Confusion matrix for training and validation sets
What is the purpose of the "newdata" argument?
Why don't we need to specify newdata = tlFLAG.t in the first case?
pred <- predict(tree1, type = "class")
confusionMatrix(pred, factor(tlFLAG.t$TERM_FLAG))
pred.v <- predict(tree1, type = "class", newdata = tlFLAG.v)
confusionMatrix(pred.v, factor(tlFLAG.v$TERM_FLAG))
r confusion-matrix
edited Nov 15 '18 at 7:50
RLave
4,76711124
asked Nov 15 '18 at 5:09
FICFIC
163
add a comment |
What is the purpose of the "newdata" argument?
Why don't we need to specify newdata = tlFLAG.t in the first case?
pred <- predict(tree1, type = "class")
confusionMatrix(pred, factor(tlFLAG.t$TERM_FLAG))
pred.v <- predict(tree1, type = "class", newdata = tlFLAG.v)
confusionMatrix(pred.v, factor(tlFLAG.v$TERM_FLAG))
r confusion-matrix
edited Nov 15 '18 at 7:50
RLave
4,76711124
asked Nov 15 '18 at 5:09
FICFIC
163
2
It's not really clear what you're asking, or at least the point you're trying to make with the code you showed. The "newdata" argument is for making predictions from a model with a new data set.
– mickey
Nov 15 '18 at 5:13
based on u r data u r going to use in confusion matrix..try to google it for more information or below thread have more info for u rdocumentation.org/packages/caret/versions/3.45/topics/… ,
– sai saran
Nov 15 '18 at 5:22
add a comment |
What is the purpose of the "newdata" argument?
Why don't we need to specify newdata = tlFLAG.t in the first case?
pred <- predict(tree1, type = "class")
confusionMatrix(pred, factor(tlFLAG.t$TERM_FLAG))
pred.v <- predict(tree1, type = "class", newdata = tlFLAG.v)
confusionMatrix(pred.v, factor(tlFLAG.v$TERM_FLAG))
r confusion-matrix
edited Nov 15 '18 at 7:50
RLave
4,76711124
asked Nov 15 '18 at 5:09
FICFIC
163
What is the purpose of the "newdata" argument?
Why don't we need to specify newdata = tlFLAG.t in the first case?
pred <- predict(tree1, type = "class")
confusionMatrix(pred, factor(tlFLAG.t$TERM_FLAG))
pred.v <- predict(tree1, type = "class", newdata = tlFLAG.v)
confusionMatrix(pred.v, factor(tlFLAG.v$TERM_FLAG))
r confusion-matrix
r confusion-matrix
edited Nov 15 '18 at 7:50
RLave
4,76711124
asked Nov 15 '18 at 5:09
FICFIC
163
edited Nov 15 '18 at 7:50
RLave
4,76711124
asked Nov 15 '18 at 5:09
FICFIC
163
edited Nov 15 '18 at 7:50
RLave
4,76711124
edited Nov 15 '18 at 7:50
RLave
4,76711124
edited Nov 15 '18 at 7:50
RLave
4,76711124
4,76711124
asked Nov 15 '18 at 5:09
FICFIC
163
asked Nov 15 '18 at 5:09
FICFIC
163
asked Nov 15 '18 at 5:09
FICFIC
163
163
2
It's not really clear what you're asking, or at least the point you're trying to make with the code you showed. The "newdata" argument is for making predictions from a model with a new data set.
– mickey
Nov 15 '18 at 5:13
based on u r data u r going to use in confusion matrix..try to google it for more information or below thread have more info for u rdocumentation.org/packages/caret/versions/3.45/topics/… ,
– sai saran
Nov 15 '18 at 5:22
add a comment |
2
It's not really clear what you're asking, or at least the point you're trying to make with the code you showed. The "newdata" argument is for making predictions from a model with a new data set.
– mickey
Nov 15 '18 at 5:13
based on u r data u r going to use in confusion matrix..try to google it for more information or below thread have more info for u rdocumentation.org/packages/caret/versions/3.45/topics/… ,
– sai saran
Nov 15 '18 at 5:22
2
2
It's not really clear what you're asking, or at least the point you're trying to make with the code you showed. The "newdata" argument is for making predictions from a model with a new data set.
– mickey
Nov 15 '18 at 5:13
It's not really clear what you're asking, or at least the point you're trying to make with the code you showed. The "newdata" argument is for making predictions from a model with a new data set.
– mickey
Nov 15 '18 at 5:13
based on u r data u r going to use in confusion matrix..try to google it for more information or below thread have more info for u rdocumentation.org/packages/caret/versions/3.45/topics/… ,
– sai saran
Nov 15 '18 at 5:22
based on u r data u r going to use in confusion matrix..try to google it for more information or below thread have more info for u rdocumentation.org/packages/caret/versions/3.45/topics/… ,
– sai saran
Nov 15 '18 at 5:22
add a comment |
1 Answer
1
active
oldest
votes
In every machine learning process (in this case a classification problem), you have to split your data in a train and a test set.
This is useful because you can train your algorithm in the first set, and test it on the second.
This has to be done, otherwise (if you use all the data) you're exposing yourself to overfitting, because almost every algorithm will try to fit best the data you feed.
You'll end up with even a perfect model for your data, but that will predict very poorly on new data, that it has not yet seen.
The predict function, because of this, lets you pick new data to "test" the goodness of your model on unseen data by the newdata= arg.
In your first case so, you "test" you performance on the already trained data by not specifying the newdata= arg, so the confusionMatrix could be over-ottimistic.
In the second case you should specify newdata=test_set, and with this your prediction will be based on test data, so the performance will be more accurate, and even more interesting on this second case.
I'll build here an example for you to see a classic approach:
data <- iris # iris dataset
# first split the data
set.seed(123) # for reproducibility
pos <- sample(100)
train <- data[pos, ] # random pick of 100 obs
test <- data[-pos, ] # remaining 50
# now you can start with your model - please not that this is a dummy example
library(rpart)
tree <- rpart(Species ~ ., data=train) # fit tree on train data
# make prediction on train data (no need to specify newclass= ) # NOT very useful
pred <- predict(tree, type = "class")
caret::confusionMatrix(pred, train$Species)
# make prediction on test data (remove the response)
pred <- predict(tree, type = "class", newdata = test[, -5]) # I removed Species (5th column in test)
# build confusion from predictions against the truth (ie the test$Species)
caret::confusionMatrix(pred, test$Species)
Note how the performance is awful on the test data, while it was almost perfect on train data.
answered Nov 15 '18 at 8:10
RLaveRLave
4,76711124
add a comment |
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In every machine learning process (in this case a classification problem), you have to split your data in a train and a test set.
This is useful because you can train your algorithm in the first set, and test it on the second.
This has to be done, otherwise (if you use all the data) you're exposing yourself to overfitting, because almost every algorithm will try to fit best the data you feed.
You'll end up with even a perfect model for your data, but that will predict very poorly on new data, that it has not yet seen.
The predict function, because of this, lets you pick new data to "test" the goodness of your model on unseen data by the newdata= arg.
In your first case so, you "test" you performance on the already trained data by not specifying the newdata= arg, so the confusionMatrix could be over-ottimistic.
In the second case you should specify newdata=test_set, and with this your prediction will be based on test data, so the performance will be more accurate, and even more interesting on this second case.
I'll build here an example for you to see a classic approach:
data <- iris # iris dataset
# first split the data
set.seed(123) # for reproducibility
pos <- sample(100)
train <- data[pos, ] # random pick of 100 obs
test <- data[-pos, ] # remaining 50
# now you can start with your model - please not that this is a dummy example
library(rpart)
tree <- rpart(Species ~ ., data=train) # fit tree on train data
# make prediction on train data (no need to specify newclass= ) # NOT very useful
pred <- predict(tree, type = "class")
caret::confusionMatrix(pred, train$Species)
# make prediction on test data (remove the response)
pred <- predict(tree, type = "class", newdata = test[, -5]) # I removed Species (5th column in test)
# build confusion from predictions against the truth (ie the test$Species)
caret::confusionMatrix(pred, test$Species)
Note how the performance is awful on the test data, while it was almost perfect on train data.
answered Nov 15 '18 at 8:10
RLaveRLave
4,76711124
add a comment |
In every machine learning process (in this case a classification problem), you have to split your data in a train and a test set.
This is useful because you can train your algorithm in the first set, and test it on the second.
This has to be done, otherwise (if you use all the data) you're exposing yourself to overfitting, because almost every algorithm will try to fit best the data you feed.
You'll end up with even a perfect model for your data, but that will predict very poorly on new data, that it has not yet seen.
The predict function, because of this, lets you pick new data to "test" the goodness of your model on unseen data by the newdata= arg.
In your first case so, you "test" you performance on the already trained data by not specifying the newdata= arg, so the confusionMatrix could be over-ottimistic.
In the second case you should specify newdata=test_set, and with this your prediction will be based on test data, so the performance will be more accurate, and even more interesting on this second case.
I'll build here an example for you to see a classic approach:
data <- iris # iris dataset
# first split the data
set.seed(123) # for reproducibility
pos <- sample(100)
train <- data[pos, ] # random pick of 100 obs
test <- data[-pos, ] # remaining 50
# now you can start with your model - please not that this is a dummy example
library(rpart)
tree <- rpart(Species ~ ., data=train) # fit tree on train data
# make prediction on train data (no need to specify newclass= ) # NOT very useful
pred <- predict(tree, type = "class")
caret::confusionMatrix(pred, train$Species)
# make prediction on test data (remove the response)
pred <- predict(tree, type = "class", newdata = test[, -5]) # I removed Species (5th column in test)
# build confusion from predictions against the truth (ie the test$Species)
caret::confusionMatrix(pred, test$Species)
Note how the performance is awful on the test data, while it was almost perfect on train data.
answered Nov 15 '18 at 8:10
RLaveRLave
4,76711124
add a comment |
In every machine learning process (in this case a classification problem), you have to split your data in a train and a test set.
This is useful because you can train your algorithm in the first set, and test it on the second.
This has to be done, otherwise (if you use all the data) you're exposing yourself to overfitting, because almost every algorithm will try to fit best the data you feed.
You'll end up with even a perfect model for your data, but that will predict very poorly on new data, that it has not yet seen.
The predict function, because of this, lets you pick new data to "test" the goodness of your model on unseen data by the newdata= arg.
In your first case so, you "test" you performance on the already trained data by not specifying the newdata= arg, so the confusionMatrix could be over-ottimistic.
In the second case you should specify newdata=test_set, and with this your prediction will be based on test data, so the performance will be more accurate, and even more interesting on this second case.
I'll build here an example for you to see a classic approach:
data <- iris # iris dataset
# first split the data
set.seed(123) # for reproducibility
pos <- sample(100)
train <- data[pos, ] # random pick of 100 obs
test <- data[-pos, ] # remaining 50
# now you can start with your model - please not that this is a dummy example
library(rpart)
tree <- rpart(Species ~ ., data=train) # fit tree on train data
# make prediction on train data (no need to specify newclass= ) # NOT very useful
pred <- predict(tree, type = "class")
caret::confusionMatrix(pred, train$Species)
# make prediction on test data (remove the response)
pred <- predict(tree, type = "class", newdata = test[, -5]) # I removed Species (5th column in test)
# build confusion from predictions against the truth (ie the test$Species)
caret::confusionMatrix(pred, test$Species)
Note how the performance is awful on the test data, while it was almost perfect on train data.
answered Nov 15 '18 at 8:10
RLaveRLave
4,76711124
In every machine learning process (in this case a classification problem), you have to split your data in a train and a test set.
This is useful because you can train your algorithm in the first set, and test it on the second.
This has to be done, otherwise (if you use all the data) you're exposing yourself to overfitting, because almost every algorithm will try to fit best the data you feed.
You'll end up with even a perfect model for your data, but that will predict very poorly on new data, that it has not yet seen.
The predict function, because of this, lets you pick new data to "test" the goodness of your model on unseen data by the newdata= arg.
In your first case so, you "test" you performance on the already trained data by not specifying the newdata= arg, so the confusionMatrix could be over-ottimistic.
In the second case you should specify newdata=test_set, and with this your prediction will be based on test data, so the performance will be more accurate, and even more interesting on this second case.
I'll build here an example for you to see a classic approach:
data <- iris # iris dataset
# first split the data
set.seed(123) # for reproducibility
pos <- sample(100)
train <- data[pos, ] # random pick of 100 obs
test <- data[-pos, ] # remaining 50
# now you can start with your model - please not that this is a dummy example
library(rpart)
tree <- rpart(Species ~ ., data=train) # fit tree on train data
# make prediction on train data (no need to specify newclass= ) # NOT very useful
pred <- predict(tree, type = "class")
caret::confusionMatrix(pred, train$Species)
# make prediction on test data (remove the response)
pred <- predict(tree, type = "class", newdata = test[, -5]) # I removed Species (5th column in test)
# build confusion from predictions against the truth (ie the test$Species)
caret::confusionMatrix(pred, test$Species)
Note how the performance is awful on the test data, while it was almost perfect on train data.
answered Nov 15 '18 at 8:10
RLaveRLave
4,76711124
edited Nov 15 '18 at 8:18
answered Nov 15 '18 at 8:10
RLaveRLave
4,76711124
answered Nov 15 '18 at 8:10
RLaveRLave
4,76711124
answered Nov 15 '18 at 8:10
RLaveRLave
4,76711124
4,76711124
add a comment |
add a comment |
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2
It's not really clear what you're asking, or at least the point you're trying to make with the code you showed. The "newdata" argument is for making predictions from a model with a new data set.
– mickey
Nov 15 '18 at 5:13
based on u r data u r going to use in confusion matrix..try to google it for more information or below thread have more info for u rdocumentation.org/packages/caret/versions/3.45/topics/… ,
– sai saran
Nov 15 '18 at 5:22