Error in root.matrix(crossprod(process)) : matrix is not positive semidefinite
0
I want to extend the RandomForest so that each leaf will contain naivebayes regression instead of average. In the following, I first tried to use mob() for adding linearModel. I got the following error:
Error in root.matrix(crossprod(process)) : matrix is not positive semidefinite
Here is my code:
require (data.table)
require (party)
set.seed(123)
data1 <- read.csv('https://archive.ics.uci.edu/ml/machine-learning-databases/car/car.data',header = TRUE)
colnames(data1)<- c("BuyingPrice","Maintenance","NumDoors","NumPersons","BootSpace","Safety","Condition")
# Split into Train and Validation sets
# Training Set : Validation Set = 70 : 30 (random)
set.seed(100)
train <- sample(nrow(data1), 0.7*nrow(data1), replace = FALSE)
TrainSet <- data1[train,]
ValidSet <- data1[-train,]
summary(TrainSet)
summary(ValidSet)
# Create a Random Forest model with default parameters
model1 <- randomForest(Condition ~ ., data = TrainSet, importance = TRUE)
model1
# Fine tuning parameters of Random Forest model
model2 <- randomForest(Condition ~ ., data = TrainSet, ntree = 500, mtry = 6, importance = TRUE)
model2
fmBH <- mob(Condition ~ BuyingPrice + Maintenance | NumDoors+ NumPersons + BootSpace + Safety ,
data = TrainSet, model = linearModel)
r random-forest naivebayes
asked Nov 18 '18 at 13:34
AviAvi
1,0161630
|
show 1 more comment
0
I want to extend the RandomForest so that each leaf will contain naivebayes regression instead of average. In the following, I first tried to use mob() for adding linearModel. I got the following error:
Error in root.matrix(crossprod(process)) : matrix is not positive semidefinite
Here is my code:
require (data.table)
require (party)
set.seed(123)
data1 <- read.csv('https://archive.ics.uci.edu/ml/machine-learning-databases/car/car.data',header = TRUE)
colnames(data1)<- c("BuyingPrice","Maintenance","NumDoors","NumPersons","BootSpace","Safety","Condition")
# Split into Train and Validation sets
# Training Set : Validation Set = 70 : 30 (random)
set.seed(100)
train <- sample(nrow(data1), 0.7*nrow(data1), replace = FALSE)
TrainSet <- data1[train,]
ValidSet <- data1[-train,]
summary(TrainSet)
summary(ValidSet)
# Create a Random Forest model with default parameters
model1 <- randomForest(Condition ~ ., data = TrainSet, importance = TRUE)
model1
# Fine tuning parameters of Random Forest model
model2 <- randomForest(Condition ~ ., data = TrainSet, ntree = 500, mtry = 6, importance = TRUE)
model2
fmBH <- mob(Condition ~ BuyingPrice + Maintenance | NumDoors+ NumPersons + BootSpace + Safety ,
data = TrainSet, model = linearModel)
r random-forest naivebayes
asked Nov 18 '18 at 13:34
AviAvi
1,0161630
I see no question marks... Also, your code doesn't really contain anything related to Naive Bayes. So, be sure to make it clear whether you care about understanding the error (as the title suggests) or something related to Naive Bayes.
– Julius Vainora
Nov 18 '18 at 13:50
Thanks, @Julius Vainora, I asked how to change the model = linearModel to naivebayes and on the way I got error for linear model.
– Avi
Nov 18 '18 at 13:57
Hi @Avi, I think you have to adapt the model to your problem. Your target variable is a factor, so you have no regression problem, it is either a classification problem. And I don't know if mob is working with multiclassification-problem. See for more details:vignette("MOB")I tried the following and your initial problem is kind of solved:fmBH <- mob(TrainSet$Condition ~ BuyingPrice + Maintenance | NumDoors+ NumPersons + BootSpace + Safety , data = TrainSet, model = glinearModel, family = binomial())
– floe
Nov 18 '18 at 14:05
@Avi, I suggest to make it clear in your question. The error comes from the fact thatConditionis a factor and you are fitting a linear model. As to use Naive Bayes, you need to create this model type, seemodelin?mob.
– Julius Vainora
Nov 18 '18 at 14:06
Thanks both. @floe - I tried your solution it works with 26 warnings I don't know why... I would like to know how can I change the model to naivebayes instead of glinearModel when the target value is not factor?
– Avi
Nov 18 '18 at 14:14
|
show 1 more comment
0
0
0
I want to extend the RandomForest so that each leaf will contain naivebayes regression instead of average. In the following, I first tried to use mob() for adding linearModel. I got the following error:
Error in root.matrix(crossprod(process)) : matrix is not positive semidefinite
Here is my code:
require (data.table)
require (party)
set.seed(123)
data1 <- read.csv('https://archive.ics.uci.edu/ml/machine-learning-databases/car/car.data',header = TRUE)
colnames(data1)<- c("BuyingPrice","Maintenance","NumDoors","NumPersons","BootSpace","Safety","Condition")
# Split into Train and Validation sets
# Training Set : Validation Set = 70 : 30 (random)
set.seed(100)
train <- sample(nrow(data1), 0.7*nrow(data1), replace = FALSE)
TrainSet <- data1[train,]
ValidSet <- data1[-train,]
summary(TrainSet)
summary(ValidSet)
# Create a Random Forest model with default parameters
model1 <- randomForest(Condition ~ ., data = TrainSet, importance = TRUE)
model1
# Fine tuning parameters of Random Forest model
model2 <- randomForest(Condition ~ ., data = TrainSet, ntree = 500, mtry = 6, importance = TRUE)
model2
fmBH <- mob(Condition ~ BuyingPrice + Maintenance | NumDoors+ NumPersons + BootSpace + Safety ,
data = TrainSet, model = linearModel)
r random-forest naivebayes
asked Nov 18 '18 at 13:34
AviAvi
1,0161630
I want to extend the RandomForest so that each leaf will contain naivebayes regression instead of average. In the following, I first tried to use mob() for adding linearModel. I got the following error:
Error in root.matrix(crossprod(process)) : matrix is not positive semidefinite
Here is my code:
require (data.table)
require (party)
set.seed(123)
data1 <- read.csv('https://archive.ics.uci.edu/ml/machine-learning-databases/car/car.data',header = TRUE)
colnames(data1)<- c("BuyingPrice","Maintenance","NumDoors","NumPersons","BootSpace","Safety","Condition")
# Split into Train and Validation sets
# Training Set : Validation Set = 70 : 30 (random)
set.seed(100)
train <- sample(nrow(data1), 0.7*nrow(data1), replace = FALSE)
TrainSet <- data1[train,]
ValidSet <- data1[-train,]
summary(TrainSet)
summary(ValidSet)
# Create a Random Forest model with default parameters
model1 <- randomForest(Condition ~ ., data = TrainSet, importance = TRUE)
model1
# Fine tuning parameters of Random Forest model
model2 <- randomForest(Condition ~ ., data = TrainSet, ntree = 500, mtry = 6, importance = TRUE)
model2
fmBH <- mob(Condition ~ BuyingPrice + Maintenance | NumDoors+ NumPersons + BootSpace + Safety ,
data = TrainSet, model = linearModel)
r random-forest naivebayes
r random-forest naivebayes
asked Nov 18 '18 at 13:34
AviAvi
1,0161630
asked Nov 18 '18 at 13:34
AviAvi
1,0161630
asked Nov 18 '18 at 13:34
AviAvi
1,0161630
asked Nov 18 '18 at 13:34
AviAvi
1,0161630
asked Nov 18 '18 at 13:34
AviAvi
1,0161630
1,0161630
I see no question marks... Also, your code doesn't really contain anything related to Naive Bayes. So, be sure to make it clear whether you care about understanding the error (as the title suggests) or something related to Naive Bayes.
– Julius Vainora
Nov 18 '18 at 13:50
Thanks, @Julius Vainora, I asked how to change the model = linearModel to naivebayes and on the way I got error for linear model.
– Avi
Nov 18 '18 at 13:57
Hi @Avi, I think you have to adapt the model to your problem. Your target variable is a factor, so you have no regression problem, it is either a classification problem. And I don't know if mob is working with multiclassification-problem. See for more details:vignette("MOB")I tried the following and your initial problem is kind of solved:fmBH <- mob(TrainSet$Condition ~ BuyingPrice + Maintenance | NumDoors+ NumPersons + BootSpace + Safety , data = TrainSet, model = glinearModel, family = binomial())
– floe
Nov 18 '18 at 14:05
@Avi, I suggest to make it clear in your question. The error comes from the fact thatConditionis a factor and you are fitting a linear model. As to use Naive Bayes, you need to create this model type, seemodelin?mob.
– Julius Vainora
Nov 18 '18 at 14:06
Thanks both. @floe - I tried your solution it works with 26 warnings I don't know why... I would like to know how can I change the model to naivebayes instead of glinearModel when the target value is not factor?
– Avi
Nov 18 '18 at 14:14
|
show 1 more comment
I see no question marks... Also, your code doesn't really contain anything related to Naive Bayes. So, be sure to make it clear whether you care about understanding the error (as the title suggests) or something related to Naive Bayes.
– Julius Vainora
Nov 18 '18 at 13:50
Thanks, @Julius Vainora, I asked how to change the model = linearModel to naivebayes and on the way I got error for linear model.
– Avi
Nov 18 '18 at 13:57
Hi @Avi, I think you have to adapt the model to your problem. Your target variable is a factor, so you have no regression problem, it is either a classification problem. And I don't know if mob is working with multiclassification-problem. See for more details:vignette("MOB")I tried the following and your initial problem is kind of solved:fmBH <- mob(TrainSet$Condition ~ BuyingPrice + Maintenance | NumDoors+ NumPersons + BootSpace + Safety , data = TrainSet, model = glinearModel, family = binomial())
– floe
Nov 18 '18 at 14:05
@Avi, I suggest to make it clear in your question. The error comes from the fact thatConditionis a factor and you are fitting a linear model. As to use Naive Bayes, you need to create this model type, seemodelin?mob.
– Julius Vainora
Nov 18 '18 at 14:06
Thanks both. @floe - I tried your solution it works with 26 warnings I don't know why... I would like to know how can I change the model to naivebayes instead of glinearModel when the target value is not factor?
– Avi
Nov 18 '18 at 14:14
I see no question marks... Also, your code doesn't really contain anything related to Naive Bayes. So, be sure to make it clear whether you care about understanding the error (as the title suggests) or something related to Naive Bayes.
– Julius Vainora
Nov 18 '18 at 13:50
I see no question marks... Also, your code doesn't really contain anything related to Naive Bayes. So, be sure to make it clear whether you care about understanding the error (as the title suggests) or something related to Naive Bayes.
– Julius Vainora
Nov 18 '18 at 13:50
Thanks, @Julius Vainora, I asked how to change the model = linearModel to naivebayes and on the way I got error for linear model.
– Avi
Nov 18 '18 at 13:57
Thanks, @Julius Vainora, I asked how to change the model = linearModel to naivebayes and on the way I got error for linear model.
– Avi
Nov 18 '18 at 13:57
Hi @Avi, I think you have to adapt the model to your problem. Your target variable is a factor, so you have no regression problem, it is either a classification problem. And I don't know if mob is working with multiclassification-problem. See for more details:
vignette("MOB") I tried the following and your initial problem is kind of solved: fmBH <- mob(TrainSet$Condition ~ BuyingPrice + Maintenance | NumDoors+ NumPersons + BootSpace + Safety , data = TrainSet, model = glinearModel, family = binomial())– floe
Nov 18 '18 at 14:05
Hi @Avi, I think you have to adapt the model to your problem. Your target variable is a factor, so you have no regression problem, it is either a classification problem. And I don't know if mob is working with multiclassification-problem. See for more details:
vignette("MOB") I tried the following and your initial problem is kind of solved: fmBH <- mob(TrainSet$Condition ~ BuyingPrice + Maintenance | NumDoors+ NumPersons + BootSpace + Safety , data = TrainSet, model = glinearModel, family = binomial())– floe
Nov 18 '18 at 14:05
@Avi, I suggest to make it clear in your question. The error comes from the fact that
Condition is a factor and you are fitting a linear model. As to use Naive Bayes, you need to create this model type, see model in ?mob.– Julius Vainora
Nov 18 '18 at 14:06
@Avi, I suggest to make it clear in your question. The error comes from the fact that
Condition is a factor and you are fitting a linear model. As to use Naive Bayes, you need to create this model type, see model in ?mob.– Julius Vainora
Nov 18 '18 at 14:06
Thanks both. @floe - I tried your solution it works with 26 warnings I don't know why... I would like to know how can I change the model to naivebayes instead of glinearModel when the target value is not factor?
– Avi
Nov 18 '18 at 14:14
Thanks both. @floe - I tried your solution it works with 26 warnings I don't know why... I would like to know how can I change the model to naivebayes instead of glinearModel when the target value is not factor?
– Avi
Nov 18 '18 at 14:14
|
show 1 more comment
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I see no question marks... Also, your code doesn't really contain anything related to Naive Bayes. So, be sure to make it clear whether you care about understanding the error (as the title suggests) or something related to Naive Bayes.
– Julius Vainora
Nov 18 '18 at 13:50
Thanks, @Julius Vainora, I asked how to change the model = linearModel to naivebayes and on the way I got error for linear model.
– Avi
Nov 18 '18 at 13:57
Hi @Avi, I think you have to adapt the model to your problem. Your target variable is a factor, so you have no regression problem, it is either a classification problem. And I don't know if mob is working with multiclassification-problem. See for more details:
vignette("MOB")I tried the following and your initial problem is kind of solved:fmBH <- mob(TrainSet$Condition ~ BuyingPrice + Maintenance | NumDoors+ NumPersons + BootSpace + Safety , data = TrainSet, model = glinearModel, family = binomial())– floe
Nov 18 '18 at 14:05
@Avi, I suggest to make it clear in your question. The error comes from the fact that
Conditionis a factor and you are fitting a linear model. As to use Naive Bayes, you need to create this model type, seemodelin?mob.– Julius Vainora
Nov 18 '18 at 14:06
Thanks both. @floe - I tried your solution it works with 26 warnings I don't know why... I would like to know how can I change the model to naivebayes instead of glinearModel when the target value is not factor?
– Avi
Nov 18 '18 at 14:14