# cluster exercise tol <- 1e-3 data(sikadeer) cu <- sikadeer_units[, 3] cu <- (cu[1]*cu[2])/cu[3] context("sika deer") #test_that("no truncation", { # Effort : 5.000000 # # samples : 37 # Width : 200.0000 # # observations: 1921 # # Model 4 # Half-normal key, k(y) = Exp(-y**2/(2*A(1)**2)) # Cosine adjustments of order(s) : 2, 3, 4 # # # Point Standard Percent Coef. 95% Percent # Parameter Estimate Error of Variation Confidence Interval # --------- ----------- ----------- -------------- ---------------------- # D 37.403 8.0972 21.65 24.355 57.441 # N 3400.0 736.04 21.65 2214.0 5221.0 # --------- ----------- ----------- -------------- ---------------------- # # Measurement Units # --------------------------------- # Density: Numbers/Sq. kilometers # ESW: centimeters # # Component Percentages of Var(D) # ------------------------------- # Detection probability : 5.1 # Encounter rate : 81.4 # Decay rate : 13.6 #}) test_that("10% truncation", { dat <- unflatten(sikadeer) dat <- dat$data # fit the exact function fitted by Distance result <- ddf(dsmodel = ~mcds(key = "hn", formula = ~1, adj.series="cos", adj.order=2), data = dat, method = "ds", meta.data = list(width = 175), control=list(initial=list(adjustment=0.08493070, scale=log(121.1144)), nofit=TRUE)) df <- result # multipliers mult <- list(creation = data.frame(rate = 25, SE = 0), decay = data.frame(rate = 163, SE = 13)) # weight by effort since we have repeats expect_warning(deer_ests <- dht2(df, flatfile=sikadeer, strat_formula=~Region.Label, convert_units=cu, multipliers=mult, stratification="effort_sum", total_area=13.9), "One or more strata have only one transect, cannot calculate empirical encounter rate variance") deer_ests$ER_CV[is.na(deer_ests$ER_CV)] <- 0 expect_equal(deer_ests$ER_CV[-nrow(deer_ests)], c(16.75, 24.13, 22.56, 46.95, 0, 49.61, 0, 0)/100, tol=tol) expect_equal(deer_ests$Abundance, c(1027.0, 383.00, 34.000, 29.000, 210.00, 126.00, 17.000, 69.000, 497), tol=1e-2) expect_equal(deer_ests$Abundance_CV, c(18.95, 25.71, 24.24, 47.78, 8.87, 50.39, 8.87, 8.87, 15.79)/100, tol=tol) expect_equal(deer_ests$df, c(19.18, 11.88, 2.74, 4.26, 47281.58, 2.07, 47281.58, 47281.58, 20.22), tol=tol) expect_equal(deer_ests$LCI, c(694, 220, 15, 9, 176, 18, 15, 58, 358), tol=1e-2) expect_equal(deer_ests$UCI, c(1521, 666, 76, 99, 249, 867, 21, 83, 689), tol=tol) })