aggregate_time {gdalcubes}

Aggregate data cube time series to lower temporal resolution

Description

Create a proxy data cube, which applies an aggregation function over pixel time series to lower temporal resolution.

Usage

aggregate_time(cube, dt, method = "mean", fact = NULL)

Arguments

cube

source data cube

dt

character; new temporal resolution, datetime period string, e.g. "P1M"

method

aggregation method, one of "mean", "min", "max", "median", "count", "sum", "prod", "var", and "sd"

fact

simple integer factor defining how many cells become aggregated to a single new cell, can be used instead of dt

Details

This function can be used to aggregate time series to lower resolution or to regularize a data cube with irregular (labeled) time axis. It is possible to change the unit of the temporal resolution (e.g. to create monthly composites from daily images). The size of the cube may be expanded automatically if the original temporal extent is not divisible by the new temporal size of pixels.

Note

This function returns a proxy object, i.e., it will not start any computations besides deriving the shape of the result.

Examples

# create image collection from example Landsat data only 
# if not already done in other examples
if (!file.exists(file.path(tempdir(), "L8.db"))) {
  L8_files <- list.files(system.file("L8NY18", package = "gdalcubes"),
                         ".TIF", recursive = TRUE, full.names = TRUE)
  create_image_collection(L8_files, "L8_L1TP", file.path(tempdir(), "L8.db"), quiet = TRUE) 
}
L8.col = image_collection(file.path(tempdir(), "L8.db"))
v = cube_view(extent=list(left=388941.2, right=766552.4, 
              bottom=4345299, top=4744931, t0="2018-01", t1="2018-12"),
              srs="EPSG:32618", nx = 497, ny=526, dt="P3M", aggregation = "median")
L8.cube = raster_cube(L8.col, v, mask=image_mask("BQA", bits=4, values=16))
L8.rgb = select_bands(L8.cube, c("B02", "B03", "B04"))
L8.two_monthly = aggregate_time(L8.rgb, "P6M", "min")
L8.two_monthly


plot(L8.two_monthly, rgb=3:1, zlim=c(5000,12000))