tests/testthat/test_z_utilities.R

In benja0x40/Barbouille: Visualisation of empirical distributions

# > utilities ==================================================================
context("utilities")

# + FiniteValues ---------------------------------------------------------------
test_that("FiniteValues", {
  tst <- log(c(0, NA, 1, Inf))
  res <- c(FALSE, FALSE, TRUE, FALSE)
  expect_equal(FiniteValues(tst), res)
  expect_equal(FiniteValues(cbind(1, tst)), res)
  expect_equal(FiniteValues(cbind(tst, 1)), res)
  expect_equal(FiniteValues(cbind(1, tst, 1)), res)
  expect_equal(FiniteValues(cbind(tst, tst)), res)
  expect_equal(FiniteValues(matrix(tst, 4, 1)), res)
  expect_equal(FiniteValues(matrix(tst, 4, 4)), res)
})

# + RankScore -------------------------------------------------------------------
test_that("RankScore", {
  expect_equal(RankScore(1), 0.5)
  expect_equal(RankScore(1:2), c(0.25, 0.75))
  expect_equal(RankScore(2:1), c(0.75, 0.25))
  expect_equal(range(RankScore(1:10)), c(0.05, 0.95))

  m <- r <- matrix(1, 5, 2, dimnames = list(NULL, c("x", "y")))
  r[] <- 0.5
  expect_identical(RankScore(m), r)
})

# + S01 ------------------------------------------------------------------------
test_that("S01", {
  expect_equal(range(S01(0:10)), c(0, 1))
  expect_equal(range(S01(-5:5)), c(0, 1))
})

# + SX2Y -----------------------------------------------------------------------
test_that("SX2Y", {
  a <- 0:10
  b <- -5:5
  expect_equal(range(SX2Y(a, b)), range(b))
  expect_equal(range(SX2Y(b, a)), range(a))
})

# + resolve.legend.position -----------------------------------------------------------------------
test_that("resolve.legend.position", {

  expect_error(resolve.legend.position(-1))
  expect_error(resolve.legend.position(10))
  expect_error(resolve.legend.position("A_Non_Existent_Position"))

  lst <- c(
    "bottomleft", "bottom", "bottomright",
    "left", "center", "right",
    "topleft", "top" ,"topright"
  )
  for(i in 1:9) {
    p <- resolve.legend.position(i)
    r <- resolve.legend.position(lst[i])
    expect_identical(colnames(p), c("id", "name", "x", "y"))
    expect_true(p$x >= -1 & p$x <= 1)
    expect_true(p$y >= -1 & p$y <= 1)
    expect_identical(p, r)
  }

  lst <- c(
    "bl", "b", "br", "l", "c", "r", "tl", "t", "tr"
  )
  for(i in 1:9) {
    p <- resolve.legend.position(i)
    r <- resolve.legend.position(lst[i])
    expect_identical(p, r)
  }

})