# Logistic Regression in R Tutorial

Logistic regression is yet another technique borrowed by machine learning from the field of statistics. It's a powerful statistical way of modeling a binomial outcome with one or more explanatory variables. It measures the relationship between the categorical dependent variable and one or more independent variables by estimating probabilities using a logistic function, which is the cumulative logistic distribution.

This R tutorial will guide you through a simple execution of logistic regression:

**Tip**: if you're interested in taking your skills with linear regression to the next level, consider also DataCamp's Multiple and Logistic Regression course!

## Regression Analysis: Introduction

As the name already indicates, logistic regression is a regression analysis technique. Regression analysis is a set of statistical processes that you can use to estimate the relationships among variables. More specifically, you use this set of techniques to model and analyze the relationship between a dependent variable and one or more independent variables. Regression analysis helps you to understand how the typical value of the dependent variable changes when one of the independent variables is adjusted and others are held fixed.

As you already read, there are various regression techniques. You can distinguish them by looking at three aspects: the number of independent variables, the type of dependent variables and the shape of regression line.

### Linear Regression

Linear regression is one of the most widely known modeling techniques. It allows you, in short, to use a linear relationship to predict the (average) numerical value of $Y$ for a given value of $X$ with a straight line. This line is called the "regression line".

As a consequence, the linear regression model is $y= ax + b$. The model assumes that the response variable $y$ is quantitative. However, in many situations, the response variable is qualitative or, in other words, categorical. For example, gender is qualitative, taking on values male or female.

Prediciting a qualitative response for an observation can be referred to as classifying that observation, since it involves assigning the observation to a category, or class. On the other hand, the methods that are often used for classification first predict the probability of each of the categories of a qualitative variable, as the basis for making the classification.

Linear regression is not capable of predicting probability. If you use linear regression to model a binary response variable, for example, the resulting model may not restrict the predicted Y values within 0 and 1. Here's where logistic regression comes into play, where you get a probaiblity score that reflects the probability of the occurrence at the event.

### Logistic Regression

Logistic regression is an instance of classification technique that you can use to predict a qualitative response. More specifically, logistic regression models the probability that $gender$ belongs to a particular category.

That means that, if you are trying to do gender classification, where the response $gender$ falls into one of the two categories, `male`

or `female`

, you'll use logistic regression models to estimate the probability that $gender$ belongs to a particular category.

For example, the probability of $gender$ given $longhair$ can be written as:

$Pr(gender = female | longhair)$

The values of $Pr(gender = female | longhair)$ (abbreviated as $p(longhair)$) will range between 0 and 1. Then, for any given value of $long hair$, a prediction can be made for $gender$.

Given $X$ as the explanatory variable and $Y$ as the response variable, how should you then model the relationship between $p(X)=Pr(Y=1|X)$ and $X$? The linear regression model represents these probabilities as:

*p*(*X*)=*β*_{0} + *β*_{1}*X*

The problem with this approach is that, any time a straight line is fit to a binary response that is coded as $0$ or $1$, in principle we can always predict $p(X) < 0$ for some values of $X$ and $p(X) > 1$ for others.

To avoid this problem, you can use the logistic function to model $p(X)$ that gives outputs between $0$ and $1$ for all values of $X$:

$$ p(X) = \frac{ e^{\beta_{0} + \beta_{1}X} }{1 + e^{\beta_{0} + \beta_{1}X} } $$

The logistic function will always produce an S-shaped curve, so regardless of the value of $X$, we will obtain a sensible prediction.

The above equation can also be reframed as:

$$ \frac{p(X)}{1 - p(X)} = e^{\beta_{0} + \beta_{1}X}$$

The quantity $$\frac{p(X)}{1 - p(X)}$$ is called the odds ratio, and can take on any value between $0$ and $\infty$. Values of the odds ratio close to $0$ and $\infty$ indicate very low and very high probabilities of $p(X)$, respectively.

By taking the logarithm of both sides from the equation above, you get:

$$ log(\frac{p(X)}{1 - p(X)}) = \beta_{0} + \beta_{1}X $$

The left-hand side is called the **logit**. In a logistic regression
model, increasing *X* by one unit changes the logit by *β*_{0}.
The amount that *p*(*X*) changes due to a one-unit change in *X* will
depend on the current value of *X*. But regardless of the value of *X*,
if *β*_{1} is positive then increasing *X* will be associated
with increasing *p*(*X*), and if *β*_{1} is negative then
increasing *X* will be associated with decreasing *p*(*X*).

The coefficients *β*_{0} and *β*_{1} are unknown, and
must be estimated based on the available training data. For logistic
regression, you can use **maximum likelihood**, a powerful statistical
technique. Let's refer back to your gender classification example.

You seek estimates for *β*_{0} and *β*_{1} such that
plugging these estimates into the model for *p*(*X*) yields a number
close to 1 for all individuals who are female, and a number close to 0
for all individuals who are not.

This intuition can be formalized using a mathematical equation called a likelihood function:

*l*(*β*_{0}, *β*_{1})=*p*(*X*)(1 − *p*(*X*))

The estimates *β*_{0} and *β*_{1} are chosen to
maximize this likelihood function. Once the coefficients have been
estimated, you can simply compute the probability of being $female$ given
any instance of having $long hair$. Overall, maximum likelihood is a
very good approach to fit non-linear models.

### Multinomial Logistic Regression

So far, this tutorial has only focused on Binomial Logistic Regression, since you were classifying instances as male or female. Multinomial Logistic Regression model is a simple extension of the binomial logistic regression model, which you use when the exploratory variable has more than two nominal (unordered) categories.

In multinomial logistic regression, the exploratory variable is dummy coded into multiple 1/0 variables. There is a variable for all categories but one, so if there are M categories, there will be $M−1$ dummy variables. Each category’s dummy variable has a value of 1 for its category and a 0 for all others. One category, the reference category, doesn’t need its own dummy variable, as it is uniquely identified by all the other variables being 0.

The mulitnomial logistic regression then estimates a separate binary logistic regression model for each of those dummy variables. The result is $M−1$ binary logistic regression models. Each model conveys the effect of predictors on the probability of success in that category, in comparison to the reference category.

### Ordinal Logistic Regression

Next to multinomial logistic regression, you also have ordinal logistic regression, which is another extension of binomial logistics regression. Ordinal regression is used to predict the dependent variable with ‘ordered’ multiple categories and independent variables. You already see this coming back in the name of this type of logistic regression, since "ordinal" means "order of the categories".

In other words, it is used to facilitate the interaction of dependent variables (having multiple ordered levels) with one or more independent variables.

For example, you are doing customer interviews to evaluate their satisfaction towards our newly released product. You are tasked to ask a question to respondent where their answer lies between $Satisfactory$ or $Unsatisfactory$. To generalize the answers well, you add levels to your responses such as $Very Unsatisfactory$, $Unsatisfactory$, $Neutral$, $Satisfactory$, $Very Satisfactory$. This helped you to observe a natural order in the categories.

## Logistic Regression in R with `glm`

In this section, you'll study an example of a binary logistic
regression, which you'll tackle with the `ISLR`

package, which will
provide you with the data set, and the `glm()`

function, which is
generally used to fit generalized linear models, will be used to fit the
logistic regression model.

### Loading Data

The first thing to do is to install and load the `ISLR`

package, which has
all the datasets you're going to use.

```
require(ISLR)
## Loading required package: ISLR
## Warning: package 'ISLR' was built under R version 3.3.2
```

For this tutorial, you're going to work with the Smarket dataset within RStudio. The dataset shows daily percentage returns for the S&P 500 stock index between 2001 and 2005.

### Exploring Data

Let's explore it for a bit. `names()`

is useful for seeing what's on the
data frame, `head()`

is a glimpse of the first few rows, and `summary()`

is
also useful.

```
names(Smarket)
## [1] "Year" "Lag1" "Lag2" "Lag3" "Lag4" "Lag5"
## [7] "Volume" "Today" "Direction"
head(Smarket)
## Year Lag1 Lag2 Lag3 Lag4 Lag5 Volume Today Direction
## 1 2001 0.381 -0.192 -2.624 -1.055 5.010 1.1913 0.959 Up
## 2 2001 0.959 0.381 -0.192 -2.624 -1.055 1.2965 1.032 Up
## 3 2001 1.032 0.959 0.381 -0.192 -2.624 1.4112 -0.623 Down
## 4 2001 -0.623 1.032 0.959 0.381 -0.192 1.2760 0.614 Up
## 5 2001 0.614 -0.623 1.032 0.959 0.381 1.2057 0.213 Up
## 6 2001 0.213 0.614 -0.623 1.032 0.959 1.3491 1.392 Up
summary(Smarket)
## Year Lag1 Lag2
## Min. :2001 Min. :-4.922000 Min. :-4.922000
## 1st Qu.:2002 1st Qu.:-0.639500 1st Qu.:-0.639500
## Median :2003 Median : 0.039000 Median : 0.039000
## Mean :2003 Mean : 0.003834 Mean : 0.003919
## 3rd Qu.:2004 3rd Qu.: 0.596750 3rd Qu.: 0.596750
## Max. :2005 Max. : 5.733000 Max. : 5.733000
## Lag3 Lag4 Lag5
## Min. :-4.922000 Min. :-4.922000 Min. :-4.92200
## 1st Qu.:-0.640000 1st Qu.:-0.640000 1st Qu.:-0.64000
## Median : 0.038500 Median : 0.038500 Median : 0.03850
## Mean : 0.001716 Mean : 0.001636 Mean : 0.00561
## 3rd Qu.: 0.596750 3rd Qu.: 0.596750 3rd Qu.: 0.59700
## Max. : 5.733000 Max. : 5.733000 Max. : 5.73300
## Volume Today Direction
## Min. :0.3561 Min. :-4.922000 Down:602
## 1st Qu.:1.2574 1st Qu.:-0.639500 Up :648
## Median :1.4229 Median : 0.038500
## Mean :1.4783 Mean : 0.003138
## 3rd Qu.:1.6417 3rd Qu.: 0.596750
## Max. :3.1525 Max. : 5.733000
```

The `summary()`

function gives you a simple summary of each of the
variables on the Smarket data frame. You can see that there's a number
of lags, volume, today's price, and direction. You will use `Direction`

as
a response vairable, as that shows whether the market went up or down
since the previous day.

### Visualizing Data

Data visualization is perhaps the fastest and most useful way to summarize and learn more about your data. You'll start by exploring the numeric variables individually.

Histograms provide a bar chart of a numeric variable split into bins with the height showing the number of instances that fall into each bin. They are useful to get an indication of the distribution of an attribute.

```
par(mfrow=c(1,8))
for(i in 1:8) {
hist(Smarket[,i], main=names(Smarket)[i])
}
```

It's extremely hard to see, but most of the variables show a Gaussian or double Gaussian distribution.

You can look at the distribution of the data a different way using box and whisker plots. The box captures the middle 50% of the data, the line shows the median and the whiskers of the plots show the reasonable extent of data. Any dots outside the whiskers are good candidates for outliers.

```
par(mfrow=c(1,8))
for(i in 1:8) {
boxplot(Smarket[,i], main=names(Smarket)[i])
}
```

You can see that the `Lag`

s and `Today`

all has a similar range. Otherwise, there's no sign of any outliers.

Missing data have have a big impact on modeling. Thus, you can use a missing plot to get a quick idea of the amount of missing data in the dataset. The x-axis shows attributes and the y-axis shows instances. Horizontal lines indicate missing data for an instance, vertical blocks represent missing data for an attribute.

```
library(Amelia)
library(mlbench)
missmap(Smarket, col=c("blue", "red"), legend=FALSE)
```

Well, lucky for me! No missing data in this dataset!

Let's start calculating the correlation between each pair of numeric variables. These pair-wise correlations can be plotted in a correlation matrix plot to given an idea of which variables change together.

```
library(corrplot)
correlations <- cor(Smarket[,1:8])
corrplot(correlations, method="circle")
```

A dot-representation was used where blue represents positive correlation and red negative. The larger the dot the larger the correlation. You can see that the matrix is symmetrical and that the diagonal are perfectly positively correlated because it shows the correlation of each variable with itself. Unfortunately, none of the variables are correlated with one another.

Let's make a plot of the data. There's a `pairs()`

function which plots the variables in `Smarket`

into a scatterplot matrix. In this case, `Direction`

, your binary response, is the color indicator:

```
pairs(Smarket, col=Smarket$Direction)
```

It looks like there's not much correlation going on here. The class variable is derived from the variable `Today`

, so `Up`

and `Down`

seems to make a division. Other than that, there's not much going on.

Let's take a look at the density distribution of each variable broken down by `Direction`

value. Like the scatterplot matrix above, the density plot by Direction can help see the separation of `Up`

and `Down`

. It can also help to understand the overlap in `Direction`

values for a variable.

```
library(caret)
x <- Smarket[,1:8]
y <- Smarket[,9]
scales <- list(x=list(relation="free"), y=list(relation="free"))
featurePlot(x=x, y=y, plot="density", scales=scales)
```

You can see that the `Direction`

values overlap for all of these variables, meaning that it's hard to predict `Up`

or `Down`

based on just one or two variables.

### Building Logistic Regression Model

Now you call `glm.fit()`

function. The first argument that you pass to
this function is an R formula. In this case, the formula indicates that
`Direction`

is the response, while the `Lag`

and `Volume`

variables are
the predictors. As you saw in the introduction, `glm`

is generally used
to fit generalized linear models.

However, in this case, you need to
make it clear that you want to fit a logistic regression model. You
resolve this by setting the `family`

argument to `binomial`

. This way,
you tell `glm()`

to put fit a logistic regression model instead of one
of the many other models that can be fit to the `glm`

.

```
# Logistics Regression
glm.fit <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Smarket, family = binomial)
```

Next, you can do a `summary()`

, which tells you something about the fit:

```
summary(glm.fit)
##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = binomial, data = Smarket)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.446 -1.203 1.065 1.145 1.326
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.126000 0.240736 -0.523 0.601
## Lag1 -0.073074 0.050167 -1.457 0.145
## Lag2 -0.042301 0.050086 -0.845 0.398
## Lag3 0.011085 0.049939 0.222 0.824
## Lag4 0.009359 0.049974 0.187 0.851
## Lag5 0.010313 0.049511 0.208 0.835
## Volume 0.135441 0.158360 0.855 0.392
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1731.2 on 1249 degrees of freedom
## Residual deviance: 1727.6 on 1243 degrees of freedom
## AIC: 1741.6
##
## Number of Fisher Scoring iterations: 3
```

As you can see, `summary()`

returns the estimate, standard errors,
z-score, and p-values on each of the coefficients. Look like none of the
coefficients are significant here. It also gives you the null deviance
(the deviance just for the mean) and the residual deviance (the deviance
for the model with all the predictors). There's a very small difference
between the 2, along with 6 degrees of freedom.

You assign the result of `predict()`

of `glm.fit()`

to `glm.probs`

, with
type equals to response. This will make predictions on the training data
that you use to fit the model and give me a vector of fitted
probabilities.

You look at the first 5 probabilities and they are very close to 50%:

```
glm.probs <- predict(glm.fit,type = "response")
glm.probs[1:5]
## 1 2 3 4 5
## 0.5070841 0.4814679 0.4811388 0.5152224 0.5107812
```

Now I am going to make a prediction of whether the market will be up or
down based on the lags and other predictors. In particular, I'll turn
the probabilities into classifications by thresholding at 0.5. In order
to do so, I use an `ifelse()`

command.

```
glm.pred <- ifelse(glm.probs > 0.5, "Up", "Down")
```

`glm.pred`

is a vector of trues and falses. If `glm.probs`

is bigger
than 0.5, `glm.pred`

calls `"Up"`

; otherwise, it calls `"False"`

.

Here, you attach the data frame `Smarket`

and make a table of `glm.pred`

,
which is the ups and downs from the previous direction. You also take the
mean of those.

```
attach(Smarket)
table(glm.pred,Direction)
## Direction
## glm.pred Down Up
## Down 145 141
## Up 457 507
mean(glm.pred == Direction)
## [1] 0.5216
```

From the table, instances on the diagonals are where you get the correct classification, and off the diagonals are where you make mistake. Looks like you made a lot of mistakes. The mean gives a proportion of 0.52.

### Creating Training and Test Samples

How can you do better? Dividing the data up into a training set and a test set is a good strategy.

```
# Make training and test set
train = Year<2005
glm.fit <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = Smarket,
family = binomial,
subset = train)
glm.probs <- predict(glm.fit,
newdata = Smarket[!train,],
type = "response")
glm.pred <- ifelse(glm.probs > 0.5, "Up", "Down")
```

Let's look at this code chunk in detail:

`train`

is equal to the year less than 2005. For all the year less than 2005, you'll get a`true`

; otherwise, I'll get a`false`

.- You then refit the model with
`glm.fit()`

, except that the subset is equal to 'train', which means that it fits to just the data in year less than 2005. - You then use the
`predict()`

function again for`glm.probs`

to predict on the remaining data in year greater or equal to 2005. For the new data, You give it Smarket, indexed by`!train`

(`!train`

is`true`

if the year is greater or equal to 2005). You set`type`

to`"response"`

to predict the probabilities. - Finally, you use the
`ifelse()`

function again for`glm.pred`

to make`Up`

and`Down`

variable.

You now make a new variable to store a new subset for the test data and
call it `Direction.2005`

. The response variable is still `Direction`

.
You make a table and compute the mean on this new test set:

```
Direction.2005 = Smarket$Direction[!train]
table(glm.pred, Direction.2005)
## Direction.2005
## glm.pred Down Up
## Down 77 97
## Up 34 44
mean(glm.pred == Direction.2005)
## [1] 0.4801587
```

Ha, you did worse than the previous case. How could this happen?

### Solving Overfitting

Well, you might have overfitted the data. In order to fix this, you're going
to fit a smaller model and use `Lag1`

, `Lag2`

, `Lag3`

as the predictors,
thereby leaving out all other variables. The rest of the code is the
same.

```
# Fit a smaller model
glm.fit = glm(Direction ~ Lag1 + Lag2 + Lag3, data = Smarket, family = binomial, subset = train)
glm.probs = predict(glm.fit, newdata = Smarket[!train,], type = "response")
glm.pred = ifelse(glm.probs > 0.5, "Up", "Down")
table(glm.pred, Direction.2005)
## Direction.2005
## glm.pred Down Up
## Down 39 31
## Up 72 110
mean(glm.pred == Direction.2005)
## [1] 0.5912698
```

Well, you got a classification rate of 59%, not too bad. Using the smaller model appears to perform better.

Lastly, you will do a `summary()`

of `glm.fit`

to see if there are any
signficant changes.

```
summary(glm.fit)
##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3, family = binomial,
## data = Smarket, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.338 -1.189 1.072 1.163 1.335
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.032230 0.063377 0.509 0.611
## Lag1 -0.055523 0.051709 -1.074 0.283
## Lag2 -0.044300 0.051674 -0.857 0.391
## Lag3 0.008815 0.051495 0.171 0.864
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1383.3 on 997 degrees of freedom
## Residual deviance: 1381.4 on 994 degrees of freedom
## AIC: 1389.4
##
## Number of Fisher Scoring iterations: 3
```

Nothing became significant, at least the P-values are better, indicating an increase in prediction of performance.

## Conclusion

So that's the end of this R tutorial on building logistic regression models using the `glm()`

function and setting family to binomial. `glm()`

does not assume a linear relationship between dependent and independent variables. However, it assumes a linear relationship between link function and independent variables in logit model I hope you have learned something valuable!