Statiscal_LearingR
Statiscal Learning with R
Chapter 2
Basics Comands
x <- c(1,6,2)
x## [1] 1 6 2y = c(1,4,3)
y## [1] 1 4 3length(x)## [1] 3length(y)## [1] 3x+y## [1] 2 10 5- The ls() function allows us to look at a list of all of the objects, such as data and functions, that we have saved so far. The rm() function can be used to delete any that we don’t want.
ls()## [1] "x" "y"rm(x,y)
rm(list = ls()) #It's also possible to remove all objects at once- Matrix: the data (the entries in the matrix), the number of rows, and the number of columns
x <- matrix(data = c(1,2,3,4), nrow = 2, ncol = 2)
x## [,1] [,2]
## [1,] 1 3
## [2,] 2 4matrix(c(1,2,3,4), 2,2, byrow = TRUE)## [,1] [,2]
## [1,] 1 2
## [2,] 3 4- The sqrt() function returns the square root of each element of a vector or matrix. The command x^2 raises each element of x to the power 2; any powers are possible, including fractional or negative powers.
sqrt(x)## [,1] [,2]
## [1,] 1.000000 1.732051
## [2,] 1.414214 2.000000x^2## [,1] [,2]
## [1,] 1 9
## [2,] 4 16- The rnorm() function generates a vector of random normal variables, with first argument n the sample size. Each time we call this function, we will get a different answer. Here we create two correlated sets of numbers, x and y, and use the cor() function to compute the correlation between them.
x <- rnorm(50)
y <- x + rnorm(50, mean = 50, sd = .1)
cor(x, y)## [1] 0.9962812- Sometimes we want our code to reproduce the exact same set of random numbers; we can use the set.seed() function to do this. The set.seed() function takes an (arbitrary) integer argument.
set.seed(1303)
rnorm(50)## [1] -1.1439763145 1.3421293656 2.1853904757 0.5363925179 0.0631929665
## [6] 0.5022344825 -0.0004167247 0.5658198405 -0.5725226890 -1.1102250073
## [11] -0.0486871234 -0.6956562176 0.8289174803 0.2066528551 -0.2356745091
## [16] -0.5563104914 -0.3647543571 0.8623550343 -0.6307715354 0.3136021252
## [21] -0.9314953177 0.8238676185 0.5233707021 0.7069214120 0.4202043256
## [26] -0.2690521547 -1.5103172999 -0.6902124766 -0.1434719524 -1.0135274099
## [31] 1.5732737361 0.0127465055 0.8726470499 0.4220661905 -0.0188157917
## [36] 2.6157489689 -0.6931401748 -0.2663217810 -0.7206364412 1.3677342065
## [41] 0.2640073322 0.6321868074 -1.3306509858 0.0268888182 1.0406363208
## [46] 1.3120237985 -0.0300020767 -0.2500257125 0.0234144857 1.6598706557- The mean() and var() functions can be used to compute the mean and variance of a vector of numbers. Applying sqrt() to the output of var() will give the standard deviation. Or we can simply use the sd() function
set.seed(3)
y <- rnorm(100)
mean(y)## [1] 0.01103557var(y)## [1] 0.7328675sqrt(var(y))## [1] 0.8560768sd(y)## [1] 0.8560768Graphics
x <- rnorm(100)
y <- rnorm(100)
plot(x,y)
plot(x, y, xlab = "this is the x-axis", ylab = "this is the y-axis", main = "Plot of X vs Y")
pdf("Figure.pdf")
plot(x, y, col = "green")
dev.off() #The function dev.off() indicates to R that we are done creating the plot## png
## 2- The function seq() can be used to create a sequence of numbers.
x <- seq(1, 10)
x## [1] 1 2 3 4 5 6 7 8 9 10x <- 1:10
x <- seq(-pi, pi, length = 50)- We will now create some more sophisticated plots. The contour() function produces a contour plot in order to represent three-dimensional data; it is like a topographical map. It takes three arguments:
- A vector of the x values (the first dimension),
- A vector of the y values (the second dimension), and
- A matrix whose elements correspond to the z value (the third dimension) for each pair of (x, y) coordinates.
y <- x
f <- outer(x, y, function(x,y) cos(y) / (1 + x^2))
contour(x, y, f)
contour(x, y, f, nlevels = 45, add = T)
fa <- (f - t(f)) / 2
contour(x, y, fa, nlevels = 15)
- The image() function works the same way as contour(), except that it produces a color-coded plot whose colors depend on the z value. This is known as a heatmap, and is sometimes used to plot temperature in weather forecasts. Alternatively, persp() can be used to produce a three-dimensional plot. The arguments theta and phi control the angles at which the plot is viewed.
image(x, y, fa)
persp(x, y, fa)
persp(x, y, fa, theta = 30)
persp(x, y, fa, theta = 30, phi = 20)
persp(x, y, fa, theta = 30, phi = 70)
persp(x, y, fa, theta = 30, phi = 40)
Indexing Data
A <- matrix(1:16, nrow = 4, ncol = 4)
A## [,1] [,2] [,3] [,4]
## [1,] 1 5 9 13
## [2,] 2 6 10 14
## [3,] 3 7 11 15
## [4,] 4 8 12 16A[2,3] #Primeiro linha e depois seleciona a coluna## [1] 10A[c(1,3), c(2,4)] #Select multiple rows and columns at a time## [,1] [,2]
## [1,] 5 13
## [2,] 7 15A[1:3, 2:4]## [,1] [,2] [,3]
## [1,] 5 9 13
## [2,] 6 10 14
## [3,] 7 11 15A[1:2,]## [,1] [,2] [,3] [,4]
## [1,] 1 5 9 13
## [2,] 2 6 10 14A[, 1:2]## [,1] [,2]
## [1,] 1 5
## [2,] 2 6
## [3,] 3 7
## [4,] 4 8A[1,]## [1] 1 5 9 13A[-c(1,3),] #The use of a negative sign - in the index tells R to keep all rows or columns except those indicated in the index.## [,1] [,2] [,3] [,4]
## [1,] 2 6 10 14
## [2,] 4 8 12 16dim(A) #Give the number of rows and columns of given matrix## [1] 4 4Loading Data
Auto <- read.csv("C:/Users/naian/Documents/UTwente/DataScience/Auto.csv", header = TRUE, na.strings = "?", stringsAsFactors = TRUE )
head(Auto)## mpg cylinders displacement horsepower weight acceleration year origin
## 1 18 8 307 130 3504 12.0 70 1
## 2 15 8 350 165 3693 11.5 70 1
## 3 18 8 318 150 3436 11.0 70 1
## 4 16 8 304 150 3433 12.0 70 1
## 5 17 8 302 140 3449 10.5 70 1
## 6 15 8 429 198 4341 10.0 70 1
## name
## 1 chevrolet chevelle malibu
## 2 buick skylark 320
## 3 plymouth satellite
## 4 amc rebel sst
## 5 ford torino
## 6 ford galaxie 500dim(Auto)## [1] 392 9Auto[1:4,]## mpg cylinders displacement horsepower weight acceleration year origin
## 1 18 8 307 130 3504 12.0 70 1
## 2 15 8 350 165 3693 11.5 70 1
## 3 18 8 318 150 3436 11.0 70 1
## 4 16 8 304 150 3433 12.0 70 1
## name
## 1 chevrolet chevelle malibu
## 2 buick skylark 320
## 3 plymouth satellite
## 4 amc rebel sstAuto <- na.omit(Auto) #Essa função retira as linhas com missing values
names(Auto)## [1] "mpg" "cylinders" "displacement" "horsepower" "weight"
## [6] "acceleration" "year" "origin" "name"Explicação dos argumentos da função read.csv A) Header = TRUE means that the first line of the file contains the variable names
Using the option na.strings tells R that any time it sees a particular character or set of characters (such as a question mark), it should be treated as a missing element of the data matrix.
The stringsAsFactors = T argument tells R that any variable containing character strings should be interpreted as a qualitative variable, and that each distinct character string represents a distinct level for that qualitative variable.
Funções adicionais de gráfico
plot(Auto$cylinders, Auto$mpg) #Não posso somente colocar os nomes das variáveis, devo colocar o nome da tabela e $ e indicar a coluna desejda
attach(Auto) #tell R to make the variables in this data frame available by name, sem usar essa função antes, o R não me retornaria as clunas
plot(cylinders, mpg)
- The cylinders variable is stored as a numeric vector, so R has treated it as quantitative. However, since there are only a small number of possible values for cylinders, one may prefer to treat it as a qualitative variable. The as.factor() function converts quantitative variables into qualitative variables.
cylinders <- as.factor(Auto$cylinders)
class(cylinders)## [1] "factor"- If the variable plotted on the x-axis is qualitative, then boxplots will boxplot automatically be produced by the plot() function
plot(cylinders, mpg, col = "red", varwidth = T, xlab = "cylinders", ylab = "mpg", main = "Gráfico Teste")
- Histogram
hist(mpg, col = 2, breaks = 15) #Col = 2 quer dizer vermelho e breaks sobre a divisão dos numéros no gráfico
- Scatterplot Matrix
pairs(Auto)
pairs( ~ mpg + displacement + horsepower + weight + acceleration ,
data = Auto
) #produce scatterplots for just a subset of the variables.
- In conjunction with the plot() function, identify() provides a useful interactive method for identifying the value of a particular variable for points on a plot. We pass in three arguments to identify(): the x-axis variable, the y-axis variable, and the variable whose values we would like to see printed for each point. Then clicking one or more points in the plot and hitting Escape will cause R to print the values of the variable of interest. The numbers printed under the identify() function correspond to the rows for the selected points.
plot(horsepower, mpg)
identify(horsepower, mpg, name)
## integer(0)summary(Auto)## mpg cylinders displacement horsepower weight
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140
##
## acceleration year origin name
## Min. : 8.00 Min. :70.00 Min. :1.000 amc matador : 5
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 ford pinto : 5
## Median :15.50 Median :76.00 Median :1.000 toyota corolla : 5
## Mean :15.54 Mean :75.98 Mean :1.577 amc gremlin : 4
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 amc hornet : 4
## Max. :24.80 Max. :82.00 Max. :3.000 chevrolet chevette: 4
## (Other) :365summary(mpg)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 9.00 17.00 22.75 23.45 29.00 46.60Chapter 3
- Packages:
- MASS package: is a very large collection of data sets and functions.
- ISLR2: includes the data sets associated with this book
library(MASS)
library(ISLR2)## Warning: package 'ISLR2' was built under R version 4.0.5##
## Attaching package: 'ISLR2'## The following object is masked _by_ '.GlobalEnv':
##
## Auto## The following object is masked from 'package:MASS':
##
## BostonSimple Linear Regression
- The ISLR2 library contains the Boston data set, which records medv (median house value) for 506 census tracts in Boston. We will seek to predict medv using 12 predictors such as rm (average number of rooms per house), age (average age of houses), and lstat (percent of households with low socioeconomic status).
head(Boston)## crim zn indus chas nox rm age dis rad tax ptratio lstat medv
## 1 0.00632 18 2.31 0 0.538 6.575 65.2 4.0900 1 296 15.3 4.98 24.0
## 2 0.02731 0 7.07 0 0.469 6.421 78.9 4.9671 2 242 17.8 9.14 21.6
## 3 0.02729 0 7.07 0 0.469 7.185 61.1 4.9671 2 242 17.8 4.03 34.7
## 4 0.03237 0 2.18 0 0.458 6.998 45.8 6.0622 3 222 18.7 2.94 33.4
## 5 0.06905 0 2.18 0 0.458 7.147 54.2 6.0622 3 222 18.7 5.33 36.2
## 6 0.02985 0 2.18 0 0.458 6.430 58.7 6.0622 3 222 18.7 5.21 28.7View(Boston)
?Boston #Mostra as informações de cada variavel## starting httpd help server ... done- We will start by using the lm() function to fit a simple linear regression model, with medv as the response and lstat as the predictor. The basic syntax is lm(y ∼ x, data), where y is the response, x is the predictor, and data is the data set in which these two variables are kept.
lm.fit <- lm(medv ~ lstat, data = Boston)
attach(Boston) #Ao escrever essa função, não vou precisar especificar nas próximas que meus dados vem de Boston
#lm.fit <- lm(medv ~ lstat)- Observar as informações obtidas
lm.fit##
## Call:
## lm(formula = medv ~ lstat, data = Boston)
##
## Coefficients:
## (Intercept) lstat
## 34.55 -0.95summary(lm.fit)##
## Call:
## lm(formula = medv ~ lstat, data = Boston)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.168 -3.990 -1.318 2.034 24.500
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 34.55384 0.56263 61.41 <2e-16 ***
## lstat -0.95005 0.03873 -24.53 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6.216 on 504 degrees of freedom
## Multiple R-squared: 0.5441, Adjusted R-squared: 0.5432
## F-statistic: 601.6 on 1 and 504 DF, p-value: < 2.2e-16names(lm.fit)## [1] "coefficients" "residuals" "effects" "rank"
## [5] "fitted.values" "assign" "qr" "df.residual"
## [9] "xlevels" "call" "terms" "model"coef(lm.fit)## (Intercept) lstat
## 34.5538409 -0.9500494- Intervalo de Confiança
In order to obtain a confidence interval for the coefficient estimates, we can use the confint() command
confint(lm.fit)## 2.5 % 97.5 %
## (Intercept) 33.448457 35.6592247
## lstat -1.026148 -0.8739505The predict() function can be used to produce confidence intervals and prediction intervals for the prediction of medv for a given value of lstat.
predict(lm.fit, data.frame(lstat = (c(5, 10, 15))), interval = "confidence")## fit lwr upr
## 1 29.80359 29.00741 30.59978
## 2 25.05335 24.47413 25.63256
## 3 20.30310 19.73159 20.87461predict(lm.fit, data.frame(lstat = (c(5,10,15))), interval = "prediction")## fit lwr upr
## 1 29.80359 17.565675 42.04151
## 2 25.05335 12.827626 37.27907
## 3 20.30310 8.077742 32.52846For instance, the 95% confidence interval associated with a lstat value of 10 is (24.47, 25.63), and the 95% prediction interval is (12.828, 37.28). As expected, the confidence and prediction intervals are centered around the same point (a predicted value of 25.05 for medv when lstat equals 10), but the latter are substantially wider.
- Plotagem do gráfico
plot(lstat, medv)
#abline(lm.fit) #Draw any line
abline(lm.fit, lwd = 3, col = "red")
plot(lstat, medv, col = "red", pch = "+")
plot(1:20, 1:20, pch = 1:20)
- Nova Plotagem de gráfico
A função par() e mfrow() pedem para o R dividir a tela em aineis separados para mostrar multiplos gráficos ao mesmo tempo
par(mfrow = c(2,2))
plot(lm.fit)
Alternatively, we can compute the residuals from a linear regression fit using the residuals() function. The function rstudent() will return the residuals()studentized residuals, and we can use this function to plot the residuals against the fitted values.
#plot(predict(lm.fit, residuals(lm.fit))) #Função não deu certo
plot(predict(lm.fit), rstudent(lm.fit))
On the basis of the residual plots, there is some evidence of non-linearity. Leverage statistics can be computed for any number of predictors using the hatvalues() function.
The which.max() function identifies the index of the largest element of a vector. In this case, it tells us which observation has the largest leverage statistic.
plot(hatvalues(lm.fit))
which.max(hatvalues(lm.fit))## 375
## 375Multiple Linear Regression
- In order to fit a multiple linear regression model using least squares, we again use the lm() function. The syntax lm(y ∼ x1 + x2 + x3) is used to fit a model with three predictors, x1, x2, and x3. The summary() function now outputs the regression coefficients for all the predictors.
lm.fit <- lm(medv ~ lstat + age, data = Boston)
summary(lm.fit)##
## Call:
## lm(formula = medv ~ lstat + age, data = Boston)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.981 -3.978 -1.283 1.968 23.158
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 33.22276 0.73085 45.458 < 2e-16 ***
## lstat -1.03207 0.04819 -21.416 < 2e-16 ***
## age 0.03454 0.01223 2.826 0.00491 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6.173 on 503 degrees of freedom
## Multiple R-squared: 0.5513, Adjusted R-squared: 0.5495
## F-statistic: 309 on 2 and 503 DF, p-value: < 2.2e-16The Boston data set contains 12 variables, and so it would be cumbersome to have to type all of these in order to perform a regression using all of the predictors. Instead, we can use the following short-hand
lm.fit <- lm(medv ~ ., data = Boston)
summary(lm.fit)##
## Call:
## lm(formula = medv ~ ., data = Boston)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.1304 -2.7673 -0.5814 1.9414 26.2526
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 41.617270 4.936039 8.431 3.79e-16 ***
## crim -0.121389 0.033000 -3.678 0.000261 ***
## zn 0.046963 0.013879 3.384 0.000772 ***
## indus 0.013468 0.062145 0.217 0.828520
## chas 2.839993 0.870007 3.264 0.001173 **
## nox -18.758022 3.851355 -4.870 1.50e-06 ***
## rm 3.658119 0.420246 8.705 < 2e-16 ***
## age 0.003611 0.013329 0.271 0.786595
## dis -1.490754 0.201623 -7.394 6.17e-13 ***
## rad 0.289405 0.066908 4.325 1.84e-05 ***
## tax -0.012682 0.003801 -3.337 0.000912 ***
## ptratio -0.937533 0.132206 -7.091 4.63e-12 ***
## lstat -0.552019 0.050659 -10.897 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.798 on 493 degrees of freedom
## Multiple R-squared: 0.7343, Adjusted R-squared: 0.7278
## F-statistic: 113.5 on 12 and 493 DF, p-value: < 2.2e-16We can access the individual components of a summary object by name (type ?summary.lm to see what is available). Hence summary(lm.fit)\(r.sq gives us the R2, and summary(lm.fit)\)sigma gives us the RSE. The vif() function, part of the car package, can be used to compute variance inflation factors. Most VIF’s are low to moderate for this data.
library(car)## Warning: package 'car' was built under R version 4.0.5## Loading required package: carDatavif(lm.fit)## crim zn indus chas nox rm age dis
## 1.767486 2.298459 3.987181 1.071168 4.369093 1.912532 3.088232 3.954037
## rad tax ptratio lstat
## 7.445301 9.002158 1.797060 2.870777What if we would like to perform a regression using all of the variables but one? For example, in the above regression output, age has a high p-value. So we may wish to run a regression excluding this predictor. The following syntax results in a regression using all predictors except age.
lm.fit <- lm(medv ~ . -age, data = Boston)
summary(lm.fit)##
## Call:
## lm(formula = medv ~ . - age, data = Boston)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.1851 -2.7330 -0.6116 1.8555 26.3838
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 41.525128 4.919684 8.441 3.52e-16 ***
## crim -0.121426 0.032969 -3.683 0.000256 ***
## zn 0.046512 0.013766 3.379 0.000785 ***
## indus 0.013451 0.062086 0.217 0.828577
## chas 2.852773 0.867912 3.287 0.001085 **
## nox -18.485070 3.713714 -4.978 8.91e-07 ***
## rm 3.681070 0.411230 8.951 < 2e-16 ***
## dis -1.506777 0.192570 -7.825 3.12e-14 ***
## rad 0.287940 0.066627 4.322 1.87e-05 ***
## tax -0.012653 0.003796 -3.333 0.000923 ***
## ptratio -0.934649 0.131653 -7.099 4.39e-12 ***
## lstat -0.547409 0.047669 -11.483 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.794 on 494 degrees of freedom
## Multiple R-squared: 0.7343, Adjusted R-squared: 0.7284
## F-statistic: 124.1 on 11 and 494 DF, p-value: < 2.2e-16Interaction Terms
It is easy to include interaction terms in a linear model using the lm() function. The syntax lstat:black tells R to include an interaction term between lstat and black. The syntax lstat * age simultaneously includes lstat, age, and the interaction term lstat×age as predictors; it is a shorthand for lstat + age + lstat:age.
summary(lm(medv ~ lstat * age, data = Boston))##
## Call:
## lm(formula = medv ~ lstat * age, data = Boston)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.806 -4.045 -1.333 2.085 27.552
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 36.0885359 1.4698355 24.553 < 2e-16 ***
## lstat -1.3921168 0.1674555 -8.313 8.78e-16 ***
## age -0.0007209 0.0198792 -0.036 0.9711
## lstat:age 0.0041560 0.0018518 2.244 0.0252 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6.149 on 502 degrees of freedom
## Multiple R-squared: 0.5557, Adjusted R-squared: 0.5531
## F-statistic: 209.3 on 3 and 502 DF, p-value: < 2.2e-16Non-Linear Transformation of the Predictors
The lm() function can also accommodate non-linear transformations of the predictors. For instance, given a predictor X, we can create a predictor X2 using I(X^2). The function I() is needed since the ^ has a special meaning in a formula object; wrapping as we do allows the standard usage in R, which is to raise X to the power 2. We now perform a regression of medv onto lstat and lstat2.
lm.fit2 <- lm(medv ~ lstat + I(lstat^2))
summary(lm.fit2)##
## Call:
## lm(formula = medv ~ lstat + I(lstat^2))
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.2834 -3.8313 -0.5295 2.3095 25.4148
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 42.862007 0.872084 49.15 <2e-16 ***
## lstat -2.332821 0.123803 -18.84 <2e-16 ***
## I(lstat^2) 0.043547 0.003745 11.63 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.524 on 503 degrees of freedom
## Multiple R-squared: 0.6407, Adjusted R-squared: 0.6393
## F-statistic: 448.5 on 2 and 503 DF, p-value: < 2.2e-16The near-zero p-value associated with the quadratic term suggests that it leads to an improved model. We use the anova() function to further quantify the extent to which the quadratic fit is superior to the linear fit
lm.fit <- lm(medv ~ lstat)
anova(lm.fit, lm.fit2)## Analysis of Variance Table
##
## Model 1: medv ~ lstat
## Model 2: medv ~ lstat + I(lstat^2)
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 504 19472
## 2 503 15347 1 4125.1 135.2 < 2.2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Here Model 1 represents the linear submodel containing only one predictor, lstat, while Model 2 corresponds to the larger quadratic model that has two predictors, lstat and lstat2. The anova() function performs a hypothesis test comparing the two models. The null hypothesis is that the two models fit the data equally well, and the alternative hypothesis is that the full model is superior. Here the F-statistic is 135 and the associated p-value is virtually zero. This provides very clear evidence that the model containing the predictors lstat and lstat2 is far superior to the model that only contains the predictor lstat. This is not surprising, since earlier we saw evidence for non-linearity in the relationship between medv and lstat. If we type
par(mfrow = c(2,2))
plot(lm.fit2)
In order to create a cubic fit, we can include a predictor of the form I(X^3). However, this approach can start to get cumbersome for higherorder polynomials. A better approach involves using the poly() function to create the polynomial within lm(). For example, the following command produces a fifth-order polynomial fit:
lm.fit5 <- lm(medv ~ poly(lstat,5))
summary(lm.fit5)##
## Call:
## lm(formula = medv ~ poly(lstat, 5))
##
## Residuals:
## Min 1Q Median 3Q Max
## -13.5433 -3.1039 -0.7052 2.0844 27.1153
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 22.5328 0.2318 97.197 < 2e-16 ***
## poly(lstat, 5)1 -152.4595 5.2148 -29.236 < 2e-16 ***
## poly(lstat, 5)2 64.2272 5.2148 12.316 < 2e-16 ***
## poly(lstat, 5)3 -27.0511 5.2148 -5.187 3.10e-07 ***
## poly(lstat, 5)4 25.4517 5.2148 4.881 1.42e-06 ***
## poly(lstat, 5)5 -19.2524 5.2148 -3.692 0.000247 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.215 on 500 degrees of freedom
## Multiple R-squared: 0.6817, Adjusted R-squared: 0.6785
## F-statistic: 214.2 on 5 and 500 DF, p-value: < 2.2e-16summary(lm(medv ~ log(rm), data = Boston))##
## Call:
## lm(formula = medv ~ log(rm), data = Boston)
##
## Residuals:
## Min 1Q Median 3Q Max
## -19.487 -2.875 -0.104 2.837 39.816
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -76.488 5.028 -15.21 <2e-16 ***
## log(rm) 54.055 2.739 19.73 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6.915 on 504 degrees of freedom
## Multiple R-squared: 0.4358, Adjusted R-squared: 0.4347
## F-statistic: 389.3 on 1 and 504 DF, p-value: < 2.2e-16Qualitative Predtors
We will attempt to predict Sales (child car seat sales) in 400 locations based on a number of predictors.
head(Carseats)## Sales CompPrice Income Advertising Population Price ShelveLoc Age Education
## 1 9.50 138 73 11 276 120 Bad 42 17
## 2 11.22 111 48 16 260 83 Good 65 10
## 3 10.06 113 35 10 269 80 Medium 59 12
## 4 7.40 117 100 4 466 97 Medium 55 14
## 5 4.15 141 64 3 340 128 Bad 38 13
## 6 10.81 124 113 13 501 72 Bad 78 16
## Urban US
## 1 Yes Yes
## 2 Yes Yes
## 3 Yes Yes
## 4 Yes Yes
## 5 Yes No
## 6 No YesView(Carseats)
?CarseatsGiven a qualitative variable such as Shelveloc, R generates dummy variables automatically. Below we fit a multiple regression model that includes some interaction terms.
lm.fit <- lm(Sales ~ . + Income:Advertising + Price:Age, data = Carseats)
summary(lm.fit)##
## Call:
## lm(formula = Sales ~ . + Income:Advertising + Price:Age, data = Carseats)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.9208 -0.7503 0.0177 0.6754 3.3413
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.5755654 1.0087470 6.519 2.22e-10 ***
## CompPrice 0.0929371 0.0041183 22.567 < 2e-16 ***
## Income 0.0108940 0.0026044 4.183 3.57e-05 ***
## Advertising 0.0702462 0.0226091 3.107 0.002030 **
## Population 0.0001592 0.0003679 0.433 0.665330
## Price -0.1008064 0.0074399 -13.549 < 2e-16 ***
## ShelveLocGood 4.8486762 0.1528378 31.724 < 2e-16 ***
## ShelveLocMedium 1.9532620 0.1257682 15.531 < 2e-16 ***
## Age -0.0579466 0.0159506 -3.633 0.000318 ***
## Education -0.0208525 0.0196131 -1.063 0.288361
## UrbanYes 0.1401597 0.1124019 1.247 0.213171
## USYes -0.1575571 0.1489234 -1.058 0.290729
## Income:Advertising 0.0007510 0.0002784 2.698 0.007290 **
## Price:Age 0.0001068 0.0001333 0.801 0.423812
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.011 on 386 degrees of freedom
## Multiple R-squared: 0.8761, Adjusted R-squared: 0.8719
## F-statistic: 210 on 13 and 386 DF, p-value: < 2.2e-16The contrasts() function returns the coding that R uses for the dummy variables.
attach(Carseats)
contrasts(ShelveLoc)## Good Medium
## Bad 0 0
## Good 1 0
## Medium 0 1Criando uma função
LoadLibraries <- function() {
library(ISLR2)
library(MASS)
print("The libraries have been loaded")
}
LoadLibraries()## [1] "The libraries have been loaded"Chapter 4
This data set consists of percentage returns for the S&P 500 stock index over 1, 250 days, from the beginning of 2001 until the end of 2005. For each date, we have recorded the percentage returns for each of the five previous trading days, Lag1 through Lag5. We have also recorded Volume (the number of shares traded on the previous day, in billions), Today (the percentage return on the date in question) and Direction (whether the market was Up or Down on this date). Our goal is to predict Direction (a qualitative response) using the other features.
- Conhecendo os meus dados
library(ISLR2)
names(Smarket)## [1] "Year" "Lag1" "Lag2" "Lag3" "Lag4" "Lag5"
## [7] "Volume" "Today" "Direction"dim(Smarket)## [1] 1250 9summary(Smarket)## Year Lag1 Lag2 Lag3
## Min. :2001 Min. :-4.922000 Min. :-4.922000 Min. :-4.922000
## 1st Qu.:2002 1st Qu.:-0.639500 1st Qu.:-0.639500 1st Qu.:-0.640000
## Median :2003 Median : 0.039000 Median : 0.039000 Median : 0.038500
## Mean :2003 Mean : 0.003834 Mean : 0.003919 Mean : 0.001716
## 3rd Qu.:2004 3rd Qu.: 0.596750 3rd Qu.: 0.596750 3rd Qu.: 0.596750
## Max. :2005 Max. : 5.733000 Max. : 5.733000 Max. : 5.733000
## Lag4 Lag5 Volume Today
## Min. :-4.922000 Min. :-4.92200 Min. :0.3561 Min. :-4.922000
## 1st Qu.:-0.640000 1st Qu.:-0.64000 1st Qu.:1.2574 1st Qu.:-0.639500
## Median : 0.038500 Median : 0.03850 Median :1.4229 Median : 0.038500
## Mean : 0.001636 Mean : 0.00561 Mean :1.4783 Mean : 0.003138
## 3rd Qu.: 0.596750 3rd Qu.: 0.59700 3rd Qu.:1.6417 3rd Qu.: 0.596750
## Max. : 5.733000 Max. : 5.73300 Max. :3.1525 Max. : 5.733000
## Direction
## Down:602
## Up :648
##
##
##
## pairs(Smarket)
The cor() function produces a matrix that contains all of the pairwise correlations among the predictors in a data set.
cor(Smarket[, -9])## Year Lag1 Lag2 Lag3 Lag4
## Year 1.00000000 0.029699649 0.030596422 0.033194581 0.035688718
## Lag1 0.02969965 1.000000000 -0.026294328 -0.010803402 -0.002985911
## Lag2 0.03059642 -0.026294328 1.000000000 -0.025896670 -0.010853533
## Lag3 0.03319458 -0.010803402 -0.025896670 1.000000000 -0.024051036
## Lag4 0.03568872 -0.002985911 -0.010853533 -0.024051036 1.000000000
## Lag5 0.02978799 -0.005674606 -0.003557949 -0.018808338 -0.027083641
## Volume 0.53900647 0.040909908 -0.043383215 -0.041823686 -0.048414246
## Today 0.03009523 -0.026155045 -0.010250033 -0.002447647 -0.006899527
## Lag5 Volume Today
## Year 0.029787995 0.53900647 0.030095229
## Lag1 -0.005674606 0.04090991 -0.026155045
## Lag2 -0.003557949 -0.04338321 -0.010250033
## Lag3 -0.018808338 -0.04182369 -0.002447647
## Lag4 -0.027083641 -0.04841425 -0.006899527
## Lag5 1.000000000 -0.02200231 -0.034860083
## Volume -0.022002315 1.00000000 0.014591823
## Today -0.034860083 0.01459182 1.000000000As one would expect, the correlations between the lag variables and today’s returns are close to zero. In other words, there appears to be little correlation between today’s returns and previous days’ returns. The only substantial correlation is between Year and Volume. By plotting the data, which is ordered chronologically, we see that Volume is increasing over time. In other words, the average number of shares traded daily increased from 2001 to 2005.
attach(Smarket)
plot(Volume)
- Logistic Regression
Next, we will fit a logistic regression model in order to predict Direction using Lag1 through Lag5 and Volume. The glm() function can be used to fit many types of generalized linear models, including logistic regression. The syntax of the glm() function is similar to that of lm(), except that we must pass in the argument family = binomial in order to tell R to run a logistic regression rather than some other type of generalized linear model.
glm.fits <- glm(
Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = Smarket, family = binomial
)
summary(glm.fits)##
## 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: 3The smallest p-value here is associated with Lag1. The negative coefficient for this predictor suggests that if the market had a positive return yesterday, then it is less likely to go up today. However, at a value of 0.15, the p-value is still relatively large, and so there is no clear evidence of a real association between Lag1 and Direction. We use the coef() function in order to access just the coefficients for this fitted model. We can also use the summary() function to access particular aspects of the fitted model, such as the p-values for the coefficients.
coef(glm.fits)## (Intercept) Lag1 Lag2 Lag3 Lag4 Lag5
## -0.126000257 -0.073073746 -0.042301344 0.011085108 0.009358938 0.010313068
## Volume
## 0.135440659summary(glm.fits)$coef## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.126000257 0.24073574 -0.5233966 0.6006983
## Lag1 -0.073073746 0.05016739 -1.4565986 0.1452272
## Lag2 -0.042301344 0.05008605 -0.8445733 0.3983491
## Lag3 0.011085108 0.04993854 0.2219750 0.8243333
## Lag4 0.009358938 0.04997413 0.1872757 0.8514445
## Lag5 0.010313068 0.04951146 0.2082966 0.8349974
## Volume 0.135440659 0.15835970 0.8552723 0.3924004summary(glm.fits)$coef[,4]#Me retorna os valores da quarta coluna (Pr)## (Intercept) Lag1 Lag2 Lag3 Lag4 Lag5
## 0.6006983 0.1452272 0.3983491 0.8243333 0.8514445 0.8349974
## Volume
## 0.3924004The predict() function can be used to predict the probability that the market will go up, given values of the predictors. The type = “response” option tells R to output probabilities of the form P(Y = 1|X), as opposed to other information such as the logit. If no data set is supplied to the predict() function, then the probabilities are computed for the training data that was used to fit the logistic regression model. Here we have printed only the first ten probabilities. We know that these values correspond to the probability of the market going up, rather than down, because the contrasts() function indicates that R has created a dummy variable with a 1 for Up.
glm.probs <- predict(glm.fits, type = "response")
glm.probs[1:10]## 1 2 3 4 5 6 7 8
## 0.5070841 0.4814679 0.4811388 0.5152224 0.5107812 0.5069565 0.4926509 0.5092292
## 9 10
## 0.5176135 0.4888378contrasts(Direction)## Up
## Down 0
## Up 1In order to make a prediction as to whether the market will go up or down on a particular day, we must convert these predicted probabilities into class labels, Up or Down. The following two commands create a vector of class predictions based on whether the predicted probability of a market increase is greater than or less than 0.5.
glm.pred <- rep("Down", 1250)
glm.pred[glm.probs > .5] = "Up"The first command creates a vector of 1,250 Down elements. The second line transforms to Up all of the elements for which the predicted probability of a market increase exceeds 0.5. Given these predictions, the table() function can be used to produce a confusion matrix in order to determine how many observations were correctly or incorrectly classified.
table(glm.pred, Direction)## Direction
## glm.pred Down Up
## Down 145 141
## Up 457 507(507 + 145) / 1250## [1] 0.5216mean(glm.pred == Direction)## [1] 0.5216The diagonal elements of the confusion matrix indicate correct predictions, while the off-diagonals represent incorrect predictions. Hence our model correctly predicted that the market would go up on 507 days and that it would go down on 145 days, for a total of 507 + 145 = 652 correct predictions. The mean() function can be used to compute the fraction of days for which the prediction was correct. In this case, logistic regression correctly predicted the movement of the market 52.2% of the time.
We will first create a vector corresponding to the observations from 2001 through 2004. We will then use this vector to create a held out data set of observations from 2005.
train <- (Year < 2005)
Smarket.2005 <- Smarket[!train,] #The ! symbol can be used to reverse all of the elements of a Boolean vector.
dim(Smarket.2005)## [1] 252 9Direction.2005 <- Direction[!train]We now fit a logistic regression model using only the subset of the observations that correspond to dates before 2005, using the subset argument. We then obtain predicted probabilities of the stock market going up for each of the days in our test set—that is, for the days in 2005.
glm.fits <- glm(
Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = Smarket, family = binomial, subset = train
)
glm.probs <- predict(glm.fits, Smarket.2005, type = "response")
glm.pred <- rep("Down", 252)
glm.pred[glm.pred > .5] <- "Up"
table(glm.pred, Direction.2005)## Direction.2005
## glm.pred Down Up
## Up 111 141mean(glm.pred == Direction.2005)## [1] 0.5595238mean(glm.pred != Direction.2005) #The != notation means not equal to, and so the last command computes the test set error rate.## [1] 0.4404762Below we have refit the logistic regression using just Lag1 and Lag2, which seemed to have the highest predictive power in the original logistic regression model.
glm.fits <- glm(Direction ~ Lag1 + Lag2, data = Smarket, family = binomial, subset = train)
glm.probs <- predict(glm.fits, Smarket.2005, type = "response")
glm.pred <- rep("Down", 252)
glm.pred[glm.probs > .5] <- "Up"
table(glm.pred, Direction.2005)## Direction.2005
## glm.pred Down Up
## Down 35 35
## Up 76 106mean(glm.pred == Direction.2005)## [1] 0.5595238106 / (106 + 76)## [1] 0.5824176Suppose that we want to predict the returns associated with particular values of Lag1 and Lag2. In particular, we want to predict Direction on a day when Lag1 and Lag2 equal 1.2 and 1.1, respectively, and on a day when they equal 1.5 and −0.8. We do this using the predict() function.
predict(glm.fits,
newdata =
data.frame(Lag1 = c(1.2, 1.5), Lag2 = c(1.1, -0.8)), type = "response"
)## 1 2
## 0.4791462 0.4960939Linear Discriminant Analysis
Now we will perform LDA on the Smarket data. In R, we fit an LDA model using the lda() function, which is part of the MASS library. Notice that the syntax for the lda() function is identical to that of lm(), and to that of glm() except for the absence of the family option. We fit the model using only the observations before 2005.
library(MASS)
lda.fit <- lda(Direction ~ Lag1 + Lag2, data = Smarket, subset = train)
lda.fit## Call:
## lda(Direction ~ Lag1 + Lag2, data = Smarket, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.491984 0.508016
##
## Group means:
## Lag1 Lag2
## Down 0.04279022 0.03389409
## Up -0.03954635 -0.03132544
##
## Coefficients of linear discriminants:
## LD1
## Lag1 -0.6420190
## Lag2 -0.5135293plot(lda.fit)
The predict() function returns a list with three elements. The first element, class, contains LDA’s predictions about the movement of the market. The second element, posterior, is a matrix whose kth column contains the posterior probability that the corresponding observation belongs to the kth class, computed from (4.15). Finally, x contains the linear discriminants, described earlier.
lda.pred <- predict(lda.fit, Smarket.2005)
names(lda.pred)## [1] "class" "posterior" "x"The LDA and logistic regression predictions are almost identical.
lda.class <- lda.pred$class
table(lda.class, Direction.2005)## Direction.2005
## lda.class Down Up
## Down 35 35
## Up 76 106mean(lda.class == Direction.2005)## [1] 0.5595238Applying a 50% threshold to the posterior probabilities allows us to recreate the predictions contained in lda.pred$class.
sum(lda.pred$posterior[,1] >= .5)## [1] 70sum(lda.pred$posterior[,1] < .5)## [1] 182Notice that the posterior probability output by the model corresponds to the probability that the market will decrease:
lda.pred$posterior[1:20, 1]## 999 1000 1001 1002 1003 1004 1005 1006
## 0.4901792 0.4792185 0.4668185 0.4740011 0.4927877 0.4938562 0.4951016 0.4872861
## 1007 1008 1009 1010 1011 1012 1013 1014
## 0.4907013 0.4844026 0.4906963 0.5119988 0.4895152 0.4706761 0.4744593 0.4799583
## 1015 1016 1017 1018
## 0.4935775 0.5030894 0.4978806 0.4886331lda.class[1:20]## [1] Up Up Up Up Up Up Up Up Up Up Up Down Up Up Up
## [16] Up Up Down Up Up
## Levels: Down Upsum(lda.pred$posterior[, 1] > .9) #For instance, suppose that we wish to predict a market decrease only if we are very certain that the market will indeed decrease on that day—say, if the posterior probability is at least 90 %.## [1] 0#No days in 2005 meet that threshold! In fact, the greatest posterior probability of decrease in all of 2005 was 52.02 %.Quadratic Discriminant Anlysis
We will now fit a QDA model to the Smarket data. QDA is implemented in R using the qda() function, which is also part of the MASS library. The syntax is identical to that of lda().
qda.fit <- qda(Direction ~ Lag1 + Lag2, data = Smarket, subset = train)
qda.fit## Call:
## qda(Direction ~ Lag1 + Lag2, data = Smarket, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.491984 0.508016
##
## Group means:
## Lag1 Lag2
## Down 0.04279022 0.03389409
## Up -0.03954635 -0.03132544qda.class <- predict(qda.fit, Smarket.2005)$class
table(qda.class, Direction.2005)## Direction.2005
## qda.class Down Up
## Down 30 20
## Up 81 121mean(qda.class == Direction.2005)## [1] 0.5992063Interestingly, the QDA predictions are accurate almost 60% of the time, even though the 2005 data was not used to fit the model.
Naive Bayes
4.7.5 Naive Bayes Next, we fit a naive Bayes model to the Smarket data. Naive Bayes is implemented in R using the naiveBayes() function, which is part of the e1071 library. The syntax is identical to that of lda() and qda(). By default, this implementation of the naive Bayes classifier models each quantitative feature using a Gaussian distribution. However, a kernel density method can also be used to estimate the distributions.
library(e1071)## Warning: package 'e1071' was built under R version 4.0.5nb.fit <- naiveBayes(Direction ~ Lag1 + Lag2, data = Smarket, subset = train)
nb.fit##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## Down Up
## 0.491984 0.508016
##
## Conditional probabilities:
## Lag1
## Y [,1] [,2]
## Down 0.04279022 1.227446
## Up -0.03954635 1.231668
##
## Lag2
## Y [,1] [,2]
## Down 0.03389409 1.239191
## Up -0.03132544 1.220765mean(Lag1[train][Direction[train]=="Down"])## [1] 0.04279022sd(Lag1[train][Direction[train]=="Down"])## [1] 1.227446The predict() function is straightforward.
nb.class <- predict(nb.fit, Smarket.2005)
table(nb.class, Direction.2005)## Direction.2005
## nb.class Down Up
## Down 28 20
## Up 83 121mean(nb.class == Direction.2005)## [1] 0.5912698Naive Bayes performs very well on this data, with accurate predictions over 59% of the time. This is slightly worse than QDA, but much better than LDA. The predict() function can also generate estimates of the probability that each observation belongs to a particular class.
nb.preds <- predict(nb.fit, Smarket.2005, type = "raw")
nb.preds[1:5,]## Down Up
## [1,] 0.4873164 0.5126836
## [2,] 0.4762492 0.5237508
## [3,] 0.4653377 0.5346623
## [4,] 0.4748652 0.5251348
## [5,] 0.4901890 0.5098110K-Nearest Neighbors
We will now perform KNN using the knn() function, which is part of the class library. This function works rather differently from the other modelfitting functions that we have encountered thus far. Rather than a two-step approach in which we first fit the model and then we use the model to make predictions, knn() forms predictions using a single command. The function requires four inputs.
- A matrix containing the predictors associated with the training data, labeled train.X below.
- A matrix containing the predictors associated with the data for which we wish to make predictions, labeled test.X below.
- A vector containing the class labels for the training observations, labeled train.Direction below.
- A value for K, the number of nearest neighbors to be used by the classifier.
We use the cbind() function, short for column bind, to bind the Lag1 and Lag2 variables together into two matrices, one for the training set and the other for the test set.
library(class)
train.x <- cbind(Lag1, Lag2)[train,]
test.x <- cbind(Lag1, Lag2)[!train,]
train.Direction <- Direction[train]set.seed(1)
knn.pred <- knn(train.x, test.x, train.Direction, k=1)
table(knn.pred, Direction.2005)## Direction.2005
## knn.pred Down Up
## Down 43 58
## Up 68 83(83+43) / 252## [1] 0.5