ROSE-package: ROSE: Random Over-Sampling Examples

In ROSE: Random Over-Sampling Examples

Description

Functions to deal with binary classification problems in the presence of imbalanced classes. Synthetic balanced samples are generated according to ROSE (Menardi and Torelli, 2014). Functions that implement more traditional remedies to the class imbalance are also provided, as well as different metrics to evaluate a learner accuracy. These are estimated by holdout, bootrstrap or cross-validation methods.

Details

The package pivots on function ROSE which generates synthetic balanced samples and thus allows to strenghten the subsequent estimation of any binary classifier. ROSE (Random Over-Sampling Examples) is a bootstrap-based technique which aids the task of binary classification in the presence of rare classes. It handles both continuous and categorical data by generating synthetic examples from a conditional density estimate of the two classes. Different metrics to evaluate a learner accuracy are supplied by functions roc.curve and accuracy.meas. Holdout, bootstrap or cross-validation estimators of these accuracy metrics are computed by means of ROSE and provided by function ROSE.eval, to be used in conjuction with virtually any binary classifier. Additionally, function ovun.sample implements more traditional remedies to the class imbalance, such as over-sampling the minority class, under-sampling the majority class, or a combination of over- and under- sampling.

Author(s)

Nicola Lunardon, Giovanna Menardi, Nicola Torelli

Maintainer: Nicola Lunardon <[e-mail]>

References

Lunardon, N., Menardi, G., and Torelli, N. (2014). ROSE: a Package for Binary Imbalanced Learning. R Jorunal, 6:82–92.

Menardi, G. and Torelli, N. (2014). Training and assessing classification rules with imbalanced data. Data Mining and Knowledge Discovery, 28:92–122.

See Also

nnet, rpart

Examples

# loading data
data(hacide)

# check imbalance
table(hacide.train$cls)

# train logistic regression on imbalanced data
log.reg.imb <- glm(cls ~ ., data=hacide.train, family=binomial)

# use the trained model to predict test data
pred.log.reg.imb <- predict(log.reg.imb, newdata=hacide.test,
                            type="response")

# generate new balanced data by ROSE
hacide.rose <- ROSE(cls ~ ., data=hacide.train, seed=123)$data

# check (im)balance of new data
table(hacide.rose$cls)

# train logistic regression on balanced data
log.reg.bal <- glm(cls ~ ., data=hacide.rose, family=binomial)

# use the trained model to predict test data
pred.log.reg.bal <- predict(log.reg.bal, newdata=hacide.test,
                            type="response")

# check accuracy of the two learners by measuring auc
roc.curve(hacide.test$cls, pred.log.reg.imb)
roc.curve(hacide.test$cls, pred.log.reg.bal, add.roc=TRUE, col=2)

# determine bootstrap distribution of the AUC of logit models
# trained on ROSE balanced samples
# B has been reduced from 100 to 10 for time saving solely
boot.auc.bal <- ROSE.eval(cls ~ ., data=hacide.train, learner= glm,
                          method.assess = "BOOT",
                          control.learner=list(family=binomial),
                          trace=TRUE, B=10)

summary(boot.auc.bal)

Example output

Loaded ROSE 0.0-3


  0   1 
980  20 

  0   1 
507 493 
Area under the curve (AUC): 0.803
Area under the curve (AUC): 0.915
Iteration: 
1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Call: 
ROSE.eval(formula = cls ~ ., data = hacide.train, learner = glm, 
    method.assess = "BOOT", B = 10, control.learner = list(family = binomial), 
    trace = TRUE)

Summary of bootstrap distribution of auc: 
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.9274  0.9277  0.9278  0.9280  0.9282  0.9291 

ROSE documentation built on June 14, 2021, 9:06 a.m.