如何使用 R 中的 idw() 函数预测特定点的值？

Question

以 this answer from Ege Rubak 为例，如何使用 idw() 中的 idw() 函数预测特定点的 pH 值，例如 lat = -23.49184 和 long = 152.07185 14=]?

我找到的最接近的答案是通过 RPubs 中的这篇文档，但我无法仅提取特定值。

library(gstat)
library(sp)

lat <-  c(-23.49174, -23.49179, -23.49182, -23.49183, -23.49185, -23.49187)
long <- c(152.0718, 152.0718, 152.0717, 152.0717, 152.0717, 152.0717)
pH <- c(8.222411, 8.19931, 8.140428, 8.100752, 8.068141, 8.048852)
sample <- data.frame(lat, long, pH)

x.range <- range(sample$long)
y.range <- range(sample$lat)

x<-seq(x.range[1], x.range[2], length.out=20)
y<-seq(y.range[1], y.range[2], length.out=20)
grd<-expand.grid(x,y)

coordinates(sample) = ~long+lat
coordinates(grd) <- ~ Var1+Var2
gridded(grd) <- TRUE

proj4string(sample) <- CRS("+proj=longlat +datum=WGS84")
proj4string(grd) <- CRS("+proj=longlat +datum=WGS84")

dat.idw <- idw(formula=pH ~ 1, locations = sample, newdata = grd, idp = 2.0)
#> [inverse distance weighted interpolation]

我没有在评论中特别询问 Ege Rubak，因为我还没有 50 个声望。

Answer 1

您可以使用 raster 包中的 extract 函数。请注意，您的点位于原始网格之外，因此我增加 1.5 以覆盖该点。

library(gstat)
library(sp)

lat <-  c(-23.49174, -23.49179, -23.49182, -23.49183, -23.49185, -23.49187)
long <- c(152.0718, 152.0718, 152.0717, 152.0717, 152.0717, 152.0717)
pH <- c(8.222411, 8.19931, 8.140428, 8.100752, 8.068141, 8.048852)
sample <- data.frame(lat, long, pH)

x.range <- range(sample$long)
y.range <- range(sample$lat)

x<-seq(x.range[1], x.range[2] * 1.5, length.out=20)
y<-seq(y.range[1], y.range[2] * 1.5, length.out=20)
grd<-expand.grid(x,y)

coordinates(sample) = ~long+lat
coordinates(grd) <- ~ Var1+Var2
gridded(grd) <- TRUE

proj4string(sample) <- CRS("+proj=longlat +datum=WGS84")
proj4string(grd) <- CRS("+proj=longlat +datum=WGS84")

dat.idw <- idw(formula=pH ~ 1, locations = sample, newdata = grd, idp = 2.0)


library(raster)

# Convert to raster
dat.r <- raster(dat.idw)

# Create a matrix showing the coordinate of interest
p <- SpatialPoints(matrix(c(152.07185, -23.49184), ncol = 2))
proj4string(p) <- projection(dat.r)

# Extract the values
extract(dat.r, p)
# 8.048852

Answer 2

您不需要网格。以一致的方式提供您的新位置，以表示您观察到的位置。

library(gstat)
library(sp)

lat <-  c(-23.49174, -23.49179, -23.49182, -23.49183, -23.49185, -23.49187)
long <- c(152.0718, 152.0718, 152.0717, 152.0717, 152.0717, 152.0717)
pH <- c(8.222411, 8.19931, 8.140428, 8.100752, 8.068141, 8.048852)

sample <- data.frame(lat, long, pH)
coordinates(sample) = ~long+lat
proj4string(sample) <- CRS("+proj=longlat +datum=WGS84")

loc <- data.frame(long = 152.07185, lat = -23.49184)
coordinates(loc)  <- ~ long + lat
proj4string(loc) <- CRS("+proj=longlat +datum=WGS84")

oo <- idw(formula=pH ~ 1, locations = sample, newdata = loc, idp = 2.0)
oo@data$var1.pred
#[1] 8.158494

如何使用 R 中的 idw() 函数预测特定点的值？

How can I predict values for a specific point using the idw() function in R?

interpolation

r

spatial

geospatial

sp