# LUCID - three omics, binary outcome test_that("check prediction of LUCID with binary outcome (K = 2,2,2)", { i <- 1008 set.seed(i) G <- matrix(rnorm(500), nrow = 100) Z1 <- matrix(rnorm(1000),nrow = 100) Z2 <- matrix(rnorm(1000), nrow = 100) Z3 <- matrix(rnorm(1000), nrow = 100) Z <- list(Z1 = Z1, Z2 = Z2, Z3 = Z3) Y <- rbinom(n=100, size =1, prob =0.65) CoY <- matrix(rnorm(200), nrow = 100) CoG <- matrix(rnorm(200), nrow = 100) # i <- sample(1:2000, 1) # cat(paste("test1 - seed =", i, "\n")) invisible(capture.output(fit1 <- estimate_lucid(G = G, Z = Z, Y = Y, K = c(2, 2, 2),CoG = CoG, CoY = CoG, lucid_model = "parallel", family = "binary", init_omic.data.model = "VVV", seed = i, useY = TRUE))) set.seed(i+1000) n_G <- matrix(rnorm(500), nrow = 100) n_Z1 <- matrix(rnorm(1000),nrow = 100) n_Z2 <- matrix(rnorm(1000), nrow = 100) n_Z3 <- matrix(rnorm(1000), nrow = 100) n_Z <- list(Z1 = n_Z1, Z2 = n_Z2, Z3 = n_Z3) n_Y <- rbinom(n=100, size =1, prob =0.25) n_CoY <- matrix(rnorm(200), nrow = 100) n_CoG <- matrix(rnorm(200), nrow = 100) #use training data pred1 <- predict_lucid(model = fit1, lucid_model = "parallel", G = G, Z = Z, Y = Y, g_computation = FALSE, CoG = CoG, CoY = CoY, response = TRUE) expect_equal(fit1$inclusion.p, pred1$inclusion.p, tolerance = 0.05) expect_equal(class(pred1$pred.x), "list") expect_equal(max(pred1$pred.y), 1) expect_equal(mean(pred1$pred.y), 0.99) expect_equal(mean(pred1$inclusion.p[[1]]), 0.5) #use new data pred2 <- predict_lucid(model = fit1, lucid_model = "parallel", G = n_G, Z = n_Z, Y = n_Y, CoG = n_CoG, CoY = n_CoY, response = TRUE) expect_equal(class(pred2$pred.x), "list") expect_equal(max(pred2$pred.y), 1) expect_equal(mean(pred2$pred.y), 1) expect_equal(mean(pred2$inclusion.p[[1]]), 0.5) #new data not using Y, and response = FALSE pred3 <- predict_lucid(model = fit1, lucid_model = "parallel", G = n_G, Z = n_Z, Y = NULL, CoG = n_CoG, CoY = n_CoY, response = FALSE) expect_equal(class(pred3$pred.x), "list") expect_equal(max(pred3$pred.y), 0.9124988, tolerance = 0.05) expect_equal(mean(pred3$pred.y), 0.7277153, tolerance = 0.05) expect_equal(mean(pred3$inclusion.p[[1]]), 0.5) })