2017-04-27 96 views
1

我有一個柵格堆棧,需要用其他堆棧中的圖層替換它包含的某些圖層。R:使用第二個柵格堆棧中的圖層替換柵格堆棧中的多個圖層

我想解決方案非常簡單直接,但到目前爲止我無法找到它。

給你重複的例子,這裏是我試圖做:

library(raster) 

# create some raster 
r1 <- raster(nrows=10, ncols=10, xmn=0, xmx=10, ymn = 0, ymx = 10) 
r1[] <- sample(1:100, 100, replace = TRUE) 

r2 <- raster(nrows=10, ncols=10, xmn=0, xmx=10, ymn = 0, ymx = 10) 
r2[] <- sample(1:100, 100, replace = TRUE) 

r3 <- raster(nrows=10, ncols=10, xmn=0, xmx=10, ymn = 0, ymx = 10) 
r3[] <- sample(1:100, 100, replace = TRUE) 

r4<- raster(nrows=10, ncols=10, xmn=0, xmx=10, ymn = 0, ymx = 10) 
r4[] <- sample(1:100, 100, replace = TRUE) 

# put the raster into a stack 
r_stack <- stack(r1, r2, r3, r4) 

#calculate the mean of the raster 
r_mean <- mean(r_stack) 

# What I would like to do is to subtract the mean from some of the 
# raster in the stack (layers_to_replace), but not from all, and to    
# replace the raster in the stack with the new difference. 
# For example I would like to replace the second and fourth layer with  
# the difference. 
l_replace <- c(2, 4) 

# Note: I place the difference into a second stack for the sake 
# of the example as my original data comes in a second stack 
rep_stack <- r_stack[[l_replace]] - r_mean 

r_stack[[l_replace]] <- rep_stack 

不幸的是這種方法是行不通的,並引發以下錯誤:

Error in v[] <- value : incompatible types (from S4 to logical) in subassignment type fix 
In addition: Warning messages: 
1: In if (i < 1) { : the condition has length > 1 and only the first element will be used 
2: In if (i > nl + 1) { : the condition has length > 1 and only the first element will be used 
3: In if (i > nl) { : the condition has length > 1 and only the first element will be used 

如何任何想法要解決這個問題將更受歡迎。

回答

0

這應該工作:

for(ind in seq(along = l_replace)) r_stack[[l_replace[ind]]][] <- rep_stack[[ind]][] 

r_stack 
class  : RasterStack 
dimensions : 10, 10, 100, 4 (nrow, ncol, ncell, nlayers) 
resolution : 1, 1 (x, y) 
extent  : 0, 10, 0, 10 (xmin, xmax, ymin, ymax) 
coord. ref. : +proj=longlat +datum=WGS84 +ellps=WGS84 +towgs84=0,0,0 
names  : layer.1, layer.2, layer.3, layer.4 
min values : 1.00, -54.25, 2.00, -54.25 
max values : 100.0, 51.5, 100.0, 51.5 

或者,不知何故,可讀性更強:

for(ind in seq(along = l_replace)){ 
    setValues(r_stack, getValues(rep_stack[[ind]]), layer = l_replace[ind]) 
} 
+0

簡單,工作液由此接受的答案。 – Improbability

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