## R Markdown

### Problem 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:

#### a) Start R and use these commands to load the data:

Load all fingerprints predictors are target variables. Confirming the number of observation are same for both predictors and target variable.

## Warning: package 'AppliedPredictiveModeling' was built under R version 4.0.3
## [1] 165
## [1] 165
## [1] 1107
## [1] 1

There are 165 observation with 1107 predictors and 1 target variable.

#### b) 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?

## Loading required package: lattice
## Loading required package: ggplot2
## [1] 388

After removing the low frequency variable, we have 388 variables in the predictor list. nearZeroVar will help in removing all variable with zero variance or which has very few unique values relative to the number of samples.

#### c) 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?

## Warning: package 'pls' was built under R version 4.0.3
##
## Attaching package: 'pls'
## The following object is masked from 'package:caret':
##
##     R2
## The following object is masked from 'package:stats':
##
##     loadings

plsmodel$results ## ncomp RMSE Rsquared MAE RMSESD RsquaredSD MAESD ## 1 1 13.55228 0.2889906 10.254144 1.970636 0.2661762 1.788984 ## 2 2 12.31313 0.4143172 8.752912 2.427410 0.2249563 1.800905 ## 3 3 11.96318 0.4350040 8.954915 2.523226 0.2328929 1.708100 ## 4 4 12.25871 0.4192916 9.427148 2.072124 0.1812841 1.433524 ## 5 5 12.03421 0.4585351 9.103037 2.433273 0.1901322 1.613929 ## 6 6 11.86129 0.4711070 8.938106 2.211652 0.1614947 1.895656 ## 7 7 11.61248 0.4945463 8.982042 1.887481 0.1305865 1.526076 ## 8 8 11.34235 0.5103917 8.972807 1.720665 0.1430526 1.395402 ## 9 9 11.27453 0.5145793 8.778225 1.652229 0.1404113 1.087510 ## 10 10 11.22447 0.5227839 8.628987 1.967255 0.1421539 1.349183 ## 11 11 11.53618 0.5061332 8.963147 2.034791 0.1384892 1.417809 ## 12 12 11.75938 0.4929538 9.040591 2.249614 0.1500689 1.602299 ## 13 13 12.12157 0.4754017 9.256365 2.268663 0.1518130 1.686388 ## 14 14 12.31703 0.4664117 9.444963 2.389078 0.1484254 1.732642 ## 15 15 12.59417 0.4428841 9.671427 2.873839 0.1694875 2.060419 ## 16 16 13.15339 0.4029050 9.902577 2.949224 0.1895255 2.065107 ## 17 17 13.38994 0.3919127 10.051180 3.398321 0.2004601 2.397669 ## 18 18 13.52542 0.3865039 10.180677 3.527901 0.2049565 2.592259 ## 19 19 13.72755 0.3689978 10.269883 3.421158 0.2163252 2.389133 ## 20 20 13.96971 0.3538878 10.368577 3.380498 0.2212317 2.443328 optimalcomp <- which.min(plsmodel$results[,'RMSE'])

The optimal number of component is

optimalcomp
## [1] 10

The RMSE and R-squared for the optimal number of component is

optimal_rmse <- plsmodel$results[optimalcomp, 'RMSE'] optimal_r2 <- plsmodel$results[optimalcomp, 'Rsquared']

optimal_rmse
## [1] 11.22447
optimal_r2
## [1] 0.5227839

#### d) Predict the response for the test set. What is the test set estimate of R2?

# Build a DF
trainingData <- as.data.frame(train_X)
trainingData$Perm <- train_Y plsFit <- plsr(Perm ~., data=trainingData, ncomp=optimalcomp) plsPred <- predict(plsFit, test_X, ncomp=optimalcomp) head(plsPred) ## , , 10 comps ## ## Perm ## 3 17.733048 ## 4 -4.657418 ## 12 4.323120 ## 13 2.679132 ## 21 14.398584 ## 26 23.222364 postResample(pred = plsPred, obs = test_Y) ## RMSE Rsquared MAE ## 12.485482 0.483895 9.965692 #### e) Try building other models discussed in this chapter. Do any have better predictive performance? I will use ridge, lasso and elastic net models. we will use glmnet package to use this models. when alpha = 0 , lasso component goes to 0 and ridge component would be active. when alpha = 1, ridge component goes to 0 and lasso component would be active. Lets start with ridge ## Warning: package 'glmnet' was built under R version 4.0.3 ## Loading required package: Matrix ## Loaded glmnet 4.0-2 ## RMSE Rsquared MAE ## 11.3731538 0.6094073 9.0394251 Next, We will try Lasso lasso.fit <- cv.glmnet(train_X, train_Y, type.measure = "mse", alpha= 1 , family="gaussian" ) lasso.pred <- predict(lasso.fit, s=lasso.fit$lambda.1se, newx = test_X)
postResample(pred = lasso.pred, obs = test_Y)
##      RMSE  Rsquared       MAE
## 12.003534  0.448561  9.330812

Elastic net. I tried different values of alpha between 0 and 1. the optimal value found to be 0.5 with R-square 0.5

elasticnet.fit <- cv.glmnet(train_X, train_Y, type.measure = "mse", alpha= 0.5 , family="gaussian" )
elasticnet.pred <- predict(lasso.fit, s=elasticnet.fit$lambda.1se, newx = test_X) postResample(pred = elasticnet.pred, obs = test_Y) ## RMSE Rsquared MAE ## 12.3168372 0.6202373 9.6740307 #### f) Would you recommend any of your models to replace the permeability laboratory experiment? Based on results from other models, ridge regression model yields a better R-square and RMSE. ### Problem 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: #### a) Start R and use these commands to load the data: Lets look at the sample data from the dataframe. We see there are 58 columns and the target variable is Yield. head(ChemicalManufacturingProcess) ## 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 2.0 ncol(ChemicalManufacturingProcess) ## [1] 58 Lets do some preprocessing and clean the data. As part of it, we will see if there is any missing values. There are 176 rows in this dataset. Out of that 24 rows has NAs. There are 106 total NA occurances in the data set. Total number of Observation. nrow(ChemicalManufacturingProcess) ## [1] 176 #### b) 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). There are 176 rows in this dataset. Out of that 24 rows has NAs. There are 106 total NA occurances in the data set. Total number of NAs length(which(is.na(ChemicalManufacturingProcess))) ## [1] 106 Total number of rows with NA length(which(!complete.cases(ChemicalManufacturingProcess))) ## [1] 24 Lets impute the missing data impute <- preProcess(ChemicalManufacturingProcess[,-c(1)], method=c('bagImpute')) imputed <- predict(impute, ChemicalManufacturingProcess[,-c(1)]) #### c) 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? Lets try some preprocessing and remove all non-significant predicators. ## Warning: package 'car' was built under R version 4.0.3 ## Loading required package: carData ## [1] "BiologicalMaterial07" BiologicalMaterial07 is having low variance. we will remove this predictor variable. now lets remove some multi colinearity by removing pair wise correlated variables. correlations <- cor(imputed_lowvar) highCorr <- findCorrelation(correlations, names=TRUE, cutoff=0.9) (highCorr) ## [1] "BiologicalMaterial02" "BiologicalMaterial04" "BiologicalMaterial12" ## [4] "ManufacturingProcess29" "ManufacturingProcess42" "ManufacturingProcess27" ## [7] "ManufacturingProcess25" "ManufacturingProcess31" "ManufacturingProcess18" ## [10] "ManufacturingProcess40" highCorr <- findCorrelation(correlations, cutoff=0.9) imputed <- imputed_lowvar[,-highCorr] highCorr <- findCorrelation(correlations, cutoff=0.9) imputed <- imputed_lowvar[,-highCorr] Splitting the dataset into training and test data sets. # Train/test plitting data, 25% testing set.seed(1) trainRow <- createDataPartition(ChemicalManufacturingProcess$Yield, p=0.75, list=FALSE)
trainData <- imputed[trainRow, ]
trainData_X <- imputed[trainRow, ]

trainData$Yield <- ChemicalManufacturingProcess[trainRow, ]$Yield
testData <- imputed[-trainRow, ]
testData_X <- imputed[-trainRow, ]
testData$Yield <- ChemicalManufacturingProcess[-trainRow, ]$Yield

Fiting a PLS model and find performance variables.

plsmodel<- train(trainData_X, trainData$Yield, method="pls", tuneLength=20, trControl = cv_cntrl) plot(plsmodel) plsmodel$results
##    ncomp     RMSE  Rsquared      MAE     RMSESD RsquaredSD     MAESD
## 1      1 1.627390 0.2012777 1.311968  0.2542385  0.2201262 0.2308777
## 2      2 1.621468 0.2050440 1.305967  0.2493284  0.2191528 0.2281001
## 3      3 2.301524 0.1669201 1.588843  2.2200509  0.2054500 0.9613340
## 4      4 5.223241 0.1775648 2.630489 10.9359216  0.1823202 4.0247246
## 5      5 8.510862 0.1782906 3.612193 14.7463940  0.1851440 5.1450703
## 6      6 7.583476 0.2029144 3.199323 13.0670376  0.2048655 4.1587077
## 7      7 5.830914 0.2908409 2.516072  9.7857047  0.2248824 2.9892242
## 8      8 4.922830 0.2980774 2.136187  7.7220214  0.2308815 2.2661297
## 9      9 4.857704 0.3890628 2.220807  7.6416609  0.2654233 2.6278505
## 10    10 4.113983 0.4732698 1.841474  6.4016219  0.2620997 1.9637427
## 11    11 3.569480 0.4835020 1.661748  5.3442612  0.2567280 1.6282935
## 12    12 3.395403 0.5266141 1.685316  4.9416988  0.2816553 1.7339094
## 13    13 3.403486 0.5321589 1.579610  5.1800569  0.2877901 1.5884963
## 14    14 2.906176 0.5558700 1.448003  4.3328315  0.2742609 1.3675248
## 15    15 2.605877 0.5557473 1.381274  3.5726329  0.2752953 1.2004795
## 16    16 2.975488 0.5578095 1.508461  4.3445101  0.2884171 1.4684815
## 17    17 3.095662 0.5529388 1.540630  4.6574418  0.2880221 1.5173538
## 18    18 3.635812 0.5320253 1.709054  5.6021668  0.2861593 1.8298733
## 19    19 3.745813 0.5365047 1.748883  5.8128208  0.2846875 1.9114805
## 20    20 3.843954 0.5323276 1.799170  5.9539847  0.2924349 2.0146958
optimalcomp <- which.min(plsmodel$results[,'RMSE']) optimal_rmse <- plsmodel$results[optimalcomp, 'RMSE']
optimal_r2 <- plsmodel$results[optimalcomp, 'Rsquared'] optimal_rmse ## [1] 1.621468 optimal_r2 ## [1] 0.205044 From the PLS model, we get the really low R-square. This indicate the model only capture 20% of the variability of the predictors. #### d) Predict the response for the test set. What is the value of the performance metric and how does this compare with the resampled performance metric on the training set? plsFit <- plsr(Yield ~., data=trainData, ncomp=optimalcomp) plsPred <- predict(plsFit, testData_X, ncomp=optimalcomp) postResample(pred = plsPred, obs = testData$Yield)
##       RMSE   Rsquared        MAE
## 1.97180761 0.06492146 1.60536298

It seems r-square is really poor when it comes to the test results. The PLS model is not that great in this case. we may need to try some other models like ridge, lasso and elastic net.

Lets try with ridge first

ridge.fit <- cv.glmnet(as.matrix(trainData_X), trainData$Yield, type.measure = "mse", alpha= 0 , family="gaussian" ) ridge.pred <- predict(ridge.fit, s=ridge.fit$lambda.1se, newx = as.matrix(testData_X))
postResample(pred = ridge.pred, obs = testData$Yield) ## RMSE Rsquared MAE ## 1.7643785 0.5897654 1.4074020 Next, lets tru with lasso lasso.fit <- cv.glmnet(as.matrix(trainData_X), trainData$Yield, type.measure = "mse", alpha=1 , family="gaussian" )

lasso.pred <- predict(lasso.fit, s=lasso.fit$lambda.1se, newx = as.matrix(testData_X)) postResample(pred = lasso.pred, obs = testData$Yield)
##      RMSE  Rsquared       MAE
## 1.3091127 0.6179137 1.0385829

Finally, we will try with elastic net

ELastic net

elasticnet.fit <- cv.glmnet(as.matrix(trainData_X), trainData$Yield, type.measure = "mse", alpha= 0.5 , family="gaussian" ) elasticnet.pred <- predict(lasso.fit, s=elasticnet.fit$lambda.1se, newx = as.matrix(testData_X))
postResample(pred = elasticnet.pred, obs = testData$Yield) ## RMSE Rsquared MAE ## 1.461456 0.587580 1.177213 ####e ) Which predictors are most important in the model you have trained? Do either the biological or process predictors dominate the list? lassoImp <- caret::getModelInfo("glmnet")$glmnet$varImp(lasso.fit, lambda = lasso.fit$lambda.1se , scale = FALSE)
index<- which(lassoImp$Overall > .001) impvariable <- as.data.frame(lassoImp[index, 0]) impvariable$overall <- lassoImp[index, 1]
impvariable
##                             overall
## BiologicalMaterial03    0.030316362
## ManufacturingProcess04  0.008431423
## ManufacturingProcess06  0.017527652
## ManufacturingProcess09  0.290011611
## ManufacturingProcess11  0.076299732
## ManufacturingProcess13  0.190783345
## ManufacturingProcess17  0.084000774
## ManufacturingProcess32  0.167682810
## ManufacturingProcess34  2.480230512
## ManufacturingProcess36 15.270745914
## ManufacturingProcess37  0.127481831
## ManufacturingProcess39  0.033635041
## ManufacturingProcess45  0.208210924

From the important variables, it looks like manufacturing variables dominates the biological predictors

####f) 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?

None of these variables are yielding a direct linear relatioship to the target variables. However it is more clear that the biological predictors dont have much significance towards the target variable.

featurePlot(x = ChemicalManufacturingProcess['ManufacturingProcess34'],
y = ChemicalManufacturingProcess$Yield, plot = "scatter", type = c("p", 'smooth'), span = .5, pch = 20) ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## pseudoinverse used at 2.6015 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## neighborhood radius 0.1015 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## reciprocal condition number 0 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## There are other near singularities as well. 0.01 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## zero-width neighborhood. make span bigger ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## pseudoinverse used at 2.6015 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## neighborhood radius 0.1015 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## reciprocal condition number 0 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## There are other near singularities as well. 0.01 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## zero-width neighborhood. make span bigger ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## pseudoinverse used at 2.6015 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## neighborhood radius 0.1015 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## reciprocal condition number 0 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## There are other near singularities as well. 0.01 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## zero-width neighborhood. make span bigger ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## pseudoinverse used at 2.6015 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## neighborhood radius 0.1015 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## reciprocal condition number 0 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## There are other near singularities as well. 0.01 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## zero-width neighborhood. make span bigger ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## pseudoinverse used at 2.6015 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## neighborhood radius 0.1015 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## reciprocal condition number 0 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## There are other near singularities as well. 0.01 ## Warning in simpleLoess(y, x, w, span, degree = degree, parametric = FALSE, : ## zero-width neighborhood. make span bigger featurePlot(x = ChemicalManufacturingProcess['ManufacturingProcess09'], y = ChemicalManufacturingProcess$Yield,
plot = "scatter",
type = c("p", 'smooth'),
span = .5,
pch = 20)

featurePlot(x = ChemicalManufacturingProcess['ManufacturingProcess13'],
y = ChemicalManufacturingProcess$Yield, plot = "scatter", type = c("p", 'smooth'), span = .5, pch = 20) featurePlot(x = ChemicalManufacturingProcess['ManufacturingProcess32'], y = ChemicalManufacturingProcess$Yield,
plot = "scatter",
type = c("p", 'smooth'),
span = .5,
pch = 20)

featurePlot(x = ChemicalManufacturingProcess['ManufacturingProcess45'],
y = ChemicalManufacturingProcess$Yield, plot = "scatter", type = c("p", 'smooth'), span = .5, pch = 20) featurePlot(x = ChemicalManufacturingProcess['ManufacturingProcess17'], y = ChemicalManufacturingProcess$Yield,
plot = "scatter",
type = c("p", 'smooth'),
span = .5,
pch = 20)