Data624_Assignment7
Christian Uriostegui
2024-11-08
Question 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.
Part A
Start R and use these commands to load the data:
The matrix fingerprints contains the 1,107 binary molecular predictors for the 165 compounds, while permeability contains permeability response.
Part 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?
Response
When using the nearZeroVar function, it reduces the predictors from 182655 to 705.
nearZeroVar(fingerprints)## [1] 7 8 9 10 13 14 17 18 19 22 23 24 30 31 32
## [16] 33 34 45 77 81 82 83 84 85 89 90 91 92 95 100
## [31] 104 105 106 107 109 110 112 113 114 115 116 117 119 120 122
## [46] 123 124 128 131 132 134 135 136 137 139 140 144 145 147 148
## [61] 149 151 155 160 161 164 165 166 216 217 218 219 220 222 243
## [76] 252 259 273 275 277 282 283 287 288 289 292 346 347 348 349
## [91] 350 351 352 353 354 363 364 365 369 375 379 384 391 393 397
## [106] 399 402 404 405 407 408 409 410 411 412 413 414 415 416 417
## [121] 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432
## [136] 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447
## [151] 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462
## [166] 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477
## [181] 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492
## [196] 493 494 495 498 500 501 502 513 523 525 526 527 528 530 531
## [211] 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546
## [226] 547 548 550 552 555 562 563 564 566 567 569 570 572 575 578
## [241] 579 580 581 582 583 584 585 586 587 588 589 596 605 606 607
## [256] 608 609 610 611 612 614 615 616 617 618 619 620 622 623 624
## [271] 625 626 627 628 629 630 631 632 633 634 635 636 637 638 639
## [286] 640 641 642 643 644 645 646 647 648 649 650 651 652 653 654
## [301] 655 656 657 658 659 660 661 662 663 664 665 666 667 668 669
## [316] 670 671 672 673 674 675 676 677 678 680 681 682 683 684 685
## [331] 686 687 688 689 690 691 692 693 694 695 696 697 706 707 708
## [346] 709 710 711 712 713 714 715 716 717 718 720 721 722 723 724
## [361] 725 726 727 728 729 730 731 734 735 736 737 738 739 740 741
## [376] 742 743 744 745 746 747 748 749 756 757 758 759 760 761 762
## [391] 763 764 765 766 767 768 769 770 771 772 777 778 779 781 783
## [406] 784 785 786 787 788 789 790 791 794 796 797 799 802 803 804
## [421] 807 808 809 810 811 814 815 816 817 818 819 820 821 822 823
## [436] 824 825 826 827 828 829 830 831 832 833 834 835 836 837 838
## [451] 839 840 841 842 843 844 845 846 847 848 849 850 851 852 853
## [466] 854 855 856 857 858 859 860 861 862 863 864 865 866 867 868
## [481] 869 870 871 872 873 874 875 876 877 878 879 880 881 882 883
## [496] 884 885 886 887 888 889 890 891 892 893 894 895 896 897 898
## [511] 899 900 901 902 903 904 905 906 907 908 909 910 911 912 913
## [526] 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928
## [541] 929 930 931 932 933 934 935 936 937 938 939 940 941 942 943
## [556] 944 945 946 947 948 949 950 951 952 953 954 955 956 957 958
## [571] 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973
## [586] 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988
## [601] 989 990 991 992 993 994 995 996 997 998 999 1000 1001 1002 1003
## [616] 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018
## [631] 1019 1020 1021 1022 1023 1024 1025 1026 1027 1028 1029 1030 1031 1032 1033
## [646] 1034 1035 1036 1037 1038 1039 1040 1041 1042 1043 1044 1045 1046 1047 1048
## [661] 1049 1050 1051 1052 1053 1054 1055 1056 1057 1058 1059 1060 1061 1062 1063
## [676] 1064 1065 1066 1067 1068 1069 1070 1071 1072 1073 1074 1075 1076 1077 1078
## [691] 1079 1080 1081 1082 1083 1084 1085 1086 1087 1088 1089 1090 1091 1092 1093
## [706] 1094 1095 1096 1097 1098 1099 1100 1101 1102 1103 1104 1105 1106 1107Part 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?
Response
Since the values in our dataset are binary, comprised of 0s and 1s, centering or scaling are not necessary.
Based on the results of our PLS model, it determined that the optimal number of latent variables is 5. The corresponding R squared is 0.5311563
permeability_data <- data.frame(permeability = permeability, fingerprints)nzv <- nearZeroVar(permeability_data)
permeability_data_filtered <- permeability_data[, -nzv]set.seed(123)
trainIndex <- createDataPartition(permeability_data_filtered$permeability, p = 0.75, list = FALSE)
trainData <- permeability_data_filtered[trainIndex, ]
testData <- permeability_data_filtered[-trainIndex, ]# Set up cross-validation 10-fold
ctrl <- trainControl(method = "cv", number = 10) # 10-fold set.seed(123)
# Train a PLS model on the training data
plsModel <- train(
permeability ~ .,
data = trainData,
method = "pls",
tuneLength = 10,
trControl = ctrl
)
# Summarize the model
print(plsModel)## Partial Least Squares
##
## 125 samples
## 388 predictors
##
## No pre-processing
## Resampling: Cross-Validated (10 fold)
## Summary of sample sizes: 111, 112, 113, 113, 113, 113, ...
## Resampling results across tuning parameters:
##
## ncomp RMSE Rsquared MAE
## 1 13.87785 0.3223096 10.642327
## 2 11.93335 0.4605485 8.632442
## 3 11.57889 0.4697451 8.606836
## 4 11.58383 0.4890124 8.852139
## 5 11.34660 0.5199843 8.451817
## 6 11.16105 0.5509412 8.091074
## 7 11.32849 0.5514611 8.261755
## 8 11.62200 0.5392276 8.876021
## 9 12.10421 0.5174443 9.181724
## 10 12.58414 0.4854590 9.611658
##
## RMSE was used to select the optimal model using the smallest value.
## The final value used for the model was ncomp = 6.Part D
Predict the response for the test set. What is the test set estimate of R2?
Response
On the test set, the PLS model performs poorly, resulting in an R squared of 0.3593983.
# Evaluate the model on the test set
plsPred <- predict(plsModel, newdata = testData)
# Calculate performance metrics (for regression)
postResample(pred = plsPred, obs = testData$permeability)## RMSE Rsquared MAE
## 11.6969547 0.3781227 8.0737779Part E
Try building other models discussed in this chapter. Do any have better predictive performance?
Response
The OLR model perfomed the best across the RMSE, R squared and MAE. The R^2 squared was 0.378, meaning that it explains more variance in the target variable compared to PLS (0.359) and the PCR model (0.263). It also has the least prediction error.
# PCR model
set.seed(123)
pcrModel <- train(
permeability ~ .,
data = trainData,
method = "pcr",
tuneLength = 10,
trControl = ctrl,
)
print(pcrModel)## Principal Component Analysis
##
## 125 samples
## 388 predictors
##
## No pre-processing
## Resampling: Cross-Validated (10 fold)
## Summary of sample sizes: 111, 112, 113, 113, 113, 113, ...
## Resampling results across tuning parameters:
##
## ncomp RMSE Rsquared MAE
## 1 15.24104 0.1770015 11.825043
## 2 15.22293 0.1775454 11.920684
## 3 14.04574 0.3542879 10.794504
## 4 12.01014 0.4465492 8.892864
## 5 12.09482 0.4472377 8.859232
## 6 12.41015 0.4294419 9.092347
## 7 12.36056 0.4288184 9.047108
## 8 12.22602 0.4379938 9.053613
## 9 12.20281 0.4375291 9.004970
## 10 12.18693 0.4332150 9.020523
##
## RMSE was used to select the optimal model using the smallest value.
## The final value used for the model was ncomp = 4.# Train an OLR model
set.seed(123)
olrModel <- plsr(
permeability ~ .,
data = trainData,
ncomp = 10,
validation = "CV",
method = "oscorespls"
)
summary(olrModel)## Data: X dimension: 125 388
## Y dimension: 125 1
## Fit method: oscorespls
## Number of components considered: 10
##
## VALIDATION: RMSEP
## Cross-validated using 10 random segments.
## (Intercept) 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps
## CV 16.09 14.01 12.09 11.66 11.98 11.83 11.62
## adjCV 16.09 13.99 12.06 11.61 11.86 11.68 11.42
## 7 comps 8 comps 9 comps 10 comps
## CV 11.74 12.03 12.68 12.94
## adjCV 11.54 11.78 12.37 12.60
##
## TRAINING: % variance explained
## 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps 7 comps
## X 32.64 44.55 51.16 57.20 65.26 67.47 70.38
## permeability 27.59 50.88 58.95 66.19 70.50 75.39 76.99
## 8 comps 9 comps 10 comps
## X 72.23 74.55 76.90
## permeability 79.24 81.01 82.47Test PCR
pcrPred <- predict(pcrModel, newdata = testData)
pcrPerformance <- postResample(pred = pcrPred, obs = testData$permeability)
print(pcrPerformance)## RMSE Rsquared MAE
## 12.2976342 0.2631962 8.6566083Test OLR
olrPred <- predict(olrModel, newdata = testData, ncomp = 6)
olrPerformance <- postResample(pred = olrPred, obs = testData$permeability)
print(olrPerformance)## RMSE Rsquared MAE
## 11.6969547 0.3781227 8.0737779Part F
Would you recommend any of your models to replace the permeability laboratory experiment?
All of the models scored relative low Rsquared values. I wouldn’t replace the laboratory experiment with any of the models I used.
Question 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), 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:
Part A
Start R and use these commands to load the data:
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 eachrun.
Part 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).
Response
cmp <- ChemicalManufacturingProcess %>%
mutate(across(where(is.numeric), ~ ifelse(is.na(.), mean(., na.rm = TRUE), .)))Part 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?
Response
set.seed(123)
trainIndex <- createDataPartition(cmp$Yield, p = 0.75, list = FALSE)
trainData <- cmp[trainIndex, ]
testData <- cmp[-trainIndex, ]# Train the PLS model
set.seed(123)
plsModel <- train(
Yield ~ .,
data = trainData,
method = "pls",
tuneLength = 10,
trControl = ctrl,
preProcess = c("center", "scale")
)
print(plsModel)## Partial Least Squares
##
## 132 samples
## 57 predictor
##
## Pre-processing: centered (57), scaled (57)
## Resampling: Cross-Validated (10 fold)
## Summary of sample sizes: 119, 118, 117, 119, 120, 118, ...
## Resampling results across tuning parameters:
##
## ncomp RMSE Rsquared MAE
## 1 1.570635 0.4913847 1.190149
## 2 2.088226 0.4999404 1.260397
## 3 1.362611 0.5958961 1.026489
## 4 1.615482 0.5276703 1.124717
## 5 1.919107 0.5195987 1.170184
## 6 2.029507 0.5041811 1.206177
## 7 2.202229 0.4949182 1.271059
## 8 2.355408 0.4912501 1.332010
## 9 2.645476 0.4824542 1.430559
## 10 2.875658 0.4682531 1.507346
##
## RMSE was used to select the optimal model using the smallest value.
## The final value used for the model was ncomp = 3.Part D
Predict the response for the test set. What is the value of the performance metric and how does this compare with the resampled performance metric on the training set?
# Predict on the test set using the optimal model
plsPred <- predict(plsModel, newdata = testData)
# Calculate performance metrics on the test set
testPerformance <- postResample(pred = plsPred, obs = testData$Yield)
print(testPerformance)## RMSE Rsquared MAE
## 1.3062242 0.4794095 1.0894618Part E
Which predictors are most important in the model you have trained? Do either the biological or process predictors dominate the list?
The most important variables tended to process predictors. Among the top five are the variables ManufacturingProcess 32, 09, 36, 17 and 13.
# Calculate variable importance
importance <- varImp(plsModel, scale = TRUE)
print(importance)## pls variable importance
##
## only 20 most important variables shown (out of 57)
##
## Overall
## ManufacturingProcess32 100.00
## ManufacturingProcess09 85.95
## ManufacturingProcess36 85.67
## ManufacturingProcess17 85.35
## ManufacturingProcess13 84.82
## ManufacturingProcess06 67.50
## BiologicalMaterial02 59.97
## ManufacturingProcess33 59.94
## BiologicalMaterial06 58.20
## BiologicalMaterial08 57.81
## BiologicalMaterial03 57.45
## ManufacturingProcess11 57.12
## BiologicalMaterial12 51.78
## BiologicalMaterial11 51.43
## BiologicalMaterial01 50.78
## BiologicalMaterial04 49.59
## ManufacturingProcess37 47.65
## ManufacturingProcess28 45.75
## ManufacturingProcess12 45.44
## ManufacturingProcess02 40.60Part 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?
Response
We can see that when using scatter plots to examine the relationship between top predictors and the Yield variable that there tends to be a positive or negative relationship. This can inform us on what part of manufacturing process are performing as expected and which aren’t.
ggplot(data = cmp, aes(x = Yield, y = ManufacturingProcess32)) +
geom_point(color = "blue") +
geom_smooth(method = "lm", color = "red", se = FALSE) + # Linear trend line
labs(title = "Scatterplot",
x = "X Variable", y = "Y Variable") +
theme_minimal()## `geom_smooth()` using formula = 'y ~ x'
ggplot(data = cmp, aes(x = Yield, y = ManufacturingProcess09)) +
geom_point(color = "blue") +
geom_smooth(method = "lm", color = "red", se = FALSE) + # Linear trend line
labs(title = "Scatterplot",
x = "X Variable", y = "Y Variable") +
theme_minimal()## `geom_smooth()` using formula = 'y ~ x'
ggplot(data = cmp, aes(x = Yield, y = ManufacturingProcess36)) +
geom_point(color = "blue") +
geom_smooth(method = "lm", color = "red", se = FALSE) + # Linear trend line
labs(title = "Scatterplot",
x = "X Variable", y = "Y Variable") +
theme_minimal()## `geom_smooth()` using formula = 'y ~ x'
ggplot(data = cmp, aes(x = Yield, y = ManufacturingProcess17)) +
geom_point(color = "blue") +
geom_smooth(method = "lm", color = "red", se = FALSE) + # Linear trend line
labs(title = "Scatterplot",
x = "X Variable", y = "Y Variable") +
theme_minimal()## `geom_smooth()` using formula = 'y ~ x'
ggplot(data = cmp, aes(x = Yield, y = ManufacturingProcess13)) +
geom_point(color = "blue") +
geom_smooth(method = "lm", color = "red", se = FALSE) + # Linear trend line
labs(title = "Scatterplot",
x = "X Variable", y = "Y Variable") +
theme_minimal()## `geom_smooth()` using formula = 'y ~ x'