如何过滤 Raster 或仅在 R 中绘制某些栅格类别值？

Question

我在 R 中有多年的土地覆盖栅格数据，其属性表如下所示（屏幕截图和代码）：

ID   COUNT   CLASSNAME
11  525631  Open Water                          0   0   255 0.000000    0.000000    1.000000
500 1176    Developed-Upland Deciduous Forest   64  61  168 0.250980    0.239216    0.658824
501 3965    Developed-Upland Evergreen Forest   68  79  137 0.266667    0.309804    0.537255
502 3619    Developed-Upland Mixed Forest   102 119 205 0.400000    0.466667    0.803922
503 49181   Developed-Upland Herbaceous 122 142 245 0.478431    0.556863    0.960784
504 16592   Developed-Upland Shrubland  158 170 215 0.619608    0.666667    0.843137
505 42867   Developed - Low Intensity   255 122 143 1.000000    0.478431    0.560784
506 21570   Developed - Medium Intensity    253 44  79  0.992157    0.172549    0.309804
507 12451   Developed - High Intensity  173 0   28  0.678431    0.000000    0.109804
25  80384   Developed-Roads 1   1   1   0.003922    0.003922    0.003922

我试图在下面的传单中只绘制以下类别。 我尝试过重新编码、空间过滤、使用 colorBin 进行分箱，并且正在寻找一种新方法，通过直接过滤数据或更改绘图。

Developed - Forested", "Developed - herbaceous or shrubland", 
"Developed - Medium Intensity", "Developed - High Intensity", "Cropland",
"Tree Cover 0-50%", "Tree Cover 50-100%", "Shrub Cover", "Herb Cover 0-50%", "Herb Cover 50-100%")

我正在寻找这样的最终产品（使用正确的值而不是下面的颜色）

Answer 1

你不能在绘制数据之前filter()你的数据吗？

没有可重现的例子，这里只是一个想法：

filter_classnames <- c("Developed - Forested", "Developed - herbaceous or shrubland", 
"Developed - Medium Intensity", "Developed - High Intensity", "Cropland",
"Tree Cover 0-50%", "Tree Cover 50-100%", "Shrub Cover", "Herb Cover 0-50%", "Herb Cover 50-100%")

filtered_data <- your_data %>%
  filter(CLASSNAMES %in% filter_classnames)