tests/testthat/test-kinshipLR.R

In forrel: Forensic Pedigree Analysis and Relatedness Inference

test_that("kinshipLR() catches input errors", {
  s = singleton(1)
  expect_error(kinshipLR(s), "The input must contain at least two pedigrees")
  expect_error(kinshipLR(list(s)), "The input must contain at least two pedigrees")
  expect_error(kinshipLR(list(s, 1)), "The input is not a list of pedigrees")
  expect_error(kinshipLR(list(s, s)), "None of the pedigrees")
  expect_error(kinshipLR(s, s), "None of the pedigrees")
  expect_error(kinshipLR(s, s, source = 1), "The source pedigree has no attached markers")
  expect_error(kinshipLR(list(s, s), source = 1), "The source pedigree has no attached markers")

  s1 = addMarker(s)
  expect_error(kinshipLR(s1, s1, ref = 3), "Invalid value for `ref`")
  expect_error(kinshipLR(s1, s1, ref = 0), "Invalid value for `ref`")
  expect_error(kinshipLR(s1, s1, ref = "unrel"), "Invalid value for `ref`")

  expect_error(kinshipLR(a=s, a=s1), "Duplicated hypothesis name")
  expect_error(kinshipLR(s, H1=s1), "Duplicated hypothesis name")

  s2 = addMarker(s1)
  expect_error(kinshipLR(s1, s2), "The pedigrees have different")
})

test_that("kinshipLR() computes correctly in paternity case", {
  H1 = nuclearPed(fa = "fa", child = "ch") |>
    addMarker(name = "M1", fa = "A/A", ch = "A/A", afreq = c(A=0.05, B=0.95)) |>
    addMarker(name = "M2", fa = "A/A", ch = "A/A", afreq = c(A=0.1, B=0.9))
  H2 = list(singleton("fa"), singleton("ch"))

  expect_equal(kinshipLR(H1, H2, source = 1)$LRtotal[[1]], 20*10)
  expect_equal(kinshipLR(H1, H2, markers = "M1")$LRtotal[[1]], 20)
  expect_equal(kinshipLR(H1, H2, markers = -1, source = 1)$LRtotal[[1]], 10)
})