tests/testthat/test_mkinpredict_SFO_SFO.R

# $Id: jranke $

# Copyright (C) 2012-2015 Johannes Ranke
# Contact: [email protected]

# 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", quiet = TRUE)
  SFO_SFO.2 <- mkinmod(parent = list(type = "SFO", to = "m1"),
         m1 = list(type = "SFO"), use_of_ff = "max", quiet = TRUE)

  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)))

})

Try the mkin package in your browser

Any scripts or data that you put into this service are public.

mkin documentation built on July 19, 2018, 9:03 a.m.