Predict from a fitted bridge object

Compute response predictions, extract coefficients, or identify nonzero predictors from a fitted bridge object.

Usage

# S3 method for class 'bridge'
predict(
  object,
  newx,
  s = c("lambda.min", "lambda.1se"),
  type = c("response", "nonzero", "coefficients"),
  ...
)

Arguments

object

Fitted object returned by bridge, with p predictors in its training design.

newx

Numeric matrix with one row per new observation and the same p columns, in the same order, as the training design. A numeric vector of length p represents one observation. Values are supplied on the original predictor scale. Required only for type = "response".

s

Numeric fitted penalty value(s). Values must match object$lambda exactly; no interpolation is used. Omitting this argument returns all fitted penalties.

The character selector lambda.min requests the smallest fitted penalty. Use a CV object for the selector lambda.1se.

type

Output to return: "response" for fitted response values, "coefficients" for intercept and slopes, or "nonzero" for predictor indices over the entire fitted path.

...

Additional arguments accepted for compatibility.

Value

For type = "response", a numeric vector for one selected penalty or a matrix with one row per new observation and one column per selected penalty. For type = "coefficients", a numeric matrix with p + 1 rows, including the intercept, even for one penalty. For type = "nonzero", a pathwise collection of 1-based predictor indices excluding the intercept. Its historical container can simplify depending on the support pattern and number of predictors.

Details

Prediction uses the fitted intercept and original-scale slopes. Columns of newx are matched by position, not reordered by their names. For type = "nonzero", all fitted penalties are inspected and s does not restrict the returned indices.

Author

Bahadir Yuzbasi
Maintainer: Bahadir Yuzbasi b.yzb@hotmail.com

Examples

# Reproducible synthetic demonstration: 100 observations, eight predictors.
set.seed(2019)
beta <- c(3, 1.5, 0, 0, 2, 0, 0, 0)
p <- length(beta)
beta <- matrix(beta, nrow = p, ncol = 1)
n <- 100
X <- matrix(rnorm(n * p), nrow = n, ncol = p)
y <- as.numeric(X %*% beta + rnorm(n))

# Five penalties keep this help example fast; the default is 100.
model1 <- bridge(X, y, q = 1, nlambda = 5)
# Predict five observations at one explicitly selected penalty.
predict(model1, newx = X[1:5, , drop = FALSE], s = model1$lambda[2])
#> [1] -1.5002196  0.2180873 -1.5417858  2.4731030 -9.8607011
# Extract coefficients using the complete type name.
predict(model1, s = model1$lambda[2], type = "coefficients")
#>             26.8035044402758
#> (Intercept)    -3.121540e-02
#> V1              2.848999e+00
#> V2              1.433841e+00
#> V3              1.282147e-04
#> V4             -1.150407e-05
#> V5              1.879705e+00
#> V6             -1.064341e-04
#> V7              8.497907e-08
#> V8             -1.609602e-05
# Predictor indices for every fitted penalty; the intercept is excluded.
predict(model1, type = "nonzero")
#> $`802.149726533951`
#> [1] 1 5
#> 
#> $`26.8035044402758`
#> [1] 1 2 3 4 5 6 7 8
#> 
#> $`0.895628118436414`
#> [1] 1 2 3 4 5 6 7 8
#> 
#> $`0.0299270466039737`
#> [1] 1 2 3 4 5 6 7 8
#> 
#> $`0.001`
#> [1] 1 2 3 4 5 6 7 8
#>