如何使用 Caret 包中的“gbm”方法生成混淆矩阵

Question

Issue:

I'm a beginner at building classification models, so I am sorry if this question might sound terminologically incorrect. I will try my best. I am having trouble interpreting the error messages that I am receiving when creating a confusion matrix using the e1071 package .

I have tried many solutions to fix the errors but I really can't comprehend how to move further to successfully produce a confusion matrix using the gbm method (see below) . I have tried my best to try and fix the error and I feel confused.

Error: `data` and `reference` should be factors with the same levels.

This exercise is part of a university assignment and I would be really grateful if anybody can help me solve this issue and explain what these error messages mean as a learning exercise.

My data has nine continuous independent variables , and one dependent variable called 'Country'.

Another post suggested that:
the error means that you need to give it factors as inputs (train[[predict]] > c is not a factor). Try using factor(ifelse(...), levels) instead).

I'm developing a gbm model using Caret package .

#install packages
library(gbm)
library(caret)
library(e1701)

set.seed(45L)

#Produce a new version of the data frame 'Clusters_Dummy' with the rows shuffled
NewClusters=Cluster_Dummy_2[sample(1:nrow(Cluster_Dummy_2)),]

#Produce a dataframe
NewCluster<-as.data.frame(NewClusters)

#Split the training and testing data 70:30
training.parameters <- Cluster_Dummy_2$Country %>% 
createDataPartition(p = 0.7, list = FALSE)
train.data <- NewClusters[training.parameters, ]
test.data <- NewClusters[-training.parameters, ]

dim(train.data)
#259  10

dim(test.data)
#108  10

#Auxiliary function for controlling model fitting
#10 fold cross validation; 10 times
fitControl <- trainControl(## 10-fold CV
                          method = "repeatedcv",
                          number = 10,
                          ## repeated ten times
                          repeats = 10,
                          classProbs = TRUE)
#Fit the model
gbmFit1 <- train(Country ~ ., data=train.data, 
                 method = "gbm", 
                 trControl = fitControl,
                 ## This last option is actually one
                 ## for gbm() that passes through
                 verbose = FALSE)
gbmFit1
summary(gbmFit1)

#Predict the model with the test data
pred_model_Tree1 = predict(gbmFit1, newdata = head(test.data$Country), type = "prob")
pred_model_Tree1

print(pred_model_Tree1)

Confusion Matrix

#Confusion Matrix
confusionMatrix(pred_model_Tree1, test.data$Country)

#Error
Error: `data` and `reference` should be factors with the same levels.

What type of objects are pred_model_Tree1 & test.data$Country

typeof(pred_model_Tree1)
#list

typeof(test.data$Country)
#"integer"

#Convert both objects into factors
test.data$Country<-as.factor(test.data$Country)

#check
str(test.data)

'data.frame':   108 obs. of  10 variables:
 $ Country    : Factor w/ 3 levels "France","Holland",..: 2 1 1 2 1 2 1 1 2 2 ...

#str(pred_model_Tree1)
#data.frame':   6 obs. of  3 variables:
 #$ France     : num  0.00311 0.98187 0.98882 0.00935 0.99632 ...
 #$ Holland    : num  9.24e-01 1.41e-03 1.58e-03 4.45e-01 1.86e-05 
 #$ Spain: num  0.073 0.01672 0.0096 0.54539 0.00366 ...

  #Differences: 
    pred_model_Tree1 (three columns; 6 obs; 3 variables); 
    test.data (11 columns; 6 obs, dependent variable - 3 levels)
    Question: How to transform both objects to follow the same structure and the same levels
    
#Check the number of rows of the test.data
nrow(test.data)
#108

#Check the number of rows of the predicted output
nrow(pred_model_Tree1)
#6

#What are the levels
levels(pred_model_Tree1)
#NULL

levels(test.data$Country)
#[1] "France"      "Holland"     "Spain"

table(test.data$Country)
#France     Holland Spain 
#35         36         37 

I found a really good Stackoverflow question here to try and solve the issue and I tried to find a solution

#If you can't get the confusion matrix to work, break it down'
#Error: data and reference data should be factors with the same levels

#confusionMatrix(predicted, actual)
table(pred_model_Tree1) #Predicted

#       France      Holland       Spain
#1 0.003110462 9.238903e-01 0.072999195
#2 0.981868172 1.408983e-03 0.016722845
#3 0.988820237 1.575354e-03 0.009604409
#4 0.009346725 4.452638e-01 0.545389520
#5 0.996322192 1.864682e-05 0.003659161
#6 0.012668621 9.803462e-01 0.006985212

table(test.data$Country) #Actual

#France     Holland Spain 
#38         46         24 

#Great, they both have the same column headings

#Do the predicted and actual data match (are they factors)

confusionMatrix(as.factor(pred_model_Tree1), as.factor(test.data$Country))

#Error in confusionMatrix.default(as.factor(pred_model_Tree1), as.factor(test.data$Country)) : 
#The data must contain some levels that overlap the reference.
#In addition: Warning message:
#  In xtfrm.data.frame(x) : cannot xtfrm data frames

#format() treats the elements of a vector as character strings using a common format. 
pred<-format(round(predict(pred_model_Tree1, test.data)))

#Error 
Error in UseMethod("predict") : 
  no applicable method for 'predict' applied to an object of class "data.frame"

#One answer contained a custom made function
#They suggest that at least one number in the test.data that is never predicted. This is what is meant why "different number of levels". 

table(factor(pred_model_Tree1, levels=min(test.data):max(test.data)), 
      factor(test.data$Country, levels=min(test.data):max(test.data)))

#Error
Error in FUN(X[[i]], ...) : 
  only defined on a data frame with all numeric-alike variables

#Lastly, I found a function on StackOverflow that can be used to fix the unequal levels problem

# Create a confusion matrix from the given outcomes, whose rows correspond
# to the actual and the columns to the predicated classes.
createConfusionMatrix <- function(act, pred) {
  # You've mentioned that neither actual nor predicted may give a complete
  # picture of the available classes, hence:
  numClasses <- max(act, pred)
  # Sort predicted and actual as it simplifies what's next. You can make this
  # faster by storing `order(act)` in a temporary variable.
  pred <- pred[order(act)]
  act  <- act[order(act)]
  sapply(split(pred, act), tabulate, nbins=numClasses)
}

act<-pred_model_Tree1
pred<-test.data$Country

print(createConfusionMatrix(act, pred))

#Error
Error in FUN(X[[i]], ...) : 
  only defined on a data frame with all numeric-alike variables

Data

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"Spain", "Spain", "Spain", "Spain", "Spain", "Spain", "Spain", 
"Spain", "Spain", "Spain", "Spain", "Holland", "Holland", "Holland", 
"Holland", "Holland", "Holland", "France", "France", "France", 
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"France", "Spain", "Spain", "Spain", "Spain", "Spain", "Spain", 
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"Holland", "Holland", "Holland", "Holland", "Holland", "Holland", 
"Spain", "Spain", "Spain", "Spain", "Spain", "Spain", "Spain", 
"Holland", "Holland", "Holland", "Holland", "France", "France", 
"France", "France", "France", "France", "France", "Spain", "Spain", 
"Spain", "Spain", "Spain", "Spain", "Spain", "Spain", "Spain", 
"Spain", "Spain", "Spain", "Spain", "Spain", "Spain", "Spain", 
"Spain", "Spain", "France", "France", "France")), row.names = c(NA, 
99L), class = "data.frame")

Answer 1

Thanks for including all the required information; I believe this is the solution to your problem:

library(magrittr)
library(gbm)
#> Loaded gbm 2.1.8
library(caret)
#> Loading required package: ggplot2
#> Loading required package: lattice
library(e1071)

set.seed(45L)

# Load in your example data to an object ("data")
#Produce a new version of the data frame 'Clusters_Dummy' with the rows shuffled
Cluster_Dummy_2 <- data
NewClusters <- Cluster_Dummy_2[sample(1:nrow(Cluster_Dummy_2)),]

NewCluster<-as.data.frame(NewClusters)

training.parameters <- Cluster_Dummy_2$Country %>% 
  createDataPartition(p = 0.7, list = FALSE)
train.data <- NewClusters[training.parameters, ]
test.data <- NewClusters[-training.parameters, ]

dim(train.data)
#> [1] 70 11
#259  10

dim(test.data)
#> [1] 29 11
#108  10

#Auxiliary function for controlling model fitting
#10 fold cross validation; 10 times
fitControl <- trainControl(## 10-fold CV
  method = "repeatedcv",
  number = 10,
  ## repeated ten times
  repeats = 10,
  classProbs = TRUE)
#Fit the model
gbmFit1 <- train(Country ~ ., data=train.data, 
                 method = "gbm", 
                 trControl = fitControl,
                 ## This last option is actually one
                 ## for gbm() that passes through
                 verbose = FALSE)
gbmFit1
#> Stochastic Gradient Boosting 
#> 
#> 70 samples
#> 10 predictors
#>  2 classes: 'France', 'Holland' 
#> 
#> No pre-processing
#> Resampling: Cross-Validated (10 fold, repeated 10 times) 
#> Summary of sample sizes: 64, 64, 63, 63, 63, 62, ... 
#> Resampling results across tuning parameters:
#> 
#>   interaction.depth  n.trees  Accuracy   Kappa    
#>   1                   50      0.7397619  0.4810245
#>   1                  100      0.7916667  0.5816756
#>   1                  150      0.8204167  0.6392434
#>   2                   50      0.7396429  0.4813670
#>   2                  100      0.7943452  0.5901254
#>   2                  150      0.8380357  0.6768166
#>   3                   50      0.7361905  0.4711780
#>   3                  100      0.7966071  0.5897921
#>   3                  150      0.8356548  0.6694202
#> 
#> Tuning parameter 'shrinkage' was held constant at a value of 0.1
#> 
#> Tuning parameter 'n.minobsinnode' was held constant at a value of 10
#> Accuracy was used to select the optimal model using the largest value.
#> The final values used for the model were n.trees = 150, interaction.depth =
#>  2, shrinkage = 0.1 and n.minobsinnode = 10.
summary(gbmFit1)

#>                     var   rel.inf
#> ID                   ID 66.517974
#> Center_Freq Center_Freq  6.624256
#> Start.Freq   Start.Freq  5.545827
#> Delta.Time   Delta.Time  5.033223
#> Peak.Time     Peak.Time  4.951384
#> End.Freq       End.Freq  3.211461
#> Delta.Freq   Delta.Freq  2.352933
#> Low.Freq       Low.Freq  2.207371
#> High.Freq     High.Freq  1.951895
#> Peak.Freq     Peak.Freq  1.603675

#Predict the model with the test data
pred_model_Tree1 <- predict(object = gbmFit1, newdata = test.data, type = "prob")
pred_model_Tree1
#>         France     Holland
#> 1  0.919393487 0.080606513
#> 2  0.095638010 0.904361990
#> 3  0.019038102 0.980961898
#> 4  0.045807668 0.954192332
#> 5  0.157809127 0.842190873
#> 6  0.987391435 0.012608565
#> 7  0.011436393 0.988563607
#> 8  0.032262438 0.967737562
#> 9  0.151393564 0.848606436
#> 10 0.993447390 0.006552610
#> 11 0.020833439 0.979166561
#> 12 0.993910239 0.006089761
#> 13 0.009170816 0.990829184
#> 14 0.010519644 0.989480356
#> 15 0.995338954 0.004661046
#> 16 0.994153479 0.005846521
#> 17 0.998099611 0.001900389
#> 18 0.056571139 0.943428861
#> 19 0.801327096 0.198672904
#> 20 0.192220458 0.807779542
#> 21 0.899189477 0.100810523
#> 22 0.766542297 0.233457703
#> 23 0.940046468 0.059953532
#> 24 0.069087397 0.930912603
#> 25 0.916674076 0.083325924
#> 26 0.023676968 0.976323032
#> 27 0.996824979 0.003175021
#> 28 0.996068088 0.003931912
#> 29 0.096807861 0.903192139

# Evaluate each prediction, i.e. if the predicted likelihood that the country is France is '0.9'
# and the likelihood it's Holland is '0.1', then the prediction is "France"
pred_model_Tree1$evaluation <- ifelse(pred_model_Tree1$France >= 0.5, "France", "Holland")

# Now you can print the confusionMatrix (make sure each factor has the same levels)
confusionMatrix(factor(pred_model_Tree1$evaluation, levels = unique(test.data$Country)),
                factor(test.data$Country, levels = unique(test.data$Country)))
#> Confusion Matrix and Statistics
#> 
#>           Reference
#> Prediction France Holland
#>    France      13       1
#>    Holland      0      15
#>                                           
#>                Accuracy : 0.9655          
#>                  95% CI : (0.8224, 0.9991)
#>     No Information Rate : 0.5517          
#>     P-Value [Acc > NIR] : 7.947e-07       
#>                                           
#>                   Kappa : 0.9308          
#>                                           
#>  Mcnemar's Test P-Value : 1               
#>                                           
#>             Sensitivity : 1.0000          
#>             Specificity : 0.9375          
#>          Pos Pred Value : 0.9286          
#>          Neg Pred Value : 1.0000          
#>              Prevalence : 0.4483          
#>          Detection Rate : 0.4483          
#>    Detection Prevalence : 0.4828          
#>       Balanced Accuracy : 0.9688          
#>                                           
#>        'Positive' Class : France          
#> 

^{Created on 2022-06-02 by the reprex package (v2.0.1)}

---

Edit

Something seems wrong - perhaps you want to remove the IDs before you train/test the model? (Maybe they weren't randomly assigned?) Eg

library(dplyr)
#> 
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#> 
#>     filter, lag
#> The following objects are masked from 'package:base':
#> 
#>     intersect, setdiff, setequal, union
library(gbm)
#> Loaded gbm 2.1.8
library(caret)
#> Loading required package: ggplot2
#> Loading required package: lattice
library(e1071)

set.seed(45L)

#Produce a new version of the data frame 'Clusters_Dummy' with the rows shuffled
Cluster_Dummy_2 <- data
NewClusters <- Cluster_Dummy_2[sample(1:nrow(Cluster_Dummy_2)),]

NewCluster<-as.data.frame(NewClusters)

training.parameters <- Cluster_Dummy_2$Country %>% 
  createDataPartition(p = 0.7, list = FALSE)
train.data <- NewClusters[training.parameters, ] %>%
  select(-ID)
test.data <- NewClusters[-training.parameters, ] %>%
  select(-ID)

dim(train.data)
#> [1] 70 10

dim(test.data)
#> [1] 29 10

#Auxiliary function for controlling model fitting
#10 fold cross validation; 10 times
fitControl <- trainControl(## 10-fold CV
  method = "repeatedcv",
  number = 10,
  ## repeated ten times
  repeats = 10,
  classProbs = TRUE)
#Fit the model
gbmFit1 <- train(Country ~ ., data=train.data, 
                 method = "gbm", 
                 trControl = fitControl,
                 ## This last option is actually one
                 ## for gbm() that passes through
                 verbose = FALSE)
gbmFit1
#> Stochastic Gradient Boosting 
#> 
#> 70 samples
#>  9 predictor
#>  2 classes: 'France', 'Holland' 
#> 
#> No pre-processing
#> Resampling: Cross-Validated (10 fold, repeated 10 times) 
#> Summary of sample sizes: 64, 64, 63, 63, 63, 62, ... 
#> Resampling results across tuning parameters:
#> 
#>   interaction.depth  n.trees  Accuracy   Kappa     
#>   1                   50      0.5515476  0.08773090
#>   1                  100      0.5908929  0.17272118
#>   1                  150      0.5958333  0.18280502
#>   2                   50      0.5386905  0.06596478
#>   2                  100      0.5767262  0.13757567
#>   2                  150      0.5785119  0.14935661
#>   3                   50      0.5575000  0.09991455
#>   3                  100      0.5585119  0.10906906
#>   3                  150      0.5780952  0.14820067
#> 
#> Tuning parameter 'shrinkage' was held constant at a value of 0.1
#> 
#> Tuning parameter 'n.minobsinnode' was held constant at a value of 10
#> Accuracy was used to select the optimal model using the largest value.
#> The final values used for the model were n.trees = 150, interaction.depth =
#>  1, shrinkage = 0.1 and n.minobsinnode = 10.
summary(gbmFit1)

#>                     var   rel.inf
#> Center_Freq Center_Freq 14.094306
#> High.Freq     High.Freq 14.060959
#> Peak.Time     Peak.Time 13.503953
#> Peak.Freq     Peak.Freq 11.358891
#> Delta.Time   Delta.Time  9.964882
#> Low.Freq       Low.Freq  9.610686
#> End.Freq       End.Freq  9.308919
#> Delta.Freq   Delta.Freq  9.097253
#> Start.Freq   Start.Freq  9.000152

#Predict the model with the test data
pred_model_Tree1 <- predict(object = gbmFit1, newdata = test.data, type = "prob")
pred_model_Tree1
#>        France    Holland
#> 1  0.75514031 0.24485969
#> 2  0.44409692 0.55590308
#> 3  0.15027904 0.84972096
#> 4  0.49861536 0.50138464
#> 5  0.95406713 0.04593287
#> 6  0.82122854 0.17877146
#> 7  0.27931450 0.72068550
#> 8  0.50113421 0.49886579
#> 9  0.61912973 0.38087027
#> 10 0.91005442 0.08994558
#> 11 0.42625105 0.57374895
#> 12 0.27339404 0.72660596
#> 13 0.14520192 0.85479808
#> 14 0.16607144 0.83392856
#> 15 0.97198722 0.02801278
#> 16 0.88614818 0.11385182
#> 17 0.65561219 0.34438781
#> 18 0.86793709 0.13206291
#> 19 0.28583233 0.71416767
#> 20 0.97002073 0.02997927
#> 21 0.74408374 0.25591626
#> 22 0.28408111 0.71591889
#> 23 0.07257257 0.92742743
#> 24 0.22724577 0.77275423
#> 25 0.32581206 0.67418794
#> 26 0.59713799 0.40286201
#> 27 0.75814205 0.24185795
#> 28 0.94018097 0.05981903
#> 29 0.51155700 0.48844300

# Evaluate each prediction, i.e. if the predicted likelihood that the country is France is '0.9'
# and the likelihood it's Holland is '0.1', then the prediction is "France"
pred_model_Tree1$evaluation <- ifelse(pred_model_Tree1$France >= 0.5, "France", "Holland")

# Now you can print the confusionMatrix (make sure each factor has the same levels)
confusionMatrix(factor(pred_model_Tree1$evaluation, levels = unique(test.data$Country)),
                factor(test.data$Country, levels = unique(test.data$Country)))
#> Confusion Matrix and Statistics
#> 
#>           Reference
#> Prediction France Holland
#>    France       9       7
#>    Holland      4       9
#>                                           
#>                Accuracy : 0.6207          
#>                  95% CI : (0.4226, 0.7931)
#>     No Information Rate : 0.5517          
#>     P-Value [Acc > NIR] : 0.2897          
#>                                           
#>                   Kappa : 0.2494          
#>                                           
#>  Mcnemar's Test P-Value : 0.5465          
#>                                           
#>             Sensitivity : 0.6923          
#>             Specificity : 0.5625          
#>          Pos Pred Value : 0.5625          
#>          Neg Pred Value : 0.6923          
#>              Prevalence : 0.4483          
#>          Detection Rate : 0.3103          
#>    Detection Prevalence : 0.5517          
#>       Balanced Accuracy : 0.6274          
#>                                           
#>        'Positive' Class : France          
#> 

^{Created on 2022-06-02 by the reprex package (v2.0.1)}

---

Edit 2

For multi-class classification (3 classes):

library(dplyr)
#> 
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#> 
#>     filter, lag
#> The following objects are masked from 'package:base':
#> 
#>     intersect, setdiff, setequal, union
library(gbm)
#> Loaded gbm 2.1.8
library(caret)
#> Loading required package: ggplot2
#> Loading required package: lattice
library(e1071)

set.seed(45L)

#Produce a new version of the data frame 'Clusters_Dummy' with the rows shuffled
Cluster_Dummy_2 <- data_updated

NewClusters <- Cluster_Dummy_2[sample(1:nrow(Cluster_Dummy_2)),]

NewCluster <- as.data.frame(NewClusters)

training.parameters <- Cluster_Dummy_2$Country %>% 
  createDataPartition(p = 0.7, list = FALSE)
train.data <- NewClusters[training.parameters, ]
test.data <- NewClusters[-training.parameters, ]

dim(train.data)
#> [1] 71 10

dim(test.data)
#> [1] 28 10

#Auxiliary function for controlling model fitting
#10 fold cross validation; 10 times
fitControl <- trainControl(## 10-fold CV
  method = "repeatedcv",
  number = 10,
  ## repeated ten times
  repeats = 10,
  classProbs = TRUE)
#Fit the model
gbmFit1 <- train(Country ~ ., data=train.data, 
                 method = "gbm", 
                 trControl = fitControl,
                 ## This last option is actually one
                 ## for gbm() that passes through
                 verbose = FALSE)
gbmFit1
#> Stochastic Gradient Boosting 
#> 
#> 71 samples
#>  9 predictor
#>  3 classes: 'France', 'Holland', 'Spain' 
#> 
#> No pre-processing
#> Resampling: Cross-Validated (10 fold, repeated 10 times) 
#> Summary of sample sizes: 63, 64, 64, 63, 63, 63, ... 
#> Resampling results across tuning parameters:
#> 
#>   interaction.depth  n.trees  Accuracy   Kappa     
#>   1                   50      0.4165476  0.07310546
#>   1                  100      0.4264683  0.09363788
#>   1                  150      0.4164683  0.08078702
#>   2                   50      0.3894048  0.03705497
#>   2                  100      0.4032341  0.06489744
#>   2                  150      0.4075794  0.06765817
#>   3                   50      0.4032341  0.05972739
#>   3                  100      0.3906944  0.04364377
#>   3                  150      0.4236905  0.10068155
#> 
#> Tuning parameter 'shrinkage' was held constant at a value of 0.1
#> 
#> Tuning parameter 'n.minobsinnode' was held constant at a value of 10
#> Accuracy was used to select the optimal model using the largest value.
#> The final values used for the model were n.trees = 100, interaction.depth =
#>  1, shrinkage = 0.1 and n.minobsinnode = 10.
summary(gbmFit1)

#>                     var   rel.inf
#> Peak.Time     Peak.Time 16.211328
#> End.Freq       End.Freq 15.001295
#> Center_Freq Center_Freq 12.583477
#> Delta.Freq   Delta.Freq 11.236692
#> Start.Freq   Start.Freq 10.692191
#> Delta.Time   Delta.Time  9.224466
#> Peak.Freq     Peak.Freq  8.772731
#> Low.Freq       Low.Freq  8.674891
#> High.Freq     High.Freq  7.602928

#Predict the model with the test data
pred_model_Tree1 <- predict(object = gbmFit1, newdata = test.data, type = "prob")
pred_model_Tree1
#>        France    Holland      Spain
#> 1  0.15839683 0.11884456 0.72275861
#> 2  0.31551164 0.62037910 0.06410925
#> 3  0.06056686 0.03289397 0.90653917
#> 4  0.22705213 0.03439780 0.73855007
#> 5  0.05455049 0.02259610 0.92285341
#> 6  0.34187929 0.25613079 0.40198992
#> 7  0.12857217 0.39860882 0.47281901
#> 8  0.08617855 0.09096950 0.82285196
#> 9  0.22635900 0.62549636 0.14814464
#> 10 0.20887256 0.64739917 0.14372826
#> 11 0.03588915 0.74148076 0.22263010
#> 12 0.03083337 0.48043152 0.48873511
#> 13 0.44698228 0.07630407 0.47671365
#> 14 0.12247065 0.01864920 0.85888015
#> 15 0.03022037 0.08301324 0.88676639
#> 16 0.18190023 0.50467449 0.31342527
#> 17 0.10173416 0.11619956 0.78206628
#> 18 0.29744577 0.31149440 0.39105983
#> 19 0.08555810 0.83492846 0.07951344
#> 20 0.67158503 0.12913684 0.19927813
#> 21 0.33985892 0.30094634 0.35919474
#> 22 0.41752286 0.43288825 0.14958889
#> 23 0.10014057 0.85848587 0.04137356
#> 24 0.02483037 0.57939110 0.39577853
#> 25 0.20376019 0.16867259 0.62756722
#> 26 0.05082254 0.11736656 0.83181090
#> 27 0.02621289 0.74597052 0.22781659
#> 28 0.37202204 0.48168272 0.14629524

# Select the most likely country (i.e. the highest prob)
pred_model_Tree1$evaluation <- factor(max.col(pred_model_Tree1[,1:3]), levels=1:3, labels = c("France", "Holland", "Spain"))

# Print the confusionMatrix (make sure each factor has the same levels)
confusionMatrix(factor(pred_model_Tree1$evaluation, levels = unique(test.data$Country)),
                factor(test.data$Country, levels = unique(test.data$Country)))
#> Confusion Matrix and Statistics
#> 
#>           Reference
#> Prediction Spain France Holland
#>    Spain      10      4       2
#>    France      0      0       1
#>    Holland     4      5       2
#> 
#> Overall Statistics
#>                                           
#>                Accuracy : 0.4286          
#>                  95% CI : (0.2446, 0.6282)
#>     No Information Rate : 0.5             
#>     P-Value [Acc > NIR] : 0.8275          
#>                                           
#>                   Kappa : 0.0968          
#>                                           
#>  Mcnemar's Test P-Value : 0.0620          
#> 
#> Statistics by Class:
#> 
#>                      Class: Spain Class: France Class: Holland
#> Sensitivity                0.7143       0.00000        0.40000
#> Specificity                0.5714       0.94737        0.60870
#> Pos Pred Value             0.6250       0.00000        0.18182
#> Neg Pred Value             0.6667       0.66667        0.82353
#> Prevalence                 0.5000       0.32143        0.17857
#> Detection Rate             0.3571       0.00000        0.07143
#> Detection Prevalence       0.5714       0.03571        0.39286
#> Balanced Accuracy          0.6429       0.47368        0.50435

#########
library(tidyverse)

^{Created on 2022-06-03 by the reprex package (v2.0.1)}