cv.bst: Cross-Validation for Binary HingeBoost

Description

Cross-validated estimation of the empirical risk for boosting parameter selection.

Usage

cv.bst(x, y, K = 10, cost = 0.5, family = c("hinge", "gaussian"), 
learner = c("tree", "ls", "sm"), ctrl = bst_control(), 
type = c("risk", "misc"), plot.it = TRUE, se = TRUE, ...)

Arguments

a data frame containing the variables in the model.

vector of responses. y must be in {1, -1} for family = "hinge".

K-fold cross-validation

cost

price to pay for false positive, 0 < cost < 1; price of false negative is 1-cost.

family

family = "hinge" for hinge loss and family="gaussian" for squared error loss. Implementing the negative gradient corresponding to the loss function to be minimized. By default, hinge loss for +1/-1

learner

a character specifying the component-wise base learner to be used: ls linear models, sm smoothing splines, tree regression trees.

ctrl

an object of class bst_control.

type

cross-validation criteria. For family="hinge", type="risk" is hinge risk and type="misc" is misclassification error. For family="gaussian", only empirical risks.

plot.it

a logical value, to plot the estimated risks if TRUE.

a logical value, to plot with standard errors.

...

additional arguments.

Value

object with
residmatempirical risks in each cross-validation at boosting iterations
fractionabscissa values at which CV curve should be computed.
cvThe CV curve at each value of fraction
cv.errorThe standard error of the CV curve
...

Examples

Run this code

x <- matrix(rnorm(100*5),ncol=5)
c <- 2*x[,1]
p <- exp(c)/(exp(c)+exp(-c))
y <- rbinom(100,1,p)
y[y != 1] <- -1
x <- as.data.frame(x)
cv.bst(x, y, ctrl = bst_control(mstop=50), family = "hinge", learner = "ls")
cv.bst(x, y, ctrl = bst_control(mstop=50), family = "hinge", learner = "ls", type="misc")

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