Evaluate number of planning units selected.

Calculate the number of planning units selected within a solution to a conservation planning problem().

eval_n_summary(x, solution)

# S3 method for default
eval_n_summary(x, solution)

# S3 method for ConservationProblem
eval_n_summary(x, solution)

Arguments

x	problem() (i.e. ConservationProblem) object.
solution	numeric, matrix, data.frame, Raster, Spatial, or sf::sf() object. The argument should be in the same format as the planning unit cost data in the argument to x. See the Solution format section for more information.

Value

tibble::tibble() object containing the number of planning units selected within a solution. It contains the following columns:

summary

character description of the summary statistic. The statistic associated with the "overall" value in this column is calculated using the entire solution (including all management zones if there are multiple zones). If multiple management zones are present, then summary statistics are also provided for each zone separately (indicated using zone names).

n

numeric number of selected planning units.

Details

This summary statistic is calculated as the sum of the values in the solution. As a consequence, this metric can produce a non-integer value (e.g. 4.3) for solutions containing proportion values (e.g. generated by solving a problem() built using the add_proportion_decisions() function).

Solution format

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 units

The 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 units

The 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 units

The 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 units

The 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 units

The 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 units

The 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.

See also

problem(), summaries.

Examples

# \dontrun{
# set seed for reproducibility
set.seed(500)

# load data
data(sim_pu_raster, sim_pu_sf, sim_features,
     sim_pu_zones_sf, sim_features_zones)

# build minimal conservation problem with raster data
p1 <- problem(sim_pu_raster, sim_features) %>%
      add_min_set_objective() %>%
      add_relative_targets(0.1) %>%
      add_binary_decisions() %>%
      add_default_solver(verbose = FALSE)

# solve the problem
s1 <- solve(p1)
#> $LogToConsole
#> [1] 0
#> 
#> $LogFile
#> [1] ""
#> 
#> $Presolve
#> [1] 2
#> 
#> $MIPGap
#> [1] 0.1
#> 
#> $TimeLimit
#> [1] 2147483647
#> 
#> $Threads
#> [1] 1
#> 
#> $NumericFocus
#> [1] 0
#> 

# print solution
print(s1)
#> 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)
#> 

# plot solution
plot(s1, main = "solution", axes = FALSE, box = FALSE)

# calculate number of selected planning units within solution
r1 <- eval_n_summary(p1, s1)
print(r1)
#> # A tibble: 1 x 2
#>   summary  cost
#>   <chr>   <dbl>
#> 1 overall    10

# build minimal conservation problem with polygon (sf) data
p2 <- problem(sim_pu_sf, sim_features, cost_column = "cost") %>%
      add_min_set_objective() %>%
      add_relative_targets(0.1) %>%
      add_binary_decisions() %>%
      add_default_solver(verbose = FALSE)

# solve the problem
s2 <- solve(p2)
#> $LogToConsole
#> [1] 0
#> 
#> $LogFile
#> [1] ""
#> 
#> $Presolve
#> [1] 2
#> 
#> $MIPGap
#> [1] 0.1
#> 
#> $TimeLimit
#> [1] 2147483647
#> 
#> $Threads
#> [1] 1
#> 
#> $NumericFocus
#> [1] 0
#> 

# plot solution
plot(s2[, "solution_1"])

# print first six rows of the attribute table
print(head(s2))
#> Simple feature collection with 6 features and 4 fields
#> Geometry type: POLYGON
#> Dimension:     XY
#> Bounding box:  xmin: 0 ymin: 0.9 xmax: 0.6 ymax: 1
#> CRS:           NA
#>       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...

# calculate number of selected planning units within solution
r2 <- eval_n_summary(p2, s2[, "solution_1"])
print(r2)
#> # A tibble: 1 x 2
#>   summary  cost
#>   <chr>   <dbl>
#> 1 overall     9

# manually calculate number of selected planning units
r2_manual <- sum(s2$solution, na.rm = TRUE)
print(r2_manual)
#> [1] 9

# build multi-zone conservation problem with polygon (sf) data
p3 <- problem(sim_pu_zones_sf, sim_features_zones,
              cost_column = c("cost_1", "cost_2", "cost_3")) %>%
      add_min_set_objective() %>%
      add_relative_targets(matrix(runif(15, 0.1, 0.2), nrow = 5,
                                  ncol = 3)) %>%
      add_binary_decisions() %>%
      add_default_solver(verbose = FALSE)

# solve the problem
s3 <- solve(p3)
#> $LogToConsole
#> [1] 0
#> 
#> $LogFile
#> [1] ""
#> 
#> $Presolve
#> [1] 2
#> 
#> $MIPGap
#> [1] 0.1
#> 
#> $TimeLimit
#> [1] 2147483647
#> 
#> $Threads
#> [1] 1
#> 
#> $NumericFocus
#> [1] 0
#> 

# print first six rows of the attribute table
print(head(s3))
#> Simple feature collection with 6 features and 9 fields
#> Geometry type: POLYGON
#> Dimension:     XY
#> Bounding box:  xmin: 0 ymin: 0.9 xmax: 0.6 ymax: 1
#> CRS:           NA
#>     cost_1   cost_2   cost_3 locked_1 locked_2 locked_3 solution_1_zone_1
#> 1 215.8638 183.3344 205.4113    FALSE    FALSE    FALSE                 0
#> 2 212.7823 189.4978 209.6404    FALSE    FALSE    FALSE                 0
#> 3 207.4962 193.6007 215.4212     TRUE    FALSE    FALSE                 0
#> 4 208.9322 197.5897 218.5241    FALSE    FALSE    FALSE                 0
#> 5 214.0419 199.8033 220.7100    FALSE    FALSE    FALSE                 0
#> 6 213.7636 203.1867 224.6809    FALSE    FALSE    FALSE                 0
#>   solution_1_zone_2 solution_1_zone_3                       geometry
#> 1                 0                 1 POLYGON ((0 1, 0.1 1, 0.1 0...
#> 2                 0                 0 POLYGON ((0.1 1, 0.2 1, 0.2...
#> 3                 0                 0 POLYGON ((0.2 1, 0.3 1, 0.3...
#> 4                 0                 0 POLYGON ((0.3 1, 0.4 1, 0.4...
#> 5                 0                 0 POLYGON ((0.4 1, 0.5 1, 0.5...
#> 6                 1                 0 POLYGON ((0.5 1, 0.6 1, 0.6...

# create new column representing the zone id that each planning unit
# was allocated to in the solution
s3$solution <- category_vector(
  s3[, c("solution_1_zone_1", "solution_1_zone_2", "solution_1_zone_3")])
s3$solution <- factor(s3$solution)

# plot solution
plot(s3[, "solution"])

# calculate number of selected planning units within solution
r3 <- eval_n_summary(
  p3, s3[, c("solution_1_zone_1", "solution_1_zone_2", "solution_1_zone_3")])
print(r3)
#> # A tibble: 4 x 2
#>   summary  cost
#>   <chr>   <dbl>
#> 1 overall    50
#> 2 zone_1     17
#> 3 zone_2     17
#> 4 zone_3     16
# }