 15 create a good linear model which predicts 65.2% of variance in the data. This informaiton will be helpfil in improving yield since adjustments to these processes could make a significant difference to the Yield outcome.