construct_question_frame {ExpertChoice}

Convert from choice_sets to a question data

Description

Convert from choice_sets to a question data

Usage

construct_question_frame(
  augmented_full_factorial,
  choice_sets,
  randomise_choice_sets = TRUE
)

Arguments

augmented_full_factorial

The augmented full factorial object.

choice_sets

The choice sets list generated from one of the methods. (See Step 6 of the tutorial)

randomise_choice_sets

A binary variable indicating if the order of the choice sets should be randomised. Some methods create choice sets which have a systematic order. Randomising the order of the choice sets does not change the alternatives within the choice sets. It simply rearranges the choice_set object in a random manner.

Value

a data.frame object

Examples

#See Step 9 of Practical Introduction to ExpertChoice vignette.

# Step 1
attrshort  = list(condition = c("0", "1", "2"),
technical =c("0", "1", "2"),
provenance = c("0", "1"))

#Step 2
# ff stands for "full fatorial"
 ff  <-  full_factorial(attrshort)
 af  <-  augment_levels(ff)
# af stands for "augmented factorial"

# Step 3
# Choose a design type: Federov or Orthogonal. Here an Orthogonal one is used.
nlevels <- unlist(purrr::map(ff, function(x){length(levels(x))}))
fractional_factorial <- DoE.base::oa.design(nlevels = nlevels, columns = "min34")

# Step 4 & 5
# The functional draws out the rows from the original augmented full factorial design.
colnames(fractional_factorial) <- colnames(ff)
fractional <- search_design(ff, fractional_factorial)
# Step 5 (skipped, but important, see vignette)

# Step 6
# Two modulators c(1,1,1) and c(0,1,1) are specified.
dce_modulo <- modulo_method(
fractional,
list(c(1,1,1),c(0,1,1))
)

# Step 7 and Step 8 are very important for the design, but skipped here.

# Step 9! -- Construct a question frame to use with your study.
# Note the use of af here.
questions <- construct_question_frame(af, dce_modulo)
levels(questions$condition) <- c("bad", "good", "excellent")
levels(questions$technical) <- c("poor", "fair", "skilled")
levels(questions$provenance) <- c("none", "present")
questions