View source: R/solution_statistics.R
solution_statistics | R Documentation |
Calculate statistics describing a solution to a project prioritization
problem()
.
solution_statistics(x, solution)
x |
project prioritization |
solution |
|
A tibble::tibble()
table containing the following
columns:
"cost"
numeric
cost of each solution.
"obj"
numeric
objective value for each solution.
This is calculated using the objective function defined for the
argument to x
.
x$project_names()
numeric
column for each
project indicating if it was completely funded (with a value of 1)
or not (with a value of 0).
x$feature_names()
numeric
column for each
feature indicating the probability that it will persist into
the future given each solution.
objectives, replacement_costs()
,
project_cost_effectiveness()
.
# load data data(sim_projects, sim_features, sim_actions) # print project data print(sim_projects) # print action data print(sim_features) # print feature data print(sim_actions) # build problem p <- problem(sim_projects, sim_actions, sim_features, "name", "success", "name", "cost", "name") %>% add_max_richness_objective(budget = 400) %>% add_feature_weights("weight") %>% add_binary_decisions() # print problem print(p) # create a table with some solutions solutions <- data.frame(F1_action = c(0, 1, 1), F2_action = c(0, 1, 0), F3_action = c(0, 1, 1), F4_action = c(0, 1, 0), F5_action = c(0, 1, 1), baseline_action = c(1, 1, 1)) # print the solutions # the first solution only has the baseline action funded # the second solution has every action funded # the third solution has only some actions funded print(solutions) # calculate statistics solution_statistics(p, solutions)
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