Sec_4.7_ISLR

Práctica

This question should be answered using the Weekly data set, which is part of the ISLR package. This data is similar in nature to the Smarket data from this chapter’s lab, except that it contains 1,089 weekly returns for 21 years, from the beginning of 1990 to the end of 2010.

Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?

summary(Weekly)

##       Year           Lag1               Lag2               Lag3         
##  Min.   :1990   Min.   :-18.1950   Min.   :-18.1950   Min.   :-18.1950  
##  1st Qu.:1995   1st Qu.: -1.1540   1st Qu.: -1.1540   1st Qu.: -1.1580  
##  Median :2000   Median :  0.2410   Median :  0.2410   Median :  0.2410  
##  Mean   :2000   Mean   :  0.1506   Mean   :  0.1511   Mean   :  0.1472  
##  3rd Qu.:2005   3rd Qu.:  1.4050   3rd Qu.:  1.4090   3rd Qu.:  1.4090  
##  Max.   :2010   Max.   : 12.0260   Max.   : 12.0260   Max.   : 12.0260  
##       Lag4               Lag5              Volume       
##  Min.   :-18.1950   Min.   :-18.1950   Min.   :0.08747  
##  1st Qu.: -1.1580   1st Qu.: -1.1660   1st Qu.:0.33202  
##  Median :  0.2380   Median :  0.2340   Median :1.00268  
##  Mean   :  0.1458   Mean   :  0.1399   Mean   :1.57462  
##  3rd Qu.:  1.4090   3rd Qu.:  1.4050   3rd Qu.:2.05373  
##  Max.   : 12.0260   Max.   : 12.0260   Max.   :9.32821  
##      Today          Direction 
##  Min.   :-18.1950   Down:484  
##  1st Qu.: -1.1540   Up  :605  
##  Median :  0.2410             
##  Mean   :  0.1499             
##  3rd Qu.:  1.4050             
##  Max.   : 12.0260

Matriz de correlación para el data set

cor(Weekly[,-9])

##               Year         Lag1        Lag2        Lag3         Lag4
## Year    1.00000000 -0.032289274 -0.03339001 -0.03000649 -0.031127923
## Lag1   -0.03228927  1.000000000 -0.07485305  0.05863568 -0.071273876
## Lag2   -0.03339001 -0.074853051  1.00000000 -0.07572091  0.058381535
## Lag3   -0.03000649  0.058635682 -0.07572091  1.00000000 -0.075395865
## Lag4   -0.03112792 -0.071273876  0.05838153 -0.07539587  1.000000000
## Lag5   -0.03051910 -0.008183096 -0.07249948  0.06065717 -0.075675027
## Volume  0.84194162 -0.064951313 -0.08551314 -0.06928771 -0.061074617
## Today  -0.03245989 -0.075031842  0.05916672 -0.07124364 -0.007825873
##                Lag5      Volume        Today
## Year   -0.030519101  0.84194162 -0.032459894
## Lag1   -0.008183096 -0.06495131 -0.075031842
## Lag2   -0.072499482 -0.08551314  0.059166717
## Lag3    0.060657175 -0.06928771 -0.071243639
## Lag4   -0.075675027 -0.06107462 -0.007825873
## Lag5    1.000000000 -0.05851741  0.011012698
## Volume -0.058517414  1.00000000 -0.033077783
## Today   0.011012698 -0.03307778  1.000000000

En forma gráfica es más fácil observar que hay una fuerte relación entre Year y Volume

M <-cor(Weekly[,-9])
corrplot(M, type="upper", order="hclust",
         col=brewer.pal(n=8, name="RdYlBu"))

Use the full data set to perform a logistic regression with Direction as the response and the five lag variables plus Volume as predictors. Use the summary function to print the results. Do any of the predictors appear to be statistically significant? If so, which ones?

modelo <-
  glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
      data = Weekly,
      family = "binomial")
summary(modelo)

## 
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + 
##     Volume, family = "binomial", data = Weekly)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.6949  -1.2565   0.9913   1.0849   1.4579  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)   
## (Intercept)  0.26686    0.08593   3.106   0.0019 **
## Lag1        -0.04127    0.02641  -1.563   0.1181   
## Lag2         0.05844    0.02686   2.175   0.0296 * 
## Lag3        -0.01606    0.02666  -0.602   0.5469   
## Lag4        -0.02779    0.02646  -1.050   0.2937   
## Lag5        -0.01447    0.02638  -0.549   0.5833   
## Volume      -0.02274    0.03690  -0.616   0.5377   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1496.2  on 1088  degrees of freedom
## Residual deviance: 1486.4  on 1082  degrees of freedom
## AIC: 1500.4
## 
## Number of Fisher Scoring iterations: 4

Compute the confusion matrix and overall fraction of correct predictions. Explain what the confusion matrix is telling you about the types of mistakes made by logistic regression.

predic <- predict (modelo, Weekly , type = "response")

glm.pred= rep("Down" ,length(predic))
glm.pred[predic > .5] = "Up"

table(glm.pred,Weekly$Direction)

##         
## glm.pred Down  Up
##     Down   54  48
##     Up    430 557

mean(glm.pred == Weekly$Direction)

## [1] 0.5610652

con estos resultados vemos que acertamos a 54 + 557 = 611 y tenemos una efectividad del 52%

Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).

train <- (Weekly$Year < 2009)
Weekly.train <- Weekly[train, ]
Weekly.test <- Weekly[!train, ]
fit.glm2 <- glm(Direction ~ Lag2, data = Weekly.train, family = binomial, subset = train)
summary(fit.glm2)

## 
## Call:
## glm(formula = Direction ~ Lag2, family = binomial, data = Weekly.train, 
##     subset = train)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.536  -1.264   1.021   1.091   1.368  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)   
## (Intercept)  0.20326    0.06428   3.162  0.00157 **
## Lag2         0.05810    0.02870   2.024  0.04298 * 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1354.7  on 984  degrees of freedom
## Residual deviance: 1350.5  on 983  degrees of freedom
## AIC: 1354.5
## 
## Number of Fisher Scoring iterations: 4

predic <- predict(fit.glm2, Weekly.test , type = "response")

glm.pred = rep("Down" ,length(predic))
glm.pred[predic > .5] = "Up"

table(glm.pred,Weekly.test$Direction)

##         
## glm.pred Down Up
##     Down    9  5
##     Up     34 56

mean(glm.pred == Weekly.test$Direction)

## [1] 0.625

using LDA

fit.glm2 <- lda(Direction ~ Lag2, data = Weekly.train, subset = train)
summary(fit.glm2)

##         Length Class  Mode     
## prior   2      -none- numeric  
## counts  2      -none- numeric  
## means   2      -none- numeric  
## scaling 1      -none- numeric  
## lev     2      -none- character
## svd     1      -none- numeric  
## N       1      -none- numeric  
## call    4      -none- call     
## terms   3      terms  call     
## xlevels 0      -none- list

predic <- predict(fit.glm2, Weekly.test)

table(predic$class,Weekly.test$Direction)

##       
##        Down Up
##   Down    9  5
##   Up     34 56

mean(predic$class == Weekly.test$Direction)

## [1] 0.625

using QDA

fit.glm2 <- qda(Direction ~ Lag2, data = Weekly.train, subset = train)
summary(fit.glm2)

##         Length Class  Mode     
## prior   2      -none- numeric  
## counts  2      -none- numeric  
## means   2      -none- numeric  
## scaling 2      -none- numeric  
## ldet    2      -none- numeric  
## lev     2      -none- character
## N       1      -none- numeric  
## call    4      -none- call     
## terms   3      terms  call     
## xlevels 0      -none- list

predic <- predict(fit.glm2, Weekly.test)

table(predic$class,Weekly.test$Direction)

##       
##        Down Up
##   Down    0  0
##   Up     43 61

mean(predic$class == Weekly.test$Direction)

## [1] 0.5865385

using KNN with k=1

library (class)
train.X=cbind(Weekly.train$Lag1 ,Weekly.train$Lag2)
test.X=cbind (Weekly.test$Lag1 ,Weekly.test$Lag2)
train.Direction =Weekly.train$Direction

set.seed (1)
knn.pred = knn(train.X, test.X, train.Direction , k = 1)
table(knn.pred , Weekly.test$Direction)

##         
## knn.pred Down Up
##     Down   18 29
##     Up     25 32

mean(knn.pred == Weekly.test$Direction)

## [1] 0.4807692

la regresion lineal y lda son los de mejor desempeño para este set de datos.

In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.

Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.

attach(Auto)
mpg01 <- rep(0, length(mpg))
mpg01[mpg > median(mpg)] <- 1
Auto <- data.frame(Auto, mpg01)

Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.

cor(Auto[, -9])

##                     mpg  cylinders displacement horsepower     weight
## mpg           1.0000000 -0.7776175   -0.8051269 -0.7784268 -0.8322442
## cylinders    -0.7776175  1.0000000    0.9508233  0.8429834  0.8975273
## displacement -0.8051269  0.9508233    1.0000000  0.8972570  0.9329944
## horsepower   -0.7784268  0.8429834    0.8972570  1.0000000  0.8645377
## weight       -0.8322442  0.8975273    0.9329944  0.8645377  1.0000000
## acceleration  0.4233285 -0.5046834   -0.5438005 -0.6891955 -0.4168392
## year          0.5805410 -0.3456474   -0.3698552 -0.4163615 -0.3091199
## origin        0.5652088 -0.5689316   -0.6145351 -0.4551715 -0.5850054
## mpg01         0.8369392 -0.7591939   -0.7534766 -0.6670526 -0.7577566
##              acceleration       year     origin      mpg01
## mpg             0.4233285  0.5805410  0.5652088  0.8369392
## cylinders      -0.5046834 -0.3456474 -0.5689316 -0.7591939
## displacement   -0.5438005 -0.3698552 -0.6145351 -0.7534766
## horsepower     -0.6891955 -0.4163615 -0.4551715 -0.6670526
## weight         -0.4168392 -0.3091199 -0.5850054 -0.7577566
## acceleration    1.0000000  0.2903161  0.2127458  0.3468215
## year            0.2903161  1.0000000  0.1815277  0.4299042
## origin          0.2127458  0.1815277  1.0000000  0.5136984
## mpg01           0.3468215  0.4299042  0.5136984  1.0000000

Split the data into a training set and a test set.

train <- (year %% 2 == 0)
Auto.train <- Auto[train, ]
Auto.test <- Auto[!train, ]
mpg01.test <- mpg01[!train]

Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?

fit.lda <- lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, subset = train)
fit.lda

## Call:
## lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, 
##     subset = train)
## 
## Prior probabilities of groups:
##         0         1 
## 0.4571429 0.5428571 
## 
## Group means:
##   cylinders   weight displacement horsepower
## 0  6.812500 3604.823     271.7396  133.14583
## 1  4.070175 2314.763     111.6623   77.92105
## 
## Coefficients of linear discriminants:
##                        LD1
## cylinders    -0.6741402638
## weight       -0.0011465750
## displacement  0.0004481325
## horsepower    0.0059035377

pred.lda <- predict(fit.lda, Auto.test)
table(pred.lda$class, mpg01.test)

##    mpg01.test
##      0  1
##   0 86  9
##   1 14 73

mean(pred.lda$class != mpg01.test)

## [1] 0.1263736

Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?

fit.qda <- qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, subset = train)
fit.qda

## Call:
## qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, 
##     subset = train)
## 
## Prior probabilities of groups:
##         0         1 
## 0.4571429 0.5428571 
## 
## Group means:
##   cylinders   weight displacement horsepower
## 0  6.812500 3604.823     271.7396  133.14583
## 1  4.070175 2314.763     111.6623   77.92105

pred.qda <- predict(fit.qda, Auto.test)
table(pred.qda$class, mpg01.test)

##    mpg01.test
##      0  1
##   0 89 13
##   1 11 69

mean(pred.qda$class != mpg01.test)

## [1] 0.1318681

Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?

fit.glm <-
  glm(
    mpg01 ~ cylinders + weight + displacement + horsepower,
    data = Auto,
    family = binomial,
    subset = train
  )
summary(fit.glm)

## 
## Call:
## glm(formula = mpg01 ~ cylinders + weight + displacement + horsepower, 
##     family = binomial, data = Auto, subset = train)
## 
## Deviance Residuals: 
##      Min        1Q    Median        3Q       Max  
## -2.48027  -0.03413   0.10583   0.29634   2.57584  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  17.658730   3.409012   5.180 2.22e-07 ***
## cylinders    -1.028032   0.653607  -1.573   0.1158    
## weight       -0.002922   0.001137  -2.569   0.0102 *  
## displacement  0.002462   0.015030   0.164   0.8699    
## horsepower   -0.050611   0.025209  -2.008   0.0447 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 289.58  on 209  degrees of freedom
## Residual deviance:  83.24  on 205  degrees of freedom
## AIC: 93.24
## 
## Number of Fisher Scoring iterations: 7

probs <- predict(fit.glm, Auto.test, type = "response")
pred.glm <- rep(0, length(probs))
pred.glm[probs > 0.5] <- 1
table(pred.glm, mpg01.test)

##         mpg01.test
## pred.glm  0  1
##        0 89 11
##        1 11 71

mean(pred.glm != mpg01.test)

## [1] 0.1208791

Sec_4.7_ISLR

Milton Godinez

27/5/2019

Práctica

Matriz de correlación para el data set