条件分析另一层基于价值观

Question

如果我有一个光栅堆栈：

require(raster) 
r_test <- stack(
    raster(ncols = 10, nrows = 10, vals = sample(c(-1, 0, 1), 100, TRUE)), 
    raster(ncols = 10, nrows = 10, vals = rnorm(100, 7, 0.4) 

我想基于在同一单元的值的函数适用于layer.2layer.1，我该怎么办？

作为一个简单的例子，在layer.2其中等效电池在layer.1等于-1乘以值。

Answer 1

一般而言，overlay()函数在许多情况下工作得很好。根据我的经验，与我第一次刺穿这个答案相反，ifelse()的效率可能更高或更低，特别是对于更大的栅格。对于1000行/列的光栅，简单的光栅代数将更快。如果我将这个数字提高到10000行/列，那么ifelse()的调用会更好。

library(raster) 
r_test <- stack(
    raster(ncols = 1000, nrows = 1000, vals = sample(c(-1, 0, 1), 1000000, TRUE)), 
    raster(ncols = 1000, nrows = 1000, vals = rnorm(1000000, 7, 0.4)) 
) 

## While overlay() works this is a more general solution and is 
##  more efficient for large raster data sets: 
system.time(
r_out <- r_test[[1]] * r_test[[2]] * (r_test[[1]] == -1) + r_test[[2]] * (r_test[[1]] != -1) 
) 
# N=1000x1000 cells: 
# user system elapsed 
# 0.05 0.01 0.06 
# N=10000x10000 cells: 
# user system elapsed 
# 48.36 19.60 77.56 

system.time(
r_out <- overlay(x = r_test[[1]], y = r_test[[2]], fun = function(x, y) ifelse(x == -1, x * y, y)) 
) 
# N=1000x1000 cells: 
# user system elapsed 
# 0.53 0.08 0.64 
# N=10000x10000 cells: 
# user system elapsed 
# 19.77 5.82 26.53