# $Id: jranke $
# Copyright (C) 2012-2015 Johannes Ranke
# Contact: jranke@uni-bremen.de
# This file is part of the R package mkin
# mkin is free software: you can redistribute it and/or modify it under the
# terms of the GNU General Public License as published by the Free Software
# Foundation, either version 3 of the License, or (at your option) any later
# version.
# This program is distributed in the hope that it will be useful, but WITHOUT
# ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS
# FOR A PARTICULAR PURPOSE. See the GNU General Public License for more
# details.
# You should have received a copy of the GNU General Public License along with
# this program. If not, see <http://www.gnu.org/licenses/>
# This was migrated from an RUnit test in inst/unitTests/runit.mkinpredict.R
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 = list(type = "SFO", to = "m1"),
m1 = list(type = "SFO"), use_of_ff = "min")
SFO_SFO.2 <- mkinmod(parent = list(type = "SFO", to = "m1"),
m1 = list(type = "SFO"), use_of_ff = "max")
ot = seq(0, 100, by = 1)
r.1.e <- subset(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(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(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(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)))
})
