如何使用R中的idw（）函数预测特定点的值？

Question

以Ege Rubak提供的答案为例，如何使用中的函数预测某个特定点（例如lat = -23.49184和）的pH值？long = 152.07185idw()R

我找到的最接近的答案是通过RPubs中的此文档，但我不能仅提取特定值。

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，因为我尚没有50个声誉。

Answer 1

您不需要网格。以表示观察到的位置的一致方式提供新位置。

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