英文:
median raster value for polygons by attribute that is not unique
问题
给定这个栅格和多边形数据集：
f <- system.file("ex/elev.tif", package="terra")
r <- rast(f)
fs <- system.file("ex/lux.shp", package="terra")
v <- vect(fs)
如何提取多边形 v 中具有相同 "NAME_1" 值的栅格单元的中位数值，并将其添加到 v 的属性中？
请注意，"NAME_1" 在 v 中不是唯一的：
v
# class : SpatVector
# geometry : polygons
# dimensions : 12, 6 (geometries, attributes)
# extent : 5.74414, 6.528252, 49.44781, 50.18162 (xmin, xmax, ymin, ymax)
# source : lux.shp
# coord. ref. : lon/lat WGS 84 (EPSG:4326)
# names : ID_1 NAME_1 ID_2 NAME_2 AREA POP
# type : <num> <chr> <num> <chr> <num> <int>
# values : 1 Diekirch 1 Clervaux 312 18081
# 1 Diekirch 2 Diekirch 218 32543
# 1 Diekirch 3 Redange 259 18664
英文:
Given this this raster and polygon dataset
f <- system.file("ex/elev.tif", package="terra")
r <- rast(f)
fs <- system.file("ex/lux.shp", package="terra")
v <- vect(fs)
How can I extract the median value of the raster cells for polygons with the same value of "NAME_1" in v, and add this to the attributes of v?
Please note that "NAME_1" is not unique in v:
v
# class : SpatVector
# geometry : polygons
# dimensions : 12, 6 (geometries, attributes)
# extent : 5.74414, 6.528252, 49.44781, 50.18162 (xmin, xmax, ymin, ymax)
# source : lux.shp
# coord. ref. : lon/lat WGS 84 (EPSG:4326)
# names : ID_1 NAME_1 ID_2 NAME_2 AREA POP
# type : <num> <chr> <num> <chr> <num> <int>
# values : 1 Diekirch 1 Clervaux 312 18081
# 1 Diekirch 2 Diekirch 218 32543
# 1 Diekirch 3 Redange 259 18664
答案1
得分: 1
为了计算所有几何体的中位数，您可以执行以下操作：
library(terra)
r <- rast(system.file("ex/elev.tif", package="terra"))
v <- vect(system.file("ex/lux.shp", package="terra"))
v$median = extract(r, v, fun=median, na.rm=TRUE, ID=FALSE)
v
# class : SpatVector
# geometry : polygons
# dimensions : 12, 7 (几何体, 属性)
# extent : 5.74414, 6.528252, 49.44781, 50.18162 (最小x, 最大x, 最小y, 最大y)
# source : lux.shp
# coord. ref. : lon/lat WGS 84 (EPSG:4326)
# names : ID_1 NAME_1 ID_2 NAME_2 AREA POP median
# type : <num> <chr> <num> <chr> <num> <int> <num>
# values : 1 Diekirch 1 Clervaux 312 18081 467.1
# 1 Diekirch 2 Diekirch 218 32543 333.9
# 1 Diekirch 3 Redange 259 18664 377.4
由于"NAME_1"对于多个几何体具有相同的值，我们需要多做一些工作。您可以选择执行以下操作之一：
a <- aggregate(v, "NAME_1")
e <- extract(r, a, fun=median, na.rm=TRUE)
e$NAME_1 <- a$NAME_1[e$ID]
e$ID <- NULL
ve <- merge(v, e, by="NAME_1")
或者
ee <- extract(r, v, na.rm=TRUE)
ee$NAME_1 <- v$NAME_1[ee$ID]
ee <- aggregate(ee[,2,drop=FALSE], ee[,"NAME_1",drop=FALSE], median, na.rm=TRUE)
vee <- merge(v, ee, by="NAME_1")
values(vee)
# NAME_1 ID_1 ID_2 NAME_2 AREA POP elevation
#1 Diekirch 1 1 Clervaux 312 18081 422.0
#2 Diekirch 1 2 Diekirch 218 32543 422.0
#3 Diekirch 1 3 Redange 259 18664 422.0
#4 Diekirch 1 4 Vianden 76 5163 422.0
#5 Diekirch 1 5 Wiltz 263 16735 422.0
#6 Grevenmacher 2 6 Echternach 188 18899 288.5
#7 Grevenmacher 2 7 Remich 129 22366 288.5
#8 Grevenmacher 2 12 Grevenmacher 210 29828 288.5
#9 Luxembourg 3 8 Capellen 185 48187 314.0
#10 Luxembourg 3 9 Esch-sur-Alzette 251 176820 314.0
#11 Luxembourg 3 10 Luxembourg 237 182607 314.0
#12 Luxembourg 3 11 Mersch 233 32112 314.0
英文:
To compute the median for all geometries, you can do
library(terra)
r <- rast(system.file("ex/elev.tif", package="terra") )
v <- vect(system.file("ex/lux.shp", package="terra"))
v$median = extract(r, v, fun=median, na.rm=TRUE, ID=FALSE)
v
# class : SpatVector
# geometry : polygons
# dimensions : 12, 7 (geometries, attributes)
# extent : 5.74414, 6.528252, 49.44781, 50.18162 (xmin, xmax, ymin, ymax)
# source : lux.shp
# coord. ref. : lon/lat WGS 84 (EPSG:4326)
# names : ID_1 NAME_1 ID_2 NAME_2 AREA POP median
# type : <num> <chr> <num> <chr> <num> <int> <num>
# values : 1 Diekirch 1 Clervaux 312 18081 467.1
# 1 Diekirch 2 Diekirch 218 32543 333.9
# 1 Diekirch 3 Redange 259 18664 377.4
Since "NAME_1" has the same value for multiple geometries, we have to do a bit more work. You can do either
a <- aggregate(v, "NAME_1")
e <- extract(r, a, fun=median, na.rm=TRUE)
e$NAME_1 <- a$NAME_1[e$ID]
e$ID <- NULL
ve <- merge(v, e, by="NAME_1")
or
ee <- extract(r, v, na.rm=TRUE)
ee$NAME_1 <- v$NAME_1[ee$ID]
ee <- aggregate(ee[,2,drop=FALSE], ee[,"NAME_1",drop=FALSE], median, na.rm=TRUE)
vee <- merge(v, ee, by="NAME_1")
values(vee)
# NAME_1 ID_1 ID_2 NAME_2 AREA POP elevation
#1 Diekirch 1 1 Clervaux 312 18081 422.0
#2 Diekirch 1 2 Diekirch 218 32543 422.0
#3 Diekirch 1 3 Redange 259 18664 422.0
#4 Diekirch 1 4 Vianden 76 5163 422.0
#5 Diekirch 1 5 Wiltz 263 16735 422.0
#6 Grevenmacher 2 6 Echternach 188 18899 288.5
#7 Grevenmacher 2 7 Remich 129 22366 288.5
#8 Grevenmacher 2 12 Grevenmacher 210 29828 288.5
#9 Luxembourg 3 8 Capellen 185 48187 314.0
#10 Luxembourg 3 9 Esch-sur-Alzette 251 176820 314.0
#11 Luxembourg 3 10 Luxembourg 237 182607 314.0
#12 Luxembourg 3 11 Mersch 233 32112 314.0
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