context("Model predictions with mkinpredict") test_that("Variants of model predictions for SFO_SFO model give equivalent results", { # Check model specification and solution types for SFO_SFO # Relative Tolerance is 0.01% # Do not use time 0, as eigenvalue based solution does not give 0 at time 0 for metabolites # and relative tolerance is thus not met tol = 0.01 SFO_SFO.1 <- mkinmod(parent = mkinsub("SFO", to = "m1"), m1 = mkinsub("SFO"), use_of_ff = "min", quiet = TRUE) SFO_SFO.2 <- mkinmod(parent = mkinsub("SFO", to = "m1"), m1 = mkinsub("SFO"), use_of_ff = "max", quiet = TRUE) ot = seq(0, 100, by = 1) r.1.e <- subset(as.data.frame(mkinpredict(SFO_SFO.1, c(k_parent_m1 = 0.1, k_parent_sink = 0.1, k_m1_sink = 0.1), c(parent = 100, m1 = 0), ot, solution_type = "eigen")), time %in% c(1, 10, 50, 100)) r.1.d <- subset(as.data.frame(mkinpredict(SFO_SFO.1, c(k_parent_m1 = 0.1, k_parent_sink = 0.1, k_m1_sink = 0.1), c(parent = 100, m1 = 0), ot, solution_type = "deSolve")), time %in% c(1, 10, 50, 100)) r.2.e <- subset(as.data.frame(mkinpredict(SFO_SFO.2, c(k_parent = 0.2, f_parent_to_m1 = 0.5, k_m1 = 0.1), c(parent = 100, m1 = 0), ot, solution_type = "eigen")), time %in% c(1, 10, 50, 100)) r.2.d <- subset(as.data.frame(mkinpredict(SFO_SFO.2, c(k_parent = 0.2, f_parent_to_m1 = 0.5, k_m1 = 0.1), c(parent = 100, m1 = 0), ot, solution_type = "deSolve")), time %in% c(1, 10, 50, 100)) # Compare eigen and deSolve for minimum use of formation fractions dev.1.e_d.percent = 100 * (r.1.e[-1] - r.1.d[-1])/r.1.e[-1] dev.1.e_d.percent = as.numeric(unlist((dev.1.e_d.percent))) dev.1.e_d.percent = ifelse(is.na(dev.1.e_d.percent), 0, dev.1.e_d.percent) expect_equivalent(dev.1.e_d.percent < tol, rep(TRUE, length(dev.1.e_d.percent))) # Compare eigen and deSolve for maximum use of formation fractions dev.2.e_d.percent = 100 * (r.1.e[-1] - r.1.d[-1])/r.1.e[-1] dev.2.e_d.percent = as.numeric(unlist((dev.2.e_d.percent))) dev.2.e_d.percent = ifelse(is.na(dev.2.e_d.percent), 0, dev.2.e_d.percent) expect_equivalent(dev.2.e_d.percent < tol, rep(TRUE, length(dev.2.e_d.percent))) # Compare minimum and maximum use of formation fractions dev.1_2.e.percent = 100 * (r.1.e[-1] - r.2.e[-1])/r.1.e[-1] dev.1_2.e.percent = as.numeric(unlist((dev.1_2.e.percent))) dev.1_2.e.percent = ifelse(is.na(dev.1_2.e.percent), 0, dev.1_2.e.percent) expect_equivalent(dev.1_2.e.percent < tol, rep(TRUE, length(dev.1_2.e.percent))) })