R-exercises – Evaluate your model with R Solutions

Below are the solutions to these exercises on error metrics.

####################
#                  #
#    Exercise 1    #
#                  #
####################
library(datasets)
data<-attitude
head(data)

##   rating complaints privileges learning raises critical advance
## 1     43         51         30       39     61       92      45
## 2     63         64         51       54     63       73      47
## 3     71         70         68       69     76       86      48
## 4     61         63         45       47     54       84      35
## 5     81         78         56       66     71       83      47
## 6     43         55         49       44     54       49      34

model.1<-lm(rating~.,data=data)
summary(model.1)

##
## Call:
## lm(formula = rating ~ ., data = data)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -10.9418  -4.3555   0.3158   5.5425  11.5990
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept) 10.78708   11.58926   0.931 0.361634
## complaints   0.61319    0.16098   3.809 0.000903 ***
## privileges  -0.07305    0.13572  -0.538 0.595594
## learning     0.32033    0.16852   1.901 0.069925 .
## raises       0.08173    0.22148   0.369 0.715480
## critical     0.03838    0.14700   0.261 0.796334
## advance     -0.21706    0.17821  -1.218 0.235577
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 7.068 on 23 degrees of freedom
## Multiple R-squared:  0.7326,	Adjusted R-squared:  0.6628
## F-statistic:  10.5 on 6 and 23 DF,  p-value: 1.24e-05

model.2<-lm(rating~complaints+privileges+learning+advance,data=data)
summary(model.2)

##
## Call:
## lm(formula = rating ~ complaints + privileges + learning + advance,
##     data = data)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -11.8976  -5.5171   0.7654   5.8086  11.5022
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept) 14.30347    7.73957   1.848   0.0765 .
## complaints   0.65338    0.13051   5.006 3.67e-05 ***
## privileges  -0.07682    0.13059  -0.588   0.5616
## learning     0.32395    0.15741   2.058   0.0502 .
## advance     -0.17151    0.14904  -1.151   0.2607
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6.821 on 25 degrees of freedom
## Multiple R-squared:  0.7293,	Adjusted R-squared:  0.686
## F-statistic: 16.84 on 4 and 25 DF,  p-value: 8.134e-07

model.3<-lm(rating~complaints+learning+advance,data=data)
summary(model.3)

##
## Call:
## lm(formula = rating ~ complaints + learning + advance, data = data)
##
## Residuals:
##     Min      1Q  Median      3Q     Max
## -12.217  -5.377   0.967   6.078  11.540
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept)  13.5777     7.5439   1.800   0.0835 .
## complaints    0.6227     0.1181   5.271 1.65e-05 ***
## learning      0.3124     0.1542   2.026   0.0532 .
## advance      -0.1870     0.1449  -1.291   0.2082
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6.734 on 26 degrees of freedom
## Multiple R-squared:  0.7256,	Adjusted R-squared:  0.6939
## F-statistic: 22.92 on 3 and 26 DF,  p-value: 1.807e-07

####################
#                  #
#    Exercise 2    #
#                  #
####################
rmse<-function(y,y_pred){
  RMSE <- sqrt(mean((y-y_pred)^2))

  return (RMSE)
}
rmse(data$rating, model.1$fitted.values)

## [1] 6.1887

rmse(data$rating, model.2$fitted.values)

## [1] 6.226319

rmse(data$rating, model.3$fitted.values)

## [1] 6.269261

####################
#                  #
#    Exercise 3    #
#                  #
####################
mae<-function(y,y_pred){

  MAE <- sum(abs(y-y_pred))/length(y)

  return (MAE)
}
mae(data$rating, model.1$fitted.values)

## [1] 5.178977

mae(data$rating, model.2$fitted.values)

## [1] 5.273396

mae(data$rating, model.3$fitted.values)

## [1] 5.316718

####################
#                  #
#    Exercise 4    #
#                  #
####################
rmlse <- function(y,y_pred) {

  RMLSE<-sqrt(1/length(y)*sum((log(y_pred +1)-log(y +1))^2))

  return(RMLSE)
}
rmlse(data$rating, model.1$fitted.values)

## [1] 0.1049955

rmlse(data$rating, model.2$fitted.values)

## [1] 0.1056288

rmlse(data$rating, model.3$fitted.values)

## [1] 0.1059711

####################
#                  #
#    Exercise 5    #
#                  #
####################
cluster.data<-iris
set.seed(42)
k.means.results.1<-kmeans(cluster.data[,1:4],centers=3,iter.max = 50, algorithm = "Lloyd")
k.means.results.1$cluster<-factor(k.means.results.1$cluster)
levels(k.means.results.1$cluster)<-list("setosa"=3,"versicolor"=1,"virginica"=2)


####################
#                  #
#    Exercise 6    #
#                  #
####################
table(cluster.data[,5],k.means.results.1$cluster)

##
##              setosa versicolor virginica
##   setosa         50          0         0
##   versicolor      0         48         2
##   virginica       0         14        36

####################
#                  #
#    Exercise 7    #
#                  #
####################
accuracy <-function (real_labels, predict_labels) {

  result.table <- table(real_labels, predict_labels)

  return(sum(diag(result.table))/sum(result.table))
}

recall <- function (real_labels, predict_labels) {

  result.table <- table(real_labels, predict_labels)
  rowsums = apply(result.table, 1, sum)


  return(diag(result.table)/rowsums)
}

precision <- function (real_labels, predict_labels) {

  result.table <- table(real_labels, predict_labels)
  colsums = apply(result.table, 2, sum)

  return(diag(result.table)/colsums)
}

accuracy(cluster.data[,5],k.means.results.1$cluster)

## [1] 0.8933333

recall(cluster.data[,5],k.means.results.1$cluster)

##     setosa versicolor  virginica
##       1.00       0.96       0.72

precision(cluster.data[,5],k.means.results.1$cluster)

##     setosa versicolor  virginica
##  1.0000000  0.7741935  0.9473684

####################
#                  #
#    Exercise 8    #
#                  #
####################
f.mesure<-function(real_labels, predict_labels,b) {

  f<-((b^2+1)*precision(real_labels, predict_labels)*recall(real_labels, predict_labels))/(b^2*precision(real_labels, predict_labels)+recall(real_labels, predict_labels))
  return(f)
}
f.mesure(cluster.data[,5],k.means.results.1$cluster,2)

##     setosa versicolor  virginica
##  1.0000000  0.9160305  0.7563025

f.mesure(cluster.data[,5],k.means.results.1$cluster,0.5)

##     setosa versicolor  virginica
##  1.0000000  0.8053691  0.8910891

####################
#                  #
#    Exercise 9    #
#                  #
####################
purity <- function (real_labels, predict_labels) {

  result.table = table(real_labels, predict_labels)
  return(sum(max(result.table[1,]),max(result.table[2,]),max(result.table[3,]) )/length(real_labels))
}
purity(cluster.data[,5],k.means.results.1$cluster)

## [1] 0.8933333

####################
#                  #
#    Exercise 10   #
#                  #
####################
library(clValid)
set.seed(42)
k.means.results.1<-kmeans(cluster.data[,1:4],centers=3,iter.max = 50, algorithm = "Lloyd")
dunn(distance =dist(iris,method="euclidean"),clusters=k.means.results.1$cluster)

## [1] 0.09880739

Please help us to improve R-exercises:

Additive Manufacturing 13 June 2022 at 13:16 on Bioinformatics Tutorial with Exercises in R (part 1)Great informative post about biostrings exercises. Thanks for sharing with us.
Fabio 6 June 2022 at 16:44 on List exercises: solutionsThanks a lot for your exercises.
AE 6 May 2022 at 14:05 on Start Here To Learn R – vol. 1: Vectors, arithmetic, and regular sequencesI found the book very useful. I think it should be more easier at the start because there are many...
João Gonçalves 22 April 2022 at 20:20 on Advanced Techniques With Raster Data – Part 3: ExercisesHello, As stated above in the text: "It also shows that (as expected, given the country geometric shape) both ‘Longitude’...
Mark 22 April 2022 at 16:00 on Advanced Techniques With Raster Data – Part 3: ExercisesWhy don't you include longitude in your GAM model?

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