简体   繁体   English

如何聚合分类 SpatRaster

[英]How to aggregate categorical SpatRaster

I am new to the terra package.我是 terra 包的新手。 I am trying to aggregate a categorical raster (or a SpatRaster, to be more precise) that has only one layer.我正在尝试aggregate只有一层的分类栅格(或 SpatRaster,更准确地说)。 The result should be a raster with as many layers as categories in the original raster;结果应该是一个与原始栅格中的类别一样多的栅格; the cells' values should have the number of original (smaller) cells in each category.单元格的值应具有每个类别中原始(较小)单元格的数量。

Here's an example showing what I trying to achieve:这是一个示例,显示了我试图实现的目标:

    library(terra)

    # the SpatRaster with 3 categories: 1, 2, 3
    set.seed(0)
    r <- rast(nrows=4, ncols=4)
    values(r) <- sample(3, ncell(r), replace=TRUE)

    # create one layer per category with binary indicators
    r1 <- subst(r, from=c(2,3), 0)
    r2 <- subst(r, from=c(1,3), 0); r2 <- subst(r2, from=2, 1)
    r3 <- subst(r, from=c(1,2), 0); r3 <- subst(r3, from=3, 1)
    
    # stack
    s <- c(r1, r2, r3)
    names(s) <- c("cat1", "cat2", "cat3")
    
    # aggregate
    a <- aggregate(s, fact = 2, fun = "sum")

This works for this example.这适用于这个例子。 But it is not practical nor efficient.但它既不实用也不高效。 It is probably(?) not feasible with large raster datasets (orders of magnitude 1GB-10GB) and many categories.对于大型栅格数据集(1GB-10GB 数量级)和许多类别,这可能(?)不可行。

So, how would a terra pro do this?那么, terra pro 将如何做到这一点?

Here is a more streamlined approach这是一个更精简的方法

Your example data:您的示例数据:

library(terra)
set.seed(0)
r <- rast(nrows=4, ncols=4)
values(r) <- sample(3, ncell(r), replace=TRUE)

solution:解决方案:

s <- segregate(r)
a <- aggregate(s, 2, sum)

It is also possible to do this:也可以这样做:

b <- list()
for (i in 1:3) {
    b[[i]] <- aggregate(r, 2, function(v) sum(v==i,na.rm=TRUE))
}
b <- rast(b)

Which you can also write like this (I wouldn't recommend it)你也可以这样写(我不推荐它)

bb <- lapply(1:3, \(i) aggregate(r, 2, \(v) sum(v==i,na.rm=TRUE))) |>
      rast()

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM