---
title: "Data Science and Predictive Analytics (UMich HS650)"
subtitle: "<h2><u>Forecasting Numeric Data using Regression Models</u></h2>"
author: "<h3>SOCR/MIDAS (Ivo Dinov)</h3>"
date: "`r format(Sys.time(), '%B %Y')`"
tags: [DSPA, SOCR, MIDAS, Big Data, Predictive Analytics] 
output:
  html_document:
    theme: spacelab
    highlight: tango
    includes:
      before_body: SOCR_header.html
      after_body: SOCR_footer_tracker.html
    toc: true
    number_sections: true
    toc_depth: 2
    toc_float:
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      smooth_scroll: true
---

# Motivation

In previous chapters ([6](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/06_LazyLearning_kNN.html), [7](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/07_NaiveBayesianClass.html), and [8](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/08_DecisionTreeClass.html)), we covered some classification methods that use mathematical formalism to address everyday life prediction problems. In this chapter, we will focus on specific model-based statistical methods providing forecasting and classification functionality. Specifically, we will (1) demonstrate the predictive power of multiple linear regression, (2) show the foundation of regression trees and model trees, and (3) examine two complementary case-studies (Baseball Players and Heart Attack).

It may be helpful to first review [Chapter 4 (Linear Algebra/Matrix Manipulations)](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/04_LinearAlgebraMatrixComputing.html) and [Chapter 6 (Intro to Machine Learning)](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/06_LazyLearning_kNN.html).

# Understanding Regression

Regression is a measurement of relationship between a *dependent variable* (value to be predicted) and a group of *independent variables* (predictors similar to features, discussed in [Chapter 6](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/06_LazyLearning_kNN.html). We assume the relationship between our dependent variable and independent variables follow a straight line.

## Simple linear regression

First recall the material in [Chapter 4: Linear Algebra & Matrix Computing](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/04_LinearAlgebraMatrixComputing.html).

The simplest case of a regression is that we only have one predictor. 
$$y=a+bx.$$ 

Does this formula appear familiar to you? In this slope-intercept formula, `a` is our intercept while `b` is the slope. That is the expression for simple linear regression. If we know `a` and `b`, for any given `x` we can calculate `y` via the above formula. If we plot `x` and `y` in a coordinate system, we will have a straight line.

However, this is the ideal case. When we plot using real world data, the pattern would be harder to recognize. Let's look at the scatter plot(you can recall [Chapter 2](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/02_ManagingData.html)) and simple linear regression line of two variables "hospital charges" or `CHARGES` (independent variable) and `length of stay` in the hospital or `LOS` (predictor). The data could be find in our class files `CaseStudy12_AdultsHeartAttack_Data`. We removed two observations that have missing data using the command `heart_attack<-heart_attack[complete.cases(heart_attack), ]`. 
```{r eval=T}
heart_attack<-read.csv("https://umich.instructure.com/files/1644953/download?download_frd=1", stringsAsFactors = F)
heart_attack$CHARGES<-as.numeric(heart_attack$CHARGES)
heart_attack<-heart_attack[complete.cases(heart_attack), ]

fit1<-lm(CHARGES~LOS, data=heart_attack)
par(cex=.8)
plot(heart_attack$LOS, heart_attack$CHARGES, xlab="LOS", ylab = "CHARGES")
abline(fit1, lwd=2)
```

It seems to be common sense that the longer you stay in the hospital, the higher the medical costs will be. However, on the scatter plot, we have only a bunch of dots showing a little bit sign of an increasing pattern. 

The estimated expression for this regression line is:
$$\hat{y}=4582.70+212.29\times x$$
or equivalently
$$CHARGES=4582.70+212.29\times LOS$$
It is simple to make predictions with this regression line. Assume we have a patient that spent 10 days in hospital, then we have `LOS=10`. The predicted charge is likely to be $\$4582.70+\$212.29\times 10=\$6705.6$. Plugging `x` into the expression equation automatically gives us an estimated value of the outcome `y`. This [chapter of the Probability and statistics EBook provides an introduction to linear modeling](http://wiki.socr.umich.edu/index.php/EBook#Chapter_X:_Correlation_and_Regression).

##Ordinary least squares estimation

How did we get the estimated expression? The most common estimating method in statistics is *ordinary least squares* (OLS). OLS estimators are obtained by minimizing sum of the squared errors - that is the sum of squared vertical distance from each dot on the scatter plot to the regression line. 

```{r echo=F, fig.width=9, fig.height=9}
plot(heart_attack$LOS, heart_attack$CHARGES, xlab="LOS", ylab = "CHARGES")
abline(fit1, lwd=2)
segments(15, 7767.05, 15, 10959, col = 'red', lwd=2)
text(18, 9363.025, "error", cex=1.5)
text(16, 9363.025, '}', cex = 9)
text(15, 10959, '.', col = "green", cex=5)
text(16, 11500, "true value(y)", cex = 1.5)
text(15, 7767.05, '.', col = "green", cex=5)
text(15, 7400.05, "estimated value(y.hat)", cex = 1.5)
```

OLS is minimizing the following formula:
$$\sum_{i=1}^{n}(y_i-\hat{y}_i)^2=\sum_{i=1}^{n}(y_i-(a+b\times x_i))^2=\sum_{i=1}^{n}e_i^2.$$
After some statistical calculations, our value `b` with the minimum squared error is:
$$b=\frac{\sum(x_i-\bar x)(y_i-\bar y)}{\sum(x_i-\bar x)^2}.$$
While the optional `a` is:
$$a=\bar y-b\bar x.$$

These expressions might seem very confusing. However, they are based on past knowledge. As we learned in [Chapter 2](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/02_ManagingData.html), the variance is obtained by averaging sum of squares ($var(x)=\frac{1}{n}\sum^{n}_{i=1} (x_i-\mu)^2$). When we use $\bar{x}$ to estimate the mean of $X$, we have the following formula for variance: $var(x)=\frac{1}{n-1}\sum^{n}_{i=1} (x_i-\bar{x})^2$. We can see that this is $\frac{1}{n-1}$ times the denominator of *b*. Similar to variance, the covariance of *x* and *y* is measuring the average sum of the deviance of *x* times the deviance of *y*:
$$Cov(x, y)=\frac{1}{n}\sum^{n}_{i=1} (x_i-\mu_x)(y_i-\mu_y).$$
If we utilize the sample averages ($\bar{x}$, $\bar{y}$), we have:  
$$Cov(x, y)=\frac{1}{n-1}\sum^{n}_{i=1} (x_i-\bar{x})(y_i-\bar{y}).$$

Excitingly, this is $\frac{1}{n-1}$ times the numerator of *b*.

Combining the above, we get an estimate of the slope coefficient (effect-size of LOS on Charge): 
$$b=\frac{Cov(x, y)}{var(x)}.$$
Let's examine these using the heart attack data.

```{r eval=T}
b<-cov(heart_attack$LOS, heart_attack$CHARGES)/var(heart_attack$LOS)
b
a<-mean(heart_attack$CHARGES)-b*mean(heart_attack$LOS)
a
```
We can see that this is exactly the same as previously stated expression.


## Model Assumptions
Regression modeling has five key assumptions:

* Linear relationship
* [Multivariate normality](https://en.wikipedia.org/wiki/Multivariate_normal_distribution),
* No or little [multicollinearity](https://en.wikipedia.org/wiki/Multicollinearity),
* No auto-correlation, independence, 
* [Homoscedasticity](https://en.wikipedia.org/wiki/Homoscedasticity)

## Correlations

*Note*: The [SOCR Interactive Scatterplot Game (requires Java enabled browser)](http://socr.umich.edu/html/gam/SOCR_Games.html) provides a dynamic interface demonstrating linear models, trends, correlations, slopes and residuals.

Based on covariance we can calculate correlation, which indicates how closely that the relationship between two variables follows a straight line.
$$\rho_{x, y}=Corr(x, y)=\frac{Cov(x, y)}{\sigma_x\sigma_y}=\frac{Cov(x, y)}{\sqrt{Var(x)Var(y)}}.$$
In R, correlation is given by `cor()` while square root of variance or standard deviation is given by `sd()`.

```{r eval=T}
r<-cov(heart_attack$LOS, heart_attack$CHARGES)/(sd(heart_attack$LOS)*sd(heart_attack$CHARGES))
r
cor(heart_attack$LOS, heart_attack$CHARGES)
```

Same outputs are obtained. This correlation is a positive number that is relatively small. We can say there is a weak positive linear association between these two variables. If we have a negative number then it is a negative linear association. We have a weak association when $0.1 \leq Cor < 0.3$, a moderate association for $0.3 \leq Cor < 0.5$, and a strong association for $0.5 \leq Cor \leq 1.0$. If the correlation is below $0.1$ then it suggests little to no linear relation between the variables.

## Multiple Linear Regression

In practice, we usually have more situations with multiple predictors and one dependent variable, which may follow a multiple linear model.  That is:
$$y=\alpha+\beta_1x_1+\beta_2x_2+...+\beta_kx_k+\epsilon,$$
or equivalently
$$y=\beta_0+\beta_1x_1+\beta_2x_2+ ... +\beta_kx_k+\epsilon .$$
We usually use the second notation method in statistics. This equation shows the linear relationship between *k* predictors and a dependent variable. In total we have *k+1* coefficients to estimate.

The matrix notation for the above equation is:
$$Y=X\beta+\epsilon,$$
where 
$$Y=\left(\begin{array}{c} 
y_1 \\
y_2\\
...\\
y_n
\end{array}\right)$$

$$X=\left(\begin{array}{ccccc} 
1 & x_{11}&x_{21}&...&x_{k1} \\
1 & x_{12}&x_{22}&...&x_{k2} \\
.&.&.&.&.\\
1 & x_{1n}&x_{2n}&...&x_{kn} 
\end{array}\right)
$$
$$\beta=\left(\begin{array}{c} 
\beta_1 \\
\beta_2\\
...\\
\beta_k
\end{array}\right)$$

and

$$\epsilon=\left(\begin{array}{c} 
\epsilon_1 \\
\epsilon_2\\
...\\
\epsilon_n
\end{array}\right)
$$ 
is the error term.

Similar to simple linear regression, our goal is to minimize sum of squared errors. After solved for $\beta$, we get:
$$\hat{\beta}=(X^TX)^{-1}X^TY .$$
The solution is in the matrix form. $X^{-1}$ is the inverse of matrix $X$ and $X^T$ is the transposed matrix.

Let's make a function of our own using this matrix formula.

```{r eval=T}
reg<-function(y, x){
  x<-as.matrix(x)
  x<-cbind(Intercept=1, x)
  solve(t(x)%*%x)%*%t(x)%*%y
}
```

`solve()` is taking the command for matrix inversion. `%*%` is matrix multiplication.

Next, we will apply this function to our heart attack dataset. To begin with, let's check if the simple linear regression output is the same as we calculated earlier.

```{r eval=T}
reg(y=heart_attack$CHARGES, x=heart_attack$LOS)
```

It works! Then, we can include more variables as predictors. As an example, we just add `age` into the model. 

```{r eval=T}
str(heart_attack)
reg(y=heart_attack$CHARGES, x=heart_attack[, c(7, 8)])
```

# Case Study 1: Baseball Players

## Step 1 - collecting data

We utilize the [mlb data "01a_data.txt"](https://umich.instructure.com/files/330381/download?download_frd=1). The dataset contains 1034 records of heights and weights for some current and recent Major League Baseball (MLB) Players. These data were obtained from different resources (e.g., IBM Many Eyes). 

Variables:

* **Name**: MLB Player Name 
* **Team**: The Baseball team the player was a member of at the time the data was acquired
* **Position**: Player field position
* **Height**: Player height in inch 
* **Weight**: Player weight in pounds
* **Age**: Player age at time of record.

## Step 2 - exploring and preparing the data
Let's load this dataset first. We use `as.is=T` to make non-numerical vectors into characters. Also, we delete the `Name` variable because we don't need players' names in this case study. 

```{r eval=T}
mlb<- read.table('https://umich.instructure.com/files/330381/download?download_frd=1', as.is=T, header=T)
str(mlb)
mlb<-mlb[, -1]
```

By looking at the `srt()` output we notice that the variable `TEAM`  and `Position` are misspecified as characters. To fix this we can use function `as.factor()` that convert numerical or character vectors to factors.

```{r eval=T}
mlb$Team<-as.factor(mlb$Team)
mlb$Position<-as.factor(mlb$Position)
```

The data is good to go. Let's explore it using some summary statistics and plots.

```{r eval=T}
summary(mlb$Weight)
hist(mlb$Weight, main = "Histogram for Weights")
```

The above plot illustrates our dependent variable `Weight`. As we learned from [Chapter 2]( http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/02_ManagingData.html), we know this is somewhat right-skewed.

Applying `GGpairs` to obtain a compact dataset summary we can mark heavy weight and light weight players (according to $light \lt median \lt heavy$) by different colors in the plot:

```{r eval=TRUE, fig.width=12, fig.height=12}
require(GGally)

mlb_binary = mlb

mlb_binary$bi_weight = as.factor(ifelse(mlb_binary$Weight>median(mlb_binary$Weight),1,0))

g_weight <- ggpairs(data=mlb_binary[-1], title="MLB Light/Heavy Weights",
            mapping=ggplot2::aes(colour = bi_weight),
            lower=list(combo=wrap("facethist",binwidth=1)),
            # upper = list(continuous = wrap("cor", size = 4.75, alignPercent = 1))
            )
g_weight
```

Next, we may also mark player positions by different colors in the plot.

```{r eval=TRUE, fig.width=12, fig.height=12}
g_position <- ggpairs(data=mlb[-1], title="MLB by Position",
              mapping=ggplot2::aes(colour = Position),
              lower=list(combo=wrap("facethist",binwidth=1)))
g_position
```

What about our potential predictors?

```{r eval=T}
table(mlb$Team)
table(mlb$Position)
summary(mlb$Height)
summary(mlb$Age)
```

Here we have two numerical predictors, two categorical predictors and $1034$ observations. Let's see how R treats these three different classes of variables.

## Exploring relationships among features - the correlation matrix

Before fitting a model, let's examine the independence of our potential predictors and the dependent variable. Multiple linear regressions assume that predictors are all independent with each other. Is this assumption valid? As we mentioned earlier, `cor()` function can answer this question in pairwise manner. Note we only look at numerical variables.

```{r eval=T}
cor(mlb[c("Weight", "Height", "Age")])
```

Here we can see $cor(y, x)=cor(x, y)$ and $cov(x, x)=1$. Also, our `Height` variable is weakly related to the players' age in a negative manner. This looks very good and wouldn't cause any multi-collinearity problem. If two of our predictors are highly correlated, they both provide almost the same information. Then that could cause multi-collinearity. A common practice is to delete one of them in the model.

## Visualizing relationships among features - the scatterplot matrix

To visualize pairwise correlations, we could use scatterplot matrix. In R, `pairs()` function will give use these outputs. 

```{r eval=T}
pairs(mlb[c("Weight", "Height", "Age")])
```

You might get a sense of it but it is difficult to see any linear pattern. We can make an more sophisticated graph using `pairs.panels()` in the `psych` package.

```{r eval=T}
# install.packages("psych")
library(psych)
pairs.panels(mlb[, c("Weight", "Height", "Age")])
```

This plot give us much more information about the three variables. Above the diagonal, we have our correlation coefficients in numerical form. On the diagonal, there are histograms of variables. Below the diagonal, more visual information are presented to help us understand the trend. This specific graph shows us height and weight are positively and strongly correlated. Also the relationships between age and height, age and weight are very weak (horizontal red line in the below diagonal graphs indicates weak relationships). 

## Step 3 - training a model on the data

The function we are going to use for this section is `lm()`. No extra package is needed when using this function.

The `lm()` function has the following components:

**m<-lm(dv ~ iv, data=mydata)**

* dv: dependent variable
* iv: independent variables. Just like `OneR()` in [Chapter 8](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/08_DecisionTreeClass.html). If we use `.` as `iv`, all of the variables, except the dependent variable ($dv$), are included as predictors.
* data: specifies the data containing both dependent viable and independent variables

```{r eval=T}
fit<-lm(Weight~., data=mlb)
fit
```
As we can see from the output, factors are included in the model by creating several indicators, one for each factor level. Each numerical variables just have one coefficient.

## Step 4 - evaluating model performance

As we did in previous case studies, let's examine model performance.

```{r fig.height=3, fig.width=6}
summary(fit)
plot(fit, which = 1:2)
```

The **summary** shows us how well does the model fits the dataset.

*Residuals*: This tells us about the residuals. If we have extremely large or extremely small residuals for some observations compared to the rest of residuals, either they are outliers due to reporting error or the model fits data poorly. We have $73.649$ as our maximum and $-48.692$ as our minimum. Their extremeness could be examined by residual diagnostic plot.

*Coefficients*: In this section, we look at the very right column that has stars. Stars or dots next to variables show if the variable is significant and should be included in the model. However, if nothing is next to a variable then it means this estimated covariance could be 0 in the linear model. Another thing we can look at is the `Pr(>|t|)` column. A number closed to 0 in this column indicates the row variable is significant, otherwise it could be deleted from the model.

Here only some of the teams and positions are not significant. `Age` and `Height` are significant.

*R-squared*: What percent in `y` is explained by included predictors. Here we have 38.58%, which indicates the model is not bad but could be improved. Usually a well-fitted linear regression would have over 70%.

The **diagnostic plots** also helps us understanding the situation.

*Residual vs Fitted*: This is the residual diagnostic plot. We can see that the residuals of observations indexed $65$, $160$ and $237$ are relatively far apart from the rest. They are potential influential points or outliers.

*Normal Q-Q*: This plot examines the normality assumption of the model. If these dots follows the line on the graph, the normality assumption is valid. In our case, it is relatively close to the line. So, we can say that our model is valid in terms of normality.

## Step 5 - improving model performance

We can employ the `step` function to perform forward or backward selection of important features/predictors. It works for both `lm` and `glm` models. In most cases, backward-selection is preferable because it tends to retain much larger models. On the other hand, there are various criteria to evaluate a model. The common used includes *AIC*, *BIC*, Adjusted $R^2$, etc. Let's compare the backward and forward model selection approaches. The `step` function argument `direction` allows this control (default is `both`, which will select the better result from either backward or forward selection).

```{r}
step(fit,direction = "backward")
step(fit,direction = "forward")
step(fit,direction = "both")
```

We can observe that forward retains the whole model. The better feature selection model uses `backward` step-wise selection.

Both backward and forward are greedy algorithms and neither guarantees an optimal model result. The optimal feature selection requires exploring every possible combination of the predictors, which is practically not feasible, due to computational complicity, $n \choose k$ combinations.

Alternatively, we can choose models based on various **information criteria**. 

```{r}
step(fit,k=2)
step(fit,k=log(nrow(mlb)))
```

$k = 2$ yields the genuine AIC criterion, and $k = log(n)$ refers to BIC. Let's try to evaluate model performance again.

```{r}
fit2 = step(fit,k=2,direction = "backward")
summary(fit2)
plot(fit2, which = 1:2)
```

Sometime, we prefer a simpler model even if there is a little bit loss of performance. In this case, we have a simpler model and $R^2=0.365$. The whole model is still very significant. We can see that observations $65$, $160$ and $237$ are relatively far from other residuals. They are potential influential points or outliers. 

Also, we can observe the leverage points. Leverage points are those either outliers or influential points or both.

```{r}
# Half-normal plot for leverages
# install.packages("faraway")
library(faraway)
halfnorm(lm.influence(fit)$hat, nlab = 2, ylab="Leverages")
mlb[c(226,879),]
summary(mlb)
```

A deeper discussion of variable selection, controlling the false discovery rate, is provided in Chapters [16](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/16_FeatureSelection.html) and [17](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/17_RegularizedLinModel_KnockoffFilter.html).

### Model specification - adding non-linear relationships
In linear regression, the relationship between independent and dependent variables is assumed to be linear. However, this might not be the case. The relationship between age and weight could be quadratic, since middle-aged people might gain weight dramatically.

```{r eval=T}
mlb$age2<-(mlb$Age)^2
fit2<-lm(Weight~., data=mlb)
summary(fit2)
```

This actually brought up the overall $R^2$ up to $0.3889$.

## Transformation - converting a numeric variable to a binary indicator

As discussed earlier, middle-aged people might have a different pattern in weight increase compared to younger people. The overall pattern could be not cumulative, but rather two separate lines for young and middle-aged people. We assume 30 is the threshold. People over 30 have a steeper line for weight increase than under 30. Here we use the `ifelse()` function that we mentioned in [Chapter 7](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/07_NaiveBayesianClass.html) to create the indicator of the threshold. 

```{r eval=T}
mlb$age30<-ifelse(mlb$Age>=30, 1, 0)
fit3<-lm(Weight~Team+Position+Age+age30+Height, data=mlb)
summary(fit3)
```
This model performs worse than the quadratic model in terms of $R^2$. Moreover, `age30` is not significant. So, we will stick with the quadratic model.

## Model specification - adding interaction effects

So far, each feature's individual effect has considered in our model. It is possible that features act in pairs to affect our independent variable. Let's examine that deeper. 

Interaction is a combined effect by two features. If we are not sure whether two features interact, we could test by adding an interaction term into the model. If the interaction is significant then we confirmed an interaction between two features.

```{r eval=T}
fit4<-lm(Weight~Team+Height+Age*Position+age2, data=mlb)
summary(fit4)
```
Here we can see the overall $R^2$ improved and some of the interactions are significant under 0.1 level.

# Understanding regression trees and model trees

## Motivation

As we saw in [Chapter 8](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/08_DecisionTreeClass.html), a decision tree builds by multiple `if-else` logical decisions and can classify observations. We could add regression into decision trees so that a decision tree can make numerical predictions.

## Adding regression to trees

Numeric prediction trees are built in the same way as classification trees. Recall the discussion in [Chapter 8](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/08_DecisionTreeClass.html) where the data are partitioned first via a  *divide-and-conquer* strategy based on features. The homogeneity of the resulting classification trees is measured by various metrics, e.g., entropy. In prediction, tree homogeneity is measured by statistics such as variance, standard deviation or absolute deviation from the mean.

A common splitting criterion for decision trees is the **standard deviation reduction (SDR)**.

$$SDR=sd(T)-\sum_{i=1}^n \left | \frac{T_i}{T} \right | \times sd(T_i),$$

where `sd(T)` is the standard deviation for the original data. After the summation of all segments, $|\frac{T_i}{T}|$ is the proportion of observations in $i^{th}$ segment compared to total number of observations. $sd(T_i)$ is the standard deviation for the $i^{th}$ segment.

An example for this would be:
$$Original\, data:\{1, 2, 3, 3, 4, 5, 6, 6, 7, 8\}$$
$$Split\, method\, 1:\{1, 2, 3|3, 4, 5, 6, 6, 7, 8\}$$
$$Split\, method\, 2:\{1, 2, 3, 3, 4, 5|6, 6, 7, 8\}$$
In split method 1, $T_1=\{1, 2, 3\}$, $T_2=\{3, 4, 5, 6, 6, 7, 8\}$.
In split method 2, $T_1=\{1, 2, 3, 3, 4, 5\}$, $T_2=\{6, 6, 7, 8\}$.

```{r eval=T}
ori<-c(1, 2, 3, 3, 4, 5, 6, 6, 7, 8)
at1<-c(1, 2, 3)
at2<-c(3, 4, 5, 6, 6, 7, 8)
bt1<-c(1, 2, 3, 3, 4, 5)
bt2<-c(6, 6, 7, 8)
sdr_a<-sd(ori)-(length(at1)/length(ori)*sd(at1)+length(at2)/length(ori)*sd(at2))
sdr_b<-sd(ori)-(length(bt1)/length(ori)*sd(bt1)+length(bt2)/length(ori)*sd(bt2))
sdr_a
sdr_b
```
`length()` is used in the above R codes to get the number of elements in a specific vector.

Larger SDR indicates greater reduction in standard deviation after splitting. Here we have split method 2 with greater SDR so the tree decide to use second method first. The second method results more homogeneous sets than the first one. 

Now, the tree will be split under `bt1` and `bt2` following same rules (greater SDR wins). Assume we cannot split further (`bt1` and `bt2` are terminal nodes). The observations classified into `bt1` will be predicted with $mean(bt1)=3$ and those classified as `bt2` with $mean(bt2)=6.75$.

# Case study 2: Baseball Players (Take 2)

## Step 2 - exploring and preparing the data

We still use the [mlb dataset](https://umich.instructure.com/files/330381/download?download_frd=1) for this section. This dataset has 1034 observations. Let's try to separate them into training and test datasets first.

```{r eval=T}
set.seed(1234)
train_index <- sample(seq_len(nrow(mlb)), size = 0.75*nrow(mlb))
mlb_train<-mlb[train_index, ]
mlb_test<-mlb[-train_index, ]
```

Here we use a randomized split (75%-25%) to divide the training and test datasets.

## Step 3 - training a model on the data

In R, function for prediction tree models is `rpart()` under `rpart` package.

**m<-rpart(dv~iv, data=mydata)**

* dv: dependent variable
* iv: independent variable
* mydata: training data containing `dv` and `iv`

We use two numerical features in the [mlb data "01a_data.txt"](https://umich.instructure.com/files/330381/download?download_frd=1) `Age` and `Height` as features.

```{r eval=T}
#install.packages("rpart")
library(rpart)
mlb.rpart<-rpart(Weight~Height+Age, data=mlb_train)
mlb.rpart
```

The output contains rich information. `split` indicates the method to split; `n` is the number of observations that falls in this segment; `yval` is the predicted value if the test data falls into a segment.

## Visualizing decision trees

A fancy way of drawing the `rpart` decision tree is by `rpart.plot()` function under `rpart.plot` package.

```{r eval=T}
# install.packages("rpart.plot")
library(rpart.plot)
rpart.plot(mlb.rpart, digits=3)
```

A more detailed graph can be obtained stating more options in the function.
```{r }
rpart.plot(mlb.rpart, digits = 4, fallen.leaves = T, type=3, extra=101)
```

Also, you can use a more fancy tree plot from package `rattle`. From the fancy plot, you can observe the order and rules of splits.

```{r}
library(rattle)
fancyRpartPlot(mlb.rpart, cex = 0.8)
```

## Step 4 - evaluating model performance

Let's make predictions with the prediction tree. `predict()` command is used in this section.

```{r eval=T}
mlb.p<-predict(mlb.rpart, mlb_test)
summary(mlb.p)
summary(mlb_test$Weight)
```

After comparing five-number statistics for the predicted and true `Weight`, we can see that the model cannot precisely identify extreme cases such as the maximum. However, within IQR, the predictions are relatively accurate. 

Correlation could be used to measure the correspondence of two equal length numeric variables. Let's use `cor()` to examine the prediction accuracy.

```{r eval=T}
cor(mlb.p, mlb_test$Weight)
```

The predicted values ($Weights$) are moderately correlated with their true value counterparts. [Chapter 13](http://www.socr.umich.edu/people/dinov/2017/Spring/DSPA_HS650/notes/13_ModelEvaluation.html) provides additional strategies for model quality assessment.

## Measuring performance with mean absolute error

To measure the distance between predicted value and the true value, we can use a measurement called mean absolute error (MAE). MAE follows the following formula:
$$MAE=\frac{1}{n}\sum_{i=1}^{n}|pred_i-obs_i|,$$
where the `pred_i` is the $i^{th}$ predicted value and `obs_i` is the $i^{th}$ observed value. Let's make a corresponding MAE function in R and evaluate our model performance.

```{r eval=T}
MAE<-function(obs, pred){
  mean(abs(obs-pred))
}
MAE(mlb_test$Weight, mlb.p)
```

This implies on average the difference between predicted value and the observed value is 14.975. Considering that the `Weight` variable in our test dataset ranges from 150 to 260, the model performs well.

What if we are using the most primitive method for prediction - the **mean**?

```{r eval=T}
mean(mlb_test$Weight)
MAE(mlb_test$Weight, 202.3643)
```

This proves that the predictive decision tree is better than using the **over all mean** to predict every observation in the test dataset. However, it is not dramatically better. There might be room for improvement.

## Step 5 - improving model performance

To improve the performance of our tree, we are going to use a model tree instead of a regression tree. `M5P()` function under package `RWeka` is used in this chapter. It stands for the `M5` algorithm. This function uses similar syntax as `rpart()`.

**m<-M5P(dv~iv, data=mydata)**

```{r eval=T}
#install.packages("RWeka")

# Sometimes RWeka installations may be off a bit, see:
# http://stackoverflow.com/questions/41878226/using-rweka-m5p-in-rstudio-yields-java-lang-noclassdeffounderror-no-uib-cipr-ma

Sys.getenv("WEKA_HOME") # where does it point to? Maybe some obscure path? 
# if yes, correct the variable:
Sys.setenv(WEKA_HOME="C:\\MY\\PATH\\WEKA_WPM")
library(RWeka)
# WPM("list-packages", "installed")

mlb.m5<-M5P(Weight~Height+Age, data=mlb_train)
mlb.m5
```

Instead of using segment averages to predict, this model uses a linear regression (`LM1`) as the terminal node. In some datasets with more variables, `M5P` could give us multiple linear models under different terminal nodes.

```{r eval=T}
summary(mlb.m5)
mlb.p.m5<-predict(mlb.m5, mlb_test)
summary(mlb.p.m5)
cor(mlb.p.m5, mlb_test$Weight)
MAE(mlb_test$Weight, mlb.p.m5)
```

`summary(mlb.m5)` give us some rough diagnostic statistics.
We can see that the correlation and MAE for this model is better than the previous model under `rpart()`.

# Practice Problem: Heart Attack Data

Let's use the heart attack dataset for practice.

```{r eval=T}
heart_attack<-read.csv("https://umich.instructure.com/files/1644953/download?download_frd=1", stringsAsFactors = F)
str(heart_attack)
```

To begin with, we need to convert the `CHARGES` (independent variable) to numerical form. NA's are created so let's remain only the complete cases as mentioned in the beginning of this chapter. Also, let's create a gender variable as an indicator for female patients using `ifelse()` and delete the previous `SEX` column.

```{r eval=T}
heart_attack$CHARGES<-as.numeric(heart_attack$CHARGES)
heart_attack<-heart_attack[complete.cases(heart_attack), ]
heart_attack$gender<-ifelse(heart_attack$SEX=="F", 1, 0)
heart_attack<-heart_attack[, -3]
```

Now we can build a model tree using `M5P()` with all the features in the model. As usual, we need to separate the `heart_attack` data in to training and test datasets (use the 75%-25% way of separation).

After using the model to predict `CHARGES` in the test dataset we can obtain the following correlation and MAE.

```{r echo=F}
set.seed(1234)
train_index <- sample(seq_len(nrow(heart_attack)), size = 0.75*nrow(heart_attack))
ha_train<-heart_attack[train_index, ]
ha_test<-heart_attack[-train_index, ]

ha.m5<-M5P(CHARGES~., data=ha_train)
ha.pred<-predict(ha.m5, ha_test)
cor(ha.pred, ha_test$CHARGES)
MAE(ha_test$CHARGES, ha.pred)
```

We can see that the predicted values and observed values are strongly correlated. In terms of MAE, it may seem very large at first glance.

```{r eval=T}
range(ha_test$CHARGES)
# 17137-701
# 3193.502/16436
```

However, the test data itself has a wide range and the MAE is within 20% of the range. With only 148 observations, the model did a fairly good job in prediction. Can you reproduce or perhaps improve these results?

Try to replicate these results with [other data from the list of our Case-Studies](https://umich.instructure.com/courses/38100/files/).
