R Under development (unstable) (2024-08-23 r87049 ucrt) -- "Unsuffered Consequences" Copyright (C) 2024 The R Foundation for Statistical Computing Platform: x86_64-w64-mingw32/x64 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. > > library("mlt") Loading required package: basefun Loading required package: variables > > set.seed(29) > n <- 1000 > y <- rnorm(n, 2, 1.5) > d <- data.frame(y = y) > > lin <- polynomial_basis(numeric_var("y", support = range(y)), coef = c(1, 1), ci = c(-Inf, 0)) > m <- ctm(lin) > > o <- mlt(m, data = d) > > s2ml <- sqrt(var(y) * (n - 1) / n) > 1 / coef(o)[2] - s2ml y 6.498866e-08 > > -coef(o)[1] / coef(o)[2] - mean(y) (Intercept) -2.102483e-07 > > x <- runif(n, max = 2 * pi) > y <- rnorm(n, sin(x), .25) > d <- data.frame(y = y, x = x) > > plot(x, y) > > Bb <- Bernstein_basis(numeric_var("x", support = c(0, 2*pi)), order = 10, ui = "zero") > m <- ctm(lin, shift = Bb) Warning message: In c.basis(bresponse = function (data, deriv = 0L) : more than one basis contains an intercept term > > o <- mlt(m, data = d) > 1 / coef(o)[2] y 0.2612713 > p <- predict(Bb, + newdata = data.frame(x = x), coef = coef(o)[-(1:2)]) > plot(x, y) > lines(sort(x), -p[order(x)] / coef(o)[2], lwd = 2, col = "red") > > x <- runif(n, max = 2 * pi) > y <- rnorm(n, 2, 1.1 + sin(x) / 2) > d <- data.frame(y = y, x = x) > > plot(x, y) > > Bb <- Bernstein_basis(numeric_var("x", support = c(0, 2*pi)), order = 10) > m <- ctm(lin, interacting = Bb) > > o <- mlt(m, data = d) > > nd <- data.frame(x = sort(x)) > layout(matrix(1:2, nr = 2)) > tmp <- matrix(coef(o), nrow = 2) > plot(nd$x, predict(Bb, nd, coef = tmp[1,])) > plot(nd$x, predict(Bb, nd, coef = tmp[2,])) > lines(nd$x, 1.1 + sin(nd$x) / 2, col = "red") > > proc.time() user system elapsed 2.42 0.42 2.81