The objective behind attempting this dataset was to understand the predictors that contribute to the life expectancy around the world. I have used Linear Regression, Decision Tree and Random Forest for this purpose. Steps Involved: - Read the csv file - Data Cleaning: - Variables Country and Status were showing as having character data types. These had to be converted to factor - 2563 missing values were encountered with Population variable having the most of the missing values i.e 652 - Missing rows were dropped before we could run the analysis. 3) Run Linear Regression - Before running linear regression, 3 variables were dropped as they were not found to be having that much of an effect on the dependent variable i.e Life Expectancy. These 3 variables were Country, Year & Status. This meant we are now working with 19 variables (1 dependent and 18 independent variables) - We run the linear regression. Multiple R squared is 83% which means that independent variables can explain 83% change or variance in the dependent variable. - OULTLIER DETECTION. We check for outliers using IQR and find 54 outliers. These outliers are then removed before we run the regression analysis once again. Multiple R squared increased from 83% to 86%. - MULTICOLLINEARITY. We check for multicollinearity using the VIF model(Variance Inflation Factor). This is being done in case when two or more independent variables showing high correlation. The thumb rule is that absolute VIF values above 5 should be removed. We find 6 variables that have a VIF value higher than 5 namely Infant.deaths, percentage.expenditure,Under.five.deaths,GDP,thinness1.19,thinness5.9. Infant deaths and Under Five deaths have strong collinearity so we drop infant deaths(which has the higher VIF value). - When we run the linear regression model again, VIF value of Under.Five.Deaths goes down from 211.46 to 2.74 while the other variable's VIF values reduce very less. Variable thinness1.19 is now dropped and we run the regression once more. - Variable thinness5.9 whose absolute VIF value was 7.61 has now dropped to 1.95. GDP and Population are still having VIF value more than 5 but I decided against dropping these as I consider them to be important independent variables. - SET THE SEED AND SPLIT THE DATA INTO TRAIN AND TEST DATA. We run the train data and get multiple R squared of 86% and p value less than that of alpha which states that it is statistically significant. We use the train data to predict the test data to find out the RMSE and MAPE. We run the library(Metrics) for this purpose. - In Linear Regression, RMSE (Root Mean Squared Error) is 3.2. This indicates that on an average, the predicted values have an error of 3.2 years as compared to the actual life expectancy values. - MAPE (Mean Absolute Percentage Error) is 0.037. This indicates an accuracy prediction of 96.20% (1-0.037). - MAE (Mean Absolute Error) is 2.55. This indicates that on an average, the predicted values deviate by approximately 2.83 years from the actual values.

We use DECISION TREE MODEL for the analysis.

• Run the required libraries (rpart, rpart.plot, RColorBrewer, rattle).
• We run the decision tree analysis using rpart and plot the tree. We use fancyRpartPlot.
• We use 5 fold cross validation method with CP (complexity parameter) being 0.01.
• In Decision Tree , RMSE (Root Mean Squared Error) is 3.06. This indicates that on an average, the predicted values have an error of 3.06 years as compared to the actual life expectancy values.
• MAPE (Mean Absolute Percentage Error) is 0.035. This indicates an accuracy prediction of 96.45% (1-0.035).
• MAE (Mean Absolute Error) is 2.35. This indicates that on an average, the predicted values deviate by approximately 2.35 years from the actual values.

We use RANDOM FOREST for the analysis.

• Run library(randomForest)
• We use varImpPlot to find out which variables are most significant and least significant. Income composition is the most important followed by adult mortality and the least relevant independent variable is Population.
• Predict Life expectancy through random forest model.
• In Random Forest , RMSE (Root Mean Squared Error) is 1.73. This indicates that on an average, the predicted values have an error of 1.73 years as compared to the actual life expectancy values.
• MAPE (Mean Absolute Percentage Error) is 0.01. This indicates an accuracy prediction of 98.27% (1-0.01).
• MAE (Mean Absolute Error) is 1.14. This indicates that on an average, the predicted values deviate by approximately 1.14 years from the actual values.

Conclusion: Random Forest is the best model for predicting the life expectancy values as it has the lowest RMSE, MAPE and MAE.

#https://www.kaggle.com/c/facial-keypoints-detection/details/getting-started-with-r #################################

###Variables for downloaded files data.dir <- ' ' train.file <- paste0(data.dir, 'training.csv') test.file <- paste0(data.dir, 'test.csv') #################################

###Load csv -- creates a data.frame matrix where each column can have a different type. d.train <- read.csv(train.file, stringsAsFactors = F) d.test <- read.csv(test.file, stringsAsFactors = F)

###In training.csv, we have 7049 rows, each one with 31 columns. ###The first 30 columns are keypoint locations, which R correctly identified as numbers. ###The last one is a string representation of the image, identified as a string.

###To look at samples of the data, uncomment this line:

head(d.train)

###Let's save the first column as another variable, and remove it from d.train: ###d.train is our dataframe, and we want the column called Image. ###Assigning NULL to a column removes it from the dataframe

im.train <- d.train$Image d.train$Image <- NULL #removes 'image' from the dataframe

im.test <- d.test$Image d.test$Image <- NULL #removes 'image' from the dataframe

################################# #The image is represented as a series of numbers, stored as a string #Convert these strings to integers by splitting them and converting the result to integer

#strsplit splits the string #unlist simplifies its output to a vector of strings #as.integer converts it to a vector of integers. as.integer(unlist(strsplit(im.train[1], " "))) as.integer(unlist(strsplit(im.test[1], " ")))

###Install and activate appropriate libraries ###The tutorial is meant for Linux and OSx, where they use a different library, so: ###Replace all instances of %dopar% with %do%.

install.packages('foreach')

library("foreach", lib.loc="~/R/win-library/3.3")

###implement parallelization im.train <- foreach(im = im.train, .combine=rbind) %do% { as.integer(unlist(strsplit(im, " "))) } im.test <- foreach(im = im.test, .combine=rbind) %do% { as.integer(unlist(strsplit(im, " "))) } #The foreach loop will evaluate the inner command for each row in im.train, and combine the results with rbind (combine by rows). #%do% instructs R to do all evaluations in parallel. #im.train is now a matrix with 7049 rows (one for each image) and 9216 columns (one for each pixel):

###Save all four variables in data.Rd file ###Can reload them at anytime with load('data.Rd')

save(d.train, im.train, d.test, im.test, file='data.Rd')

load('data.Rd')

#each image is a vector of 96*96 pixels (96*96 = 9216). #convert these 9216 integers into a 96x96 matrix: im <- matrix(data=rev(im.train[1,]), nrow=96, ncol=96)

#im.train[1,] returns the first row of im.train, which corresponds to the first training image. #rev reverse the resulting vector to match the interpretation of R's image function #(which expects the origin to be in the lower left corner).

#To visualize the image we use R's image function: image(1:96, 1:96, im, col=gray((0:255)/255))

#Let’s color the coordinates for the eyes and nose points(96-d.train$nose_tip_x[1], 96-d.train$nose_tip_y[1], col="red") points(96-d.train$left_eye_center_x[1], 96-d.train$left_eye_center_y[1], col="blue") points(96-d.train$right_eye_center_x[1], 96-d.train$right_eye_center_y[1], col="green")

#Another good check is to see how variable is our data. #For example, where are the centers of each nose in the 7049 images? (this takes a while to run): for(i in 1:nrow(d.train)) { points(96-d.train$nose_tip_x[i], 96-d.train$nose_tip_y[i], col="red") }

#there are quite a few outliers -- they could be labeling errors. Looking at one extreme example we get this: #In this case there's no labeling error, but this shows that not all faces are centralized idx <- which.max(d.train$nose_tip_x) im <- matrix(data=rev(im.train[idx,]), nrow=96, ncol=96) image(1:96, 1:96, im, col=gray((0:255)/255)) points(96-d.train$nose_tip_x[idx], 96-d.train$nose_tip_y[idx], col="red")

#One of the simplest things to try is to compute the mean of the coordinates of each keypoint in the training set and use that as a prediction for all images colMeans(d.train, na.rm=T)

#To build a submission file we need to apply these computed coordinates to the test instances: p <- matrix(data=colMeans(d.train, na.rm=T), nrow=nrow(d.test), ncol=ncol(d.train), byrow=T) colnames(p) <- names(d.train) predictions <- data.frame(ImageId = 1:nrow(d.test), p) head(predictions)

#The expected submission format has one one keypoint per row, but we can easily get that with the help of the reshape2 library:

install.packages('reshape2')

library(...