Nothing
# 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)
})
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