Aligning with CRAN suggestions.

DESCRIPTION

@@ -17,11 +17,16 @@ Authors@R: c(person("Tomasz", "Zoltak",
 Maintainer: Tomasz Zoltak <[email protected]>
 Description: Package allows to generate simulated datasets using algorithms
     that mimic different response styles to survey questions using: 1) IRTree
-    approach, 2) (G)PCM (and rating scale version of a partial credit model)
-    approach, 3) user provided function that (with some probability) chooses
-    response using information about previous responses. This allows to cover
-    wide range of potential response styles like: extreme and middle (ERS, MRS),
-    acquiesce (ARS) and also careless/inattentive responding (CR, IR).
+    approach (Bockenholt (2012) <doi:10.1037/a0028111>,
+    (2017) <doi:10.1037/met0000106>), 2) (G)PCM (and rating scale version of
+    a partial credit model) random-thresholds approach (Falk & Cai (2016)
+    <doi:10.1037/met0000059>; Henninger & Meiser (2020a)
+    <doi:10.1037/met0000249>, (2020b) <doi:10.1037/met0000268>; Plieninger
+    (2017) <doi:10.1177/0013164416636655>), 3) user provided function that (with
+    some probability) chooses response using information about previous
+    responses. This allows to cover wide range of potential response styles
+    like: extreme and middle (ERS, MRS), acquiesce (ARS) and also
+    careless/inattentive responding (CR, IR).
 License: MIT + file LICENSE
 Depends: R (>= 4.0.0)
 Suggests:

R/log_normal_pars.R

@@ -8,6 +8,10 @@
 #' @param sdlog standard deviation of the distribution on the log scale
 #' @param m mean (expected value) of the distribution on the exponential scale
 #' @param sd standard deviation of the distribution on the exponential scale
+#' @return \code{lnorm_mean()} and \code{lnorm_sd()} return numeric vectors;
+#' \code{find_pars_lnorm()} returns named numeric vector with elements named
+#' \emph{meanlog} describing means and elements named \emph{sdlog} describing
+#' standard deviations.
 #' @examples
 #' # actual expected value of the log-normal distribution with the log scale
 #' # parameters of 3 and 1.5 respectively

cran-comments.md

@@ -1,4 +1,16 @@
-There was 1 NOTE:
+# Resubmission
+
+This is a resubmission. In this version I have:
+
+- Added references to the documents describing the methods that are used in the package in the DESCRIPTION file.
+
+- Added missing \value to log-normal.Rd.
+
+- Removed `rm(list = ls())` usage from tests.
+
+# NOTES
+
+There is 1 NOTE:
 
 Possibly mis-spelled words in DESCRIPTION:
      ARS (25:16)

tests/testthat/test_expand_responses.R

@@ -15,4 +15,3 @@ test_that("", {
                                             paste0("e_", colnames(responses)))))
   expect_identical(respExp, pattern)
 })
-rm(list = ls())

tests/testthat/test_sequential_mae5.R

@@ -56,5 +56,3 @@ test_that("Item parameters of sequential M, A, E RS (with 5-point scale) recover
 #         dm = mean((intercepts[, 1] - estItemPars[, 4])^2),
 #         da = mean((intercepts[, 2] - estItemPars[, 5])^2),
 #         de = mean((intercepts[, 3] - estItemPars[, 6])^2)), 3)
-
-rm(list = ls())

tests/testthat/test_sequential_mae6.R

@@ -63,5 +63,3 @@ test_that("Item parameters of sequential M, A, E RS (with 6-point scale) recover
 #         da1 = mean((intercepts[, 2] - estItemPars[, 6])^2),
 #         da2 = mean((intercepts[, 3] - estItemPars[, 7])^2),
 #         de = mean((intercepts[, 4] - estItemPars[, 8])^2)), 3)
-
-rm(list = ls())

tests/testthat/test_simultaneous_aem.R

@@ -35,6 +35,7 @@ mSml <- suppressMessages(mirt(resp,
                          gpcm_mats = lapply(1:ncol(resp), function(x) sM),
                          method = "EM", TOL = 0.1, verbose = FALSE))
 estItemPars <- coef(mSml, simplify = TRUE)$items
+intercepts <- intercepts[, -1]
 test_that("Item parameters of simultaneous A, E, M RS (with 5-point scale) recovers in estimation with reasonable MSEs.", {
   expect_lt(mean((slopes[, 1] - estItemPars[, 1])^2), 0.2)
   expect_lt(mean((slopes[, 2] - estItemPars[, 2])^2), 0.02)
@@ -62,5 +63,3 @@ test_that("Item parameters of simultaneous A, E, M RS (with 5-point scale) recov
 #         d2 = mean((intercepts[, 2] - estItemPars[, 27])^2),
 #         d3 = mean((intercepts[, 3] - estItemPars[, 28])^2),
 #         d4 = mean((intercepts[, 4] - estItemPars[, 29])^2)), 3)
-
-rm(list = ls())