grid2poly: Converts a gridded map to a polygon map

In plotKML: Visualization of Spatial and Spatio-Temporal Objects in Google Earth

Description

Converts a "SpatialGridDataFrame" object to a polygon map with each available grid node represented with a polygon. To allow further export to KML, grid2poly will, by default, convert any projected coordinates to the lat-lon system (geographic coordinates in the WGS84 system).

Usage

grid2poly(obj, var.name = names(obj)[1], reproject = TRUE, 
    method = c("sp", "raster", "RSAGA")[1], tmp.file = TRUE, 
    saga_lib = "shapes_grid", saga_module = 3, silent = FALSE, ...)

Arguments

obj	"SpatialGridDataFrame" object
var.name	target variable column name
reproject	logical; reproject coordinates to lat lon system?
method	decide to convert grids to polygons either using "sp", "raster" or "RSAGA" packages
tmp.file	logical; specify whether to create a temporary file, or to actually write to the workding directory (in the case of SAGA GIS is used to convert grids)
saga_lib	string; SAGA GIS library name
saga_module	SAGA GIS module number; see ?rsaga_get_modules for more details
silent	logical; specifies whether to print the SAGA GIS output
...	additional arguments that can be parsed to the rasterToPolygons command

Details

grid2poly is not recommended for large grids (>>10e4 pixels). Consider splitting large input grids into tiles before running grid2poly. For converting large grids to polygons consider using SAGA GIS (method = "RSAGA") instead of using the default sp method.

Author(s)

Tomislav Hengl

See Also

vect2rast, raster::rasterToPolygons

Examples

data(eberg_grid)
library(sp)
coordinates(eberg_grid) <- ~x+y
gridded(eberg_grid) <- TRUE
proj4string(eberg_grid) <- CRS("+init=epsg:31467")
data(SAGA_pal)
## Not run: # compare various methods:
system.time(dem_poly <- grid2poly(eberg_grid, "DEMSRT6", method = "raster"))
system.time(dem_poly <- grid2poly(eberg_grid, "DEMSRT6", method = "sp"))
system.time(dem_poly <- grid2poly(eberg_grid, "DEMSRT6", method = "RSAGA"))
## plotting large polygons in R -> not a good idea
# spplot(dem_poly, col.regions = SAGA_pal[[1]])
## visualize the data in Google Earth:
kml(dem_poly, colour_scale = SAGA_pal[[1]], colour = DEMSRT6, kmz = TRUE)

## End(Not run)

plotKML documentation built on April 27, 2021, 3:01 p.m.