R/eval_ferrier_importance.R
eval_ferrier_importance.Rd
Calculate importance scores for planning units selected in a solution following Ferrier et al. (2000). Please note that the mathematical formulation for computing these scores needs verification, and so this functionality should be considered experimental at this point in time.
eval_ferrier_importance(x, solution) # S4 method for ConservationProblem,numeric eval_ferrier_importance(x, solution) # S4 method for ConservationProblem,matrix eval_ferrier_importance(x, solution) # S4 method for ConservationProblem,data.frame eval_ferrier_importance(x, solution) # S4 method for ConservationProblem,Spatial eval_ferrier_importance(x, solution) # S4 method for ConservationProblem,sf eval_ferrier_importance(x, solution) # S4 method for ConservationProblem,Raster eval_ferrier_importance(x, solution)
x 


solution 

A matrix
, tibble::tibble()
,
RasterLayer
, or
Spatial
object containing the scores for each
planning unit selected in the solution.
Specifically, the returned object is in the
same format (except if the planning units are a numeric
vector) as the
planning unit data in the argument to x
.
Importance scores are reported separately for each feature within each planning unit. Additionally, a total importance score is also calculated as the sum of the scores for each feature. Note that this function only works for problems with a minimum set objective and a single zone. It will throw an error for other types of problems that do not meet this specification.
Broadly speaking, the argument to solution
must be in the same format as
the planning unit data in the argument to x
.
Further details on the correct format are listed separately
for each of the different planning unit data formats:
x
has numeric
planning unitsThe argument to solution
must be a
numeric
vector with each element corresponding to a different planning
unit. It should have the same number of planning units as those
in the argument to x
. Additionally, any planning units missing
cost (NA
) values should also have missing (NA
) values in the
argument to solution
.
x
has matrix
planning unitsThe argument to solution
must be a
matrix
vector with each row corresponding to a different planning
unit, and each column correspond to a different management zone.
It should have the same number of planning units and zones
as those in the argument to x
. Additionally, any planning units
missing cost (NA
) values for a particular zone should also have a
missing (NA
) values in the argument to solution
.
x
has Raster
planning unitsThe argument to solution
be a Raster
object where different grid cells (pixels) correspond
to different planning units and layers correspond to
a different management zones. It should have the same dimensionality
(rows, columns, layers), resolution, extent, and coordinate reference
system as the planning units in the argument to x
. Additionally,
any planning units missing cost (NA
) values for a particular zone
should also have missing (NA
) values in the argument to solution
.
x
has data.frame
planning unitsThe argument to solution
must
be a data.frame
with each column corresponding to a different zone,
each row corresponding to a different planning unit, and cell values
corresponding to the solution value. This means that if a data.frame
object containing the solution also contains additional columns, then
these columns will need to be subsetted prior to using this function
(see below for example with sf::sf()
data).
Additionally, any planning units missing cost
(NA
) values for a particular zone should also have missing (NA
)
values in the argument to solution
.
x
has Spatial
planning unitsThe argument to solution
must be a Spatial
object with each column corresponding to a
different zone, each row corresponding to a different planning unit, and
cell values corresponding to the solution value. This means that if the
Spatial
object containing the solution also contains additional
columns, then these columns will need to be subsetted prior to using this
function (see below for example with sf::sf()
data).
Additionally, the argument to solution
must also have the same
coordinate reference system as the planning unit data.
Furthermore, any planning units missing cost
(NA
) values for a particular zone should also have missing (NA
)
values in the argument to solution
.
x
has sf::sf()
planning unitsThe argument to solution
must be
a sf::sf()
object with each column corresponding to a different
zone, each row corresponding to a different planning unit, and cell values
corresponding to the solution value. This means that if the
sf::sf()
object containing the solution also contains additional
columns, then these columns will need to be subsetted prior to using this
function (see below for example).
Additionally, the argument to solution
must also have the same
coordinate reference system as the planning unit data.
Furthermore, any planning units missing cost
(NA
) values for a particular zone should also have missing (NA
)
values in the argument to solution
.
Ferrier S, Pressey RL, and Barrett TW (2000) A new predictor of the irreplaceability of areas for achieving a conservation goal, its application to realworld planning, and a research agenda for further refinement. Biological Conservation, 93: 303325.
# seed seed for reproducibility set.seed(600) # load data data(sim_pu_raster, sim_features) # create minimal problem with binary decisions p1 < problem(sim_pu_raster, sim_features) %>% add_min_set_objective() %>% add_relative_targets(0.1) %>% add_binary_decisions() %>% add_default_solver(gap = 0, verbose = FALSE) # \dontrun{ # solve problem s1 < solve(p1)#> $LogToConsole #> [1] 0 #> #> $LogFile #> [1] "" #> #> $Presolve #> [1] 2 #> #> $MIPGap #> [1] 0 #> #> $TimeLimit #> [1] 2147483647 #> #> $Threads #> [1] 1 #> #> $NumericFocus #> [1] 0 #>#> class : RasterLayer #> dimensions : 10, 10, 100 (nrow, ncol, ncell) #> resolution : 0.1, 0.1 (x, y) #> extent : 0, 1, 0, 1 (xmin, xmax, ymin, ymax) #> crs : NA #> source : memory #> names : layer #> values : 0, 1 (min, max) #># calculate importance scores using Ferrier et al. 2000 method fs1 < eval_ferrier_importance(p1, s1) # print importance scores, # each planning unit has an importance score for each feature # (as indicated by the column names) and each planning unit also # has an overall total importance score (in the "total" column) print(fs1)#> class : RasterStack #> dimensions : 10, 10, 100, 6 (nrow, ncol, ncell, nlayers) #> resolution : 0.1, 0.1 (x, y) #> extent : 0, 1, 0, 1 (xmin, xmax, ymin, ymax) #> crs : NA #> names : layer.1, layer.2, layer.3, layer.4, layer.5, total #> min values : 0, 0, 0, 0, 0, 0 #> max values : 0.01579546, 0.01722246, 0.01575790, 0.01793544, 0.01585958, 0.06844167 #># create minimal problem with polygon (sf) planning units p2 < problem(sim_pu_sf, sim_features, cost_column = "cost") %>% add_min_set_objective() %>% add_relative_targets(0.05) %>% add_binary_decisions() %>% add_default_solver(gap = 0, verbose = FALSE) # solve problem s2 < solve(p2)#> $LogToConsole #> [1] 0 #> #> $LogFile #> [1] "" #> #> $Presolve #> [1] 2 #> #> $MIPGap #> [1] 0 #> #> $TimeLimit #> [1] 2147483647 #> #> $Threads #> [1] 1 #> #> $NumericFocus #> [1] 0 #>#> Simple feature collection with 90 features and 4 fields #> Geometry type: POLYGON #> Dimension: XY #> Bounding box: xmin: 0 ymin: 0 xmax: 1 ymax: 1 #> CRS: NA #> First 10 features: #> cost locked_in locked_out solution_1 geometry #> 1 215.8638 FALSE FALSE 0 POLYGON ((0 1, 0.1 1, 0.1 0... #> 2 212.7823 FALSE FALSE 0 POLYGON ((0.1 1, 0.2 1, 0.2... #> 3 207.4962 FALSE FALSE 0 POLYGON ((0.2 1, 0.3 1, 0.3... #> 4 208.9322 FALSE TRUE 0 POLYGON ((0.3 1, 0.4 1, 0.4... #> 5 214.0419 FALSE FALSE 0 POLYGON ((0.4 1, 0.5 1, 0.5... #> 6 213.7636 FALSE FALSE 0 POLYGON ((0.5 1, 0.6 1, 0.6... #> 7 210.4612 FALSE FALSE 0 POLYGON ((0.6 1, 0.7 1, 0.7... #> 8 211.0424 FALSE TRUE 0 POLYGON ((0.7 1, 0.8 1, 0.8... #> 9 210.3878 FALSE FALSE 0 POLYGON ((0.8 1, 0.9 1, 0.9... #> 10 204.3971 FALSE FALSE 0 POLYGON ((0.9 1, 1 1, 1 0.9...# calculate importance scores fs2 < eval_ferrier_importance(p2, s2[, "solution_1"]) # plot importance scores plot(fs2, main = "Ferrier scores")# }