Manipulate ahp problem by code

script.R

library(ahp)
library(data.tree)
vacation <- Load("vacation.ahp")

# look at the structure

vacation


# look at some preference dfs

vacation$preferences$Dad$pairwise$preferences

# manipulate preferences by code
newprefs <- structure(list(c1 = list("Costs", "Costs", "Fun"), c2 = list(
  "Fun", "Spa", "Spa"), preference = c(7, 2, 3)), .Names = c("c1", 
                                                             "c2", "preference"), row.names = c(NA, -3L), class = "data.frame")
vacation$preferences$Dad$pairwise$preferences <- newprefs

# reset voting powers by code (power to the children!):
vacation$`decision-makers` <- c(Dad = 0.1, Mom = 0.2, Kid = 0.7)
# ...except maybe for spa, that's still Mom's territory
vacation$Spa$`decision-makers` <- c(Dad = 0.2, Mom = 0.7, Kid = 0.1)

Calculate(vacation)


ahp::Analyze(vacation)

vacation.ahp

Version: 2.0
Alternatives: &alternatives
  Beach:
    cost: 10000
  Mountains:
    cost: 5000
Goal:
  name: Vacation
  decision-makers:
    #optional node, needed only if not all decision-makers have equal voting power
    - Dad: 0.4
    - Mom: 2/5
    - Kid: 0.2
  preferences:
    Dad:
      pairwise:
      - [Costs, Fun, 4]
      - [Costs, Spa, 9]
      - [Fun, Spa, 4]
    Mom:
      pairwise:
      - [Costs, Fun, 1/4]
      - [Costs, Spa, 1/9]
      - [Fun, Spa, 1/5]
    Kid:
      pairwise:
      - [Costs, Fun, 1/9]
      - [Costs, Spa, 1]
      - [Fun, Spa, 9]
  children:
    Costs:
      preferences:
        Dad:
          pairwiseFunction: 
            function(a1, a2) min(9, max(1/9, a2$cost/a1$cost))
        Mom:
          scoreFunction:
            function(a) 1/a$cost
        Kid:
          priority:
            - Beach: 1/2
            - Mountains: 0.5
      children: *alternatives
    Fun:
      preferences:
        Dad:
          pairwise:
            - [Beach, Mountains, 1/6]
        Mom:
          pairwise:
            - [Beach, Mountains, 2]
        Kid:
        # Often, entering pairwise preferences is lengthy, especially if you
        # have multiple alternatives. Instead, you can enter scores, i.e. rate
        # each alternative on a scale. The scale can be chosen freely. The 
        # priorities are derived as score / sum(scores)
          score:
            - Beach: 5
            - Mountains: 0
      children: *alternatives
    Spa:
      preferences:
        Dad:
          pairwise:
            - [Beach, Mountains, 2]
        Mom:
          pairwise:
            - [Beach, Mountains, 6]
        Kid:
          pairwise:
            - [Beach, Mountains, 1/2]
      children: *alternatives