raster

My question is pretty similar to an unfortunately unanswered issue recently posted on Stackoverflow . I am dealing with a RasterStack object consisting of twelve layers (one for each month of the year), and I would like to replicate the layers ten times, ending up with a RasterStack consisting of 120 layers, with every 12th layer being similar (i.e., layer 1 is the same like layer 13 is the same like layer 25 and so on). For replication purposes, let's take an example from the raster package: library(raster) file <- system.file("external/test.grd", package = "raster") s <- stack(file, file,

I have a RasterBrick consisting of monthly rainfall data over 7 years, so it has 7 layers with 12 slots each: rainfall <- brick("Rainfall.tif") > rainfall class : RasterBrick dimensions : 575, 497, 285775, 7 (nrow, ncol, ncell, nlayers) resolution : 463.3127, 463.3127 (x, y) extent : 3763026, 3993292, -402618.8, -136213.9 (xmin, xmax, ymin, ymax) coord. ref. : +proj=sinu +lon_0=0 +x_0=0 +y_0=0 +a=6371007.181 +b=6371007.181 +units=m +no_defs data source : in memory names : layer.1.1, layer.2.1, layer.1.2, layer.2.2, layer.1, layer.2, layer min values : 239.6526, 499.8343, 521.0316, 617.2896,

The quadratic/cubic bézier curve code I find via google mostly works by subdividing the line into a series of points and connects them with straight lines. The rasterization happens in the line algorithm, not in the bézier one. Algorithms like Bresenham's work pixel-by-pixel to rasterize a line, and can be optimized (see Po-Han Lin's solution ). What is a quadratic bézier curve algorithm that works pixel-by-pixel like line algorithms instead of by plotting a series of points? A variation of Bresenham's Algorithm works with quadratic functions like circles, ellipses, and parabolas, so it should

I am working in C on a physics experiment, Young's interference experiment and i made a program who prints to file a huge bunch of pixels : for (i=0; i < width*width; i++) { fwrite(hue(raster_matrix[i]), 1, 3, file); } Where hue , when given a value [0..255], gives back a char * with 3 bytes, R,G,B. I would like to put a minimal header in my image file in order to make this raw file a valid image file. More concise : Switching from : offset 0000 : height * width : data } my data, 24bit RGB pixels to offset 0000 : dword : magic \ : /* ?? */ \ 0012 : dword : height } Header <--> common image

I am using {raster} to clip (or crop) a raster based on an irregular shapefile (the Amazon biome) but the output always has a rectangular extent. However, I need the output in the exact same geometry of the shapefile. Any tips? Cheers. library(raster) library(rgdal) myshp <- readOGR("Amazon.shp", layer="Amazon") e <- extent(myshp) myraster <- raster("Temperature.tif") myraster.crop <- crop(myraster, e, snap="out", filename="myoutput.tif") One option is to use raster::mask() library(maptools) ## For wrld_simpl library(raster) ## Example SpatialPolygonsDataFrame data(wrld_simpl) SPDF <- subset

I'm trying to take an image and store it in an array of 16x16 subimages. The image I am using is 512x512 pixels. However, while iterating through the loop, getSubimage() is stopped by a Raster exception. Here is the code: public class TileList extends JPanel { private static final int width = 16; //width of a tile private static final int height = width; private int col = 1; private int row = 1; private BufferedImage image; File tilesetImage = new File("image.png"); BufferedImage tileset[]; public void loadAndSplitImage (File loadImage) { try{ image = ImageIO.read(loadImage); }catch(Exception

I have a data frame in which values (l) are specified for Cartesian coordinates (x, y) as in the following minimal working example. set.seed(2013) df <- data.frame( x = rep( 0:1, each=2 ), y = rep( 0:1, 2), l = rnorm( 4 )) df # x y l # 1 0 0 -0.09202453 # 2 0 1 0.78901912 # 3 1 0 -0.66744232 # 4 1 1 1.36061149 I want to create a raster using the raster package, but my reading of the documentation has not revealed a simple method for loading data in the form that I have it into the raster cells. I've come up with a couple ways to do it using for loops, but I suspect that there's a much more