Are you doing right your exercises?
CWerneck - Claudia Werneck
january, 25, 2018
Abstract
This analysis corresponds to the Project Assignment for the Practical Machine Learning course of the Johns Hopkins Bloomberg School of Public Health Data Science Specialization at Coursera.
Using devices such as Jawbone Up, Nike FuelBand, and Fitbit it is now possible to collect a large amount of data about personal activity relatively inexpensively.
These type of devices are part of the quantified self movement - a group of enthusiasts who take measurements about themselves regularly to improve their health, to find patterns in their behavior, or because they are tech geeks.
One thing that people regularly do is quantify how much of a particular activity they do, but they rarely quantify how well they do it.
In this project, the goal is: using data from accelerometers on the belt, forearm, arm, and dumbell of 6 participants, predict the manner in which they did the exercises. They were asked to perform barbell lifts correctly and incorrectly in 5 different ways: A - the correct way and B, C, D e E, four different wrong ways of do the exercise. This is the “classe” variable in the training set. It will be select any of the other variables to predict with.
More information is available from the website here: http://web.archive.org/web/20161224072740/http:/groupware.les.inf.puc-rio.br/har (see the section on the Weight Lifting Exercise Dataset) and if you use the document you create for this class, for any purpose, please cite them as they have been very generous in allowing their data to be used for this kind of assignment.
The training and test data for this project are available in this two url’s:
Data Processing
library(caret); library(rattle); library(rpart); library(rpart.plot)
library(randomForest); library(corrplot)
#Load the imported data from local
trainRead<-read.csv("C:/Coursera/08_Practical_Machine_learning/pml-training.csv", na.strings=c("NA","#DIV/0!", ""))
testRead<-read.csv("C:/Coursera/08_Practical_Machine_learning/pml-testing.csv", na.strings=c("NA","#DIV/0!", ""))
dim(trainRead);dim(testRead)## [1] 19622 160## [1] 20 160#Once we want use the columns as predictors, we must eliminate all the columns that do not have information.
trainClean<-trainRead[,colSums(is.na(trainRead))==0]
testClean<-testRead[,colSums(is.na(testRead))==0]
#that reduces the columns to only 60 columns
dim(trainClean);dim(testClean)## [1] 19622 60## [1] 20 60#Investigating the data we can see that the seven first columns have a sequencial number (the first)
#and variations of the timestamp that we are not using for this analysis so we will eliminate those columns remaining 53
trainOK<-trainClean[,-c(1:7)]
testOK<-testClean[,-c(1:7)]
dim(trainOK);dim(testOK)## [1] 19622 53## [1] 20 53#And now we are with the Dataset to proceed the study and will see if there are correlation among the variables used.
exerCorrmatrix<-cor(trainOK[sapply(trainOK, is.numeric)])
png(file="C:/Coursera/08_Practical_Machine_learning/corrpng.png", res=96, width=1000, height=1000)
corrplot(exerCorrmatrix,order="FPC", method="circle", tl.cex=0.45, tl.col="black", number.cex=0.25)
title("Correlation Matrix of the variables used", line = 1)
dev.off()## png
## 2
###Create the datasets
inTrain<-createDataPartition(trainOK$classe, p=3/4, list=FALSE)
train<-trainOK[inTrain,]
valid<-trainOK[-inTrain,] analysing the principal components, we got that 25 components are necessary to capture .95 of the variance. But it demands alot of machine processing so, we decided by a .80 thresh to capture 80% of the variance using 12 components
set.seed(2018)
PropPCA<-preProcess(train,method="pca", thresh=0.8)
PropPCA## Created from 14718 samples and 53 variables
##
## Pre-processing:
## - centered (52)
## - ignored (1)
## - principal component signal extraction (52)
## - scaled (52)
##
## PCA needed 12 components to capture 80 percent of the variance#create the preProc object, excluding the response (classe)
preProc <- preProcess(train[,-53],
method = "pca",
pcaComp = 12, thresh=0.8)
#Apply the processing to the train and test data, and add the response
#to the dataframes
train_pca <- predict(preProc, train[,-53])
train_pca$classe <- train$classe
#train_pca has only 12 principal components plus classe
valid_pca <- predict(preProc, valid[,-53])
valid_pca$classe <- valid$classe
#valid_pca has only 12 principal components plus classe
###**Choose algorithms to predict**
#####Two methods will be tested, gbm=Generalized Boosted Regression and rf=Random Forest
### GBM
fitControl<-trainControl(method="cv", number=5, allowParallel=TRUE)
fit_gbm<-train(classe ~., data=train_pca, method="gbm", trControl=fitControl)## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.0580
## 2 1.5733 nan 0.1000 0.0444
## 3 1.5457 nan 0.1000 0.0309
## 4 1.5254 nan 0.1000 0.0306
## 5 1.5061 nan 0.1000 0.0245
## 6 1.4912 nan 0.1000 0.0216
## 7 1.4780 nan 0.1000 0.0196
## 8 1.4659 nan 0.1000 0.0158
## 9 1.4558 nan 0.1000 0.0145
## 10 1.4465 nan 0.1000 0.0145
## 20 1.3735 nan 0.1000 0.0084
## 40 1.2847 nan 0.1000 0.0037
## 60 1.2285 nan 0.1000 0.0032
## 80 1.1863 nan 0.1000 0.0014
## 100 1.1536 nan 0.1000 0.0013
## 120 1.1264 nan 0.1000 0.0015
## 140 1.1038 nan 0.1000 0.0000
## 150 1.0928 nan 0.1000 0.0011
##
## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.0877
## 2 1.5554 nan 0.1000 0.0729
## 3 1.5109 nan 0.1000 0.0550
## 4 1.4782 nan 0.1000 0.0377
## 5 1.4533 nan 0.1000 0.0359
## 6 1.4309 nan 0.1000 0.0351
## 7 1.4104 nan 0.1000 0.0293
## 8 1.3918 nan 0.1000 0.0245
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## 10 1.3603 nan 0.1000 0.0197
## 20 1.2524 nan 0.1000 0.0116
## 40 1.1299 nan 0.1000 0.0083
## 60 1.0472 nan 0.1000 0.0042
## 80 0.9869 nan 0.1000 0.0038
## 100 0.9353 nan 0.1000 0.0014
## 120 0.8920 nan 0.1000 0.0016
## 140 0.8576 nan 0.1000 0.0008
## 150 0.8409 nan 0.1000 0.0029
##
## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.1096
## 2 1.5432 nan 0.1000 0.0800
## 3 1.4952 nan 0.1000 0.0577
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## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.0589
## 2 1.5727 nan 0.1000 0.0454
## 3 1.5443 nan 0.1000 0.0353
## 4 1.5224 nan 0.1000 0.0298
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## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.0899
## 2 1.5550 nan 0.1000 0.0647
## 3 1.5153 nan 0.1000 0.0611
## 4 1.4787 nan 0.1000 0.0388
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## 6 1.4303 nan 0.1000 0.0296
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## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.1136
## 2 1.5404 nan 0.1000 0.0923
## 3 1.4867 nan 0.1000 0.0615
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## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.0571
## 2 1.5729 nan 0.1000 0.0445
## 3 1.5446 nan 0.1000 0.0324
## 4 1.5249 nan 0.1000 0.0311
## 5 1.5056 nan 0.1000 0.0244
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## 40 1.2847 nan 0.1000 0.0035
## 60 1.2286 nan 0.1000 0.0029
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## 100 1.1531 nan 0.1000 0.0012
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## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.0880
## 2 1.5549 nan 0.1000 0.0600
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## 4 1.4819 nan 0.1000 0.0407
## 5 1.4567 nan 0.1000 0.0378
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## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.1110
## 2 1.5413 nan 0.1000 0.0805
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## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.0603
## 2 1.5728 nan 0.1000 0.0413
## 3 1.5464 nan 0.1000 0.0370
## 4 1.5233 nan 0.1000 0.0297
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## 40 1.2839 nan 0.1000 0.0032
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## 120 1.1247 nan 0.1000 0.0007
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## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.0895
## 2 1.5552 nan 0.1000 0.0618
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## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.1080
## 2 1.5411 nan 0.1000 0.0760
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## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.0554
## 2 1.5753 nan 0.1000 0.0406
## 3 1.5502 nan 0.1000 0.0390
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## 5 1.5079 nan 0.1000 0.0241
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## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.0953
## 2 1.5511 nan 0.1000 0.0653
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## 5 1.4558 nan 0.1000 0.0397
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## 7 1.4117 nan 0.1000 0.0289
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## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.1069
## 2 1.5418 nan 0.1000 0.0904
## 3 1.4879 nan 0.1000 0.0671
## 4 1.4472 nan 0.1000 0.0484
## 5 1.4175 nan 0.1000 0.0454
## 6 1.3893 nan 0.1000 0.0381
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## 60 0.9328 nan 0.1000 0.0029
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## Iter TrainDeviance ValidDeviance StepSize Improve
## 1 1.6094 nan 0.1000 0.1111
## 2 1.5427 nan 0.1000 0.0799
## 3 1.4941 nan 0.1000 0.0702
## 4 1.4518 nan 0.1000 0.0488
## 5 1.4219 nan 0.1000 0.0481
## 6 1.3922 nan 0.1000 0.0391
## 7 1.3681 nan 0.1000 0.0359
## 8 1.3455 nan 0.1000 0.0327
## 9 1.3242 nan 0.1000 0.0285
## 10 1.3061 nan 0.1000 0.0298
## 20 1.1709 nan 0.1000 0.0135
## 40 1.0337 nan 0.1000 0.0090
## 60 0.9384 nan 0.1000 0.0040
## 80 0.8695 nan 0.1000 0.0032
## 100 0.8102 nan 0.1000 0.0019
## 120 0.7613 nan 0.1000 0.0038
## 140 0.7182 nan 0.1000 0.0015
## 150 0.6992 nan 0.1000 0.0015print(fit_gbm, digits=4) ## Stochastic Gradient Boosting
##
## 14718 samples
## 12 predictor
## 5 classes: 'A', 'B', 'C', 'D', 'E'
##
## No pre-processing
## Resampling: Cross-Validated (5 fold)
## Summary of sample sizes: 11774, 11774, 11775, 11775, 11774
## Resampling results across tuning parameters:
##
## interaction.depth n.trees Accuracy Kappa
## 1 50 0.5069 0.3660
## 1 100 0.5590 0.4373
## 1 150 0.5798 0.4652
## 2 50 0.6087 0.5026
## 2 100 0.6694 0.5804
## 2 150 0.7008 0.6207
## 3 50 0.6540 0.5609
## 3 100 0.7199 0.6449
## 3 150 0.7550 0.6896
##
## Tuning parameter 'shrinkage' was held constant at a value of 0.1
##
## Tuning parameter 'n.minobsinnode' was held constant at a value of 10
## Accuracy was used to select the optimal model using the largest value.
## The final values used for the model were n.trees = 150,
## interaction.depth = 3, shrinkage = 0.1 and n.minobsinnode = 10.predict_gbm<-predict(fit_gbm,valid_pca)
(conf_gbm<-confusionMatrix(valid_pca$classe, predict_gbm))## Confusion Matrix and Statistics
##
## Reference
## Prediction A B C D E
## A 1180 56 71 70 18
## B 109 659 103 44 34
## C 77 57 674 28 19
## D 52 43 75 604 30
## E 59 65 74 43 660
##
## Overall Statistics
##
## Accuracy : 0.7702
## 95% CI : (0.7582, 0.7819)
## No Information Rate : 0.3012
## P-Value [Acc > NIR] : < 2.2e-16
##
## Kappa : 0.7088
## Mcnemar's Test P-Value : < 2.2e-16
##
## Statistics by Class:
##
## Class: A Class: B Class: C Class: D Class: E
## Sensitivity 0.7989 0.7489 0.6760 0.7655 0.8673
## Specificity 0.9373 0.9279 0.9537 0.9514 0.9418
## Pos Pred Value 0.8459 0.6944 0.7883 0.7512 0.7325
## Neg Pred Value 0.9154 0.9441 0.9202 0.9549 0.9748
## Prevalence 0.3012 0.1794 0.2033 0.1609 0.1552
## Detection Rate 0.2406 0.1344 0.1374 0.1232 0.1346
## Detection Prevalence 0.2845 0.1935 0.1743 0.1639 0.1837
## Balanced Accuracy 0.8681 0.8384 0.8149 0.8585 0.9046(accuracy_gbm<-conf_gbm$overall['Accuracy'])## Accuracy
## 0.7701876###rf
fitControl<-trainControl(method="cv", number=5, allowParallel=TRUE)
fit_rf<-train(classe ~., data=train_pca, method="rf", trControl=fitControl)
print(fit_rf, digits=4) ## Random Forest
##
## 14718 samples
## 12 predictor
## 5 classes: 'A', 'B', 'C', 'D', 'E'
##
## No pre-processing
## Resampling: Cross-Validated (5 fold)
## Summary of sample sizes: 11775, 11775, 11774, 11775, 11773
## Resampling results across tuning parameters:
##
## mtry Accuracy Kappa
## 2 0.9543 0.9422
## 7 0.9471 0.9331
## 12 0.9403 0.9246
##
## Accuracy was used to select the optimal model using the largest value.
## The final value used for the model was mtry = 2.predict_rf<-predict(fit_rf,valid_pca)
(conf_rf<-confusionMatrix(valid_pca$classe, predict_rf))## Confusion Matrix and Statistics
##
## Reference
## Prediction A B C D E
## A 1361 8 16 7 3
## B 25 898 23 2 1
## C 10 14 818 11 2
## D 4 8 30 756 6
## E 2 6 4 5 884
##
## Overall Statistics
##
## Accuracy : 0.9619
## 95% CI : (0.9561, 0.9671)
## No Information Rate : 0.2859
## P-Value [Acc > NIR] : < 2.2e-16
##
## Kappa : 0.9518
## Mcnemar's Test P-Value : 0.0008301
##
## Statistics by Class:
##
## Class: A Class: B Class: C Class: D Class: E
## Sensitivity 0.9708 0.9615 0.9181 0.9680 0.9866
## Specificity 0.9903 0.9872 0.9908 0.9884 0.9958
## Pos Pred Value 0.9756 0.9463 0.9567 0.9403 0.9811
## Neg Pred Value 0.9883 0.9909 0.9820 0.9939 0.9970
## Prevalence 0.2859 0.1905 0.1817 0.1593 0.1827
## Detection Rate 0.2775 0.1831 0.1668 0.1542 0.1803
## Detection Prevalence 0.2845 0.1935 0.1743 0.1639 0.1837
## Balanced Accuracy 0.9805 0.9743 0.9544 0.9782 0.9912(accuracy_rf<-conf_rf$overall['Accuracy'])## Accuracy
## 0.9618679We can now say that for this dataset, random forest method is better than Generalized Boosted Regression and the accuracy obtained would be greather than 0.95
Results - Prediction on Testing Set
Applying the Random Forest to predict the outcome variable classe for the test set
test_pca <- predict(preProc, testOK[,-53])
test_pca$problem_id <- testOK$problem_id
(predict(fit_rf, test_pca))## [1] B A A A A E D B A A B C B A E E A B B B
## Levels: A B C D E