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$Country
print(createConfusionMatrix(act, pred))
#Error
Error in FUN(X[[i]], ...) :
only defined on a data frame with all numeric-alike variables
Data
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"Spain", "Spain", "Spain", "Spain", "Spain", "Spain", "Spain",
"Spain", "Spain", "Spain", "Spain", "Holland", "Holland", "Holland",
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"France", "France", "France", "France", "France", "Spain", "Spain",
"Spain", "Spain", "Spain", "Spain", "Spain", "Spain", "Spain",
"Spain", "Spain", "Spain", "Spain", "Spain", "Spain", "Spain",
"Spain", "Spain", "France", "France", "France")), row.names = c(NA,
99L), class = "data.frame")
CodePudding user response：
Thanks for including all the required information; I believe this is the solution to your problem:
library(magrittr)
library(gbm)
#> Loaded gbm 2.1.8
library(caret)
#> Loading required package: ggplot2
#> Loading required package: lattice
library(e1071)
set.seed(45L)
# Load in your example data to an object ("data")
#Produce a new version of the data frame 'Clusters_Dummy' with the rows shuffled
Cluster_Dummy_2 <- data
NewClusters <- Cluster_Dummy_2[sample(1:nrow(Cluster_Dummy_2)),]
NewCluster<-as.data.frame(NewClusters)
training.parameters <- Cluster_Dummy_2$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?) E.g.
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)