问题
I'm using Gaussian Process model for prediction, and I'm now at the point where I need to use Grid file based on the coordinates I have in my data but I don't have one and I don't know how to create it.
I followed the post on this link , but it shows the grid on Pennsylvania not Chicago where my data coordinates located!
So I'm confused which will be the ideal way to create grid file including the other columns in the data.
station <- data.frame(lat = c(41.997946, 41.960669, 41.960669, 41.960669,41.909269,41.931841,41.909269,41.910561,41.866129,41.866129), long = c(-87.654561, -87.747456, -87.67459, -87.646438,-87.747456,-87.67459,-87.67459,-87.619112,-87.747456,-87.691617),station = 1:10)
station
lat long station
1 41.99795 -87.65456 1
2 41.96067 -87.74746 2
3 41.96067 -87.67459 3
4 41.96067 -87.64644 4
5 41.90927 -87.74746 5
6 41.93184 -87.67459 6
7 41.90927 -87.67459 7
8 41.91056 -87.61911 8
9 41.86613 -87.74746 9
10 41.86613 -87.69162 10
The data include more columns such, Hour, Day, Moths, Year, Speed, and these observations are for the 10 locations over 2 months period, but I only put the coordinates here to get an idea how to create the grid.
Here's my steps in creating the grid following the link above:
# Set the projection. They were latitude and longitude, so use WGS84 long-lat projection
proj4string(station) <- CRS("+init=epsg:4326")
# View the station location using the mapview function
mapview(station)
#3. Determine the origin
# Set the origin
ori_t <- SpatialPoints(cbind(-87.67459, 41.99795), proj4string = CRS("+init=epsg:4326"))
# Convert the projection of ori
# Use EPSG: 3857 (Spherical Mercator)
ori_t <- spTransform(ori, CRSobj = CRS("+init=epsg:3857"))
coordinates(ori_t)
#ori_t <- spTransform(ori, CRSobj = CRS("+init=epsg:3857"))
#coordinates(ori_t)
# The origin has been rounded to the nearest 100
x_ori <- round(coordinates(ori_t)[1, 1]/100) * 100
y_ori <- round(coordinates(ori_t)[1, 2]/100) * 100
# Define how many cells for x and y axis
x_cell <- 250
y_cell <- 200
# Define the resolution to be 1000 meters
cell_size <- 1000
# Create the extent
ext <- extent(x_ori, x_ori + (x_cell * cell_size), y_ori, y_ori + (y_cell * cell_size))
# Initialize a raster layer
ras <- raster(ext)
# Set the resolution to be
res(ras) <- c(cell_size, cell_size)
ras[] <- 0
# Project the raster
projection(ras) <- CRS("+init=epsg:3857")
# Create interactive map
mapview(station) + mapview(ras)
But when I view the map, the grid is located in Pennsylvania area not Chicago! Do you have an idea why? I picked for my lat : 41.99795, and my long:-87.67459 , and when I put them on Google map, it shows Chicago area , but not showing the same on R!!
# Convert to spatial pixel
st_grid <- rasterToPoints(ras, spatial = TRUE)
gridded(st_grid) <- TRUE
st_grid <- as(st_grid, "SpatialPixels")
Also, when I save the grid file, how can I clue the other columns with the grid coordinates? Because it's only shows the new long and lat columns
write.csv(st_grid, file = "st_grid.csv")
回答1:
I'm not sure what happened with your code, but it seems like your origin was set incorrectly. I've updated the code above to produce a grid over Chicago. I've picked a random starting point from Google Maps and modified x_cell and y_cell to produce a reasonably sized map over the city.
library(sp)
library(rgdal)
library(raster)
library(leaflet)
library(mapview)
station <- data.frame(lat = c(41.997946, 41.960669, 41.960669, 41.960669,41.909269,41.931841,41.909269,41.910561,41.866129,41.866129),
long = c(-87.654561, -87.747456, -87.67459, -87.646438,-87.747456,-87.67459,-87.67459,-87.619112,-87.747456,-87.691617),
station = 1:10)
coordinates(station) <- ~long + lat
# Set the projection. They were latitude and longitude, so use WGS84 long-lat projection
proj4string(station) <- CRS("+init=epsg:4326")
# View the station location using the mapview function
mapview(station)
#3. Determine the origin
# Set the origin
ori <- SpatialPoints(cbind(-87.872660, 41.619136), proj4string = CRS("+init=epsg:4326"))
# Convert the projection of ori
# Use EPSG: 3857 (Spherical Mercator)
ori_t <- spTransform(ori, CRSobj = CRS("+init=epsg:3857"))
# The origin has been rounded to the nearest 100
x_ori <- round(coordinates(ori_t)[1, 1]/100) * 100
y_ori <- round(coordinates(ori_t)[1, 2]/100) * 100
# Define how many cells for x and y axis
x_cell <- 60
y_cell <- 80
# Define the resolution to be 1000 meters
cell_size <- 1000
# Create the extent
ext <- extent(x_ori, x_ori + (x_cell * cell_size), y_ori, y_ori + (y_cell * cell_size))
# Initialize a raster layer
ras <- raster(ext)
# Set the resolution to be
res(ras) <- c(cell_size, cell_size)
ras[] <- 0
# Project the raster
projection(ras) <- CRS("+init=epsg:3857")
# Create interactive map
mapview(station) + mapview(ras)
This is the image I get in the end:
As for your other question, I'm not sure you're supposed to combine the grid with your data. According to the tutorial linked in the question you've mentioned, krige for example uses both the data meuse and the grid meuse.grid as arguments: lzn.kriged <- krige(log(zinc) ~ 1, meuse, meuse.grid, model=lzn.fit). Check whether this is the case also for the function and the package you're using.
EDIT: How to pick the origin? The origin in this particular code is the bottom left corner of the grid, so I went to Google Maps and picked a random point that was slightly outside the city limits (based on Google's data), so a bit below and a bit on the left from the limits.
回答2:
If these are your points
library(sp)
station <- data.frame(lat = c(41.997946, 41.960669, 41.960669, 41.960669,41.909269,41.931841,41.909269,41.910561,41.866129,41.866129), long = c(-87.654561, -87.747456, -87.67459, -87.646438,-87.747456,-87.67459,-87.67459,-87.619112,-87.747456,-87.691617),station = 1:10)
coordinates(station) = ~ long+lat
proj4string(station) <- CRS("+proj=longlat +datum=WGS84")
stp <- spTransform(station, CRSobj = CRS("+proj=merc +a=6378137 +b=6378137 +lat_ts=0.0 +lon_0=0.0 +x_0=0.0 +y_0=0 +k=1.0 +units=m"))
You can do
library(raster)
r <- raster(stp, res=250)
You can further manipulate it the extent with extend or like this (expand with 10 km, and then round, without changing the resolution)
rr <- setExtent(r, round(extent(r)+10000,-3), keepres=TRUE)
来源：https://stackoverflow.com/questions/57873549/how-to-create-grid-for-coordinates-for-prediction-in-r