以较粗的分辨率聚合栅格数据

Question

我的数据集的网格分辨率为0.25 * 0.25度

lon <- c(-53.615449969, -53.365449969, -53.115449969, -53.365449969, 
     -53.115449969, -52.865449969, -53.365449969, -53.115449969, -52.865449969, 
     -52.615449969, -53.365449969, -53.115449969, -52.865449969, -52.615449969, 
     -52.365449969, -53.365449969, -53.115449969, -52.865449969, -52.615449969, 
     -52.365449969, -53.615449969, -53.365449969, -53.115449969, -52.865449969, 
     -52.615449969, -52.365449969, -52.115449969, -53.865449969, -53.615449969, 
     -53.365449969, -53.115449969, -52.865449969, -52.615449969, -52.365449969, 
     -52.115449969, -51.865449969, -54.365449969, -54.115449969, -53.865449969, 
     -53.615449969, -53.365449969, -53.115449969, -52.865449969, -52.615449969, 
     -52.365449969, -52.115449969, -51.865449969, -51.615449969, -54.615449969, 
     -54.365449969, -54.115449969, -53.865449969, -53.615449969, -53.365449969, 
     -53.115449969, -52.865449969, -52.615449969, -52.365449969, -52.115449969, 
     -51.615449969)


lat <- c(-33.627081271, -33.627081271, -33.627081271, -33.377081271, 
     -33.377081271, -33.377081271, -33.127081271, -33.127081271, -33.127081271, 
     -33.127081271, -32.877081271, -32.877081271, -32.877081271, -32.877081271, 
     -32.877081271, -32.627081271, -32.627081271, -32.627081271, -32.627081271, 
     -32.627081271, -32.377081271, -32.377081271, -32.377081271, -32.377081271, 
     -32.377081271, -32.377081271, -32.377081271, -32.127081271, -32.127081271, 
     -32.127081271, -32.127081271, -32.127081271, -32.127081271, -32.127081271, 
     -32.127081271, -32.127081271, -31.877081271, -31.877081271, -31.877081271, 
     -31.877081271, -31.877081271, -31.877081271, -31.877081271, -31.877081271, 
     -31.877081271, -31.877081271, -31.877081271, -31.877081271, -31.627081271, 
     -31.627081271, -31.627081271, -31.627081271, -31.627081271, -31.627081271, 
     -31.627081271, -31.627081271, -31.627081271, -31.627081271, -31.627081271, 
    -31.627081271)

 df <- as.data.frame(cbind(lon, lat))
 df$ID <- 1:nrow(df)
 coordinates(df) <- c(1,2)

 library(raster)
 elev <- getData('alt', country='BRA')

 plot(elev)
 plot(df, add = T)

对于我的0.25 * 0.25度网格的每一个，我想计算平均高程。如何在R中做到这一点？

Answer 1

鉴于res(elev)为0.008333333（即1/120），您希望以因子.25 * 120 = 30进行汇总。最简单的方法是

a <- aggregate(elev, 30, fun=mean, na.rm=TRUE)

现在，分辨率是所需的，但不是与您指定的栅格的对齐。图解：

r <- rasterFromXYZ(df)
s <- as(r, 'SpatialPolygons')
z <- crop(a, r, snap='out')
plot(z)
plot(s, add=TRUE)

相反，您可以这样做：

x <- resample(elev, r)

或（更精确，但更慢）

e <- extract(elev, s, fun=mean, na.rm=TRUE)
s$elev <- e

比较值

z <- rasterize(r, s)
plot(z, e)