R代码adehabitatHR -网格太小，不适合kernelUD /getverticeshr/adehabitatHR家程估计

Question

抱歉，我的新问题。我还在学习如何在R中进行空间分析，我意识到这个问题已经被问到了(here)。

目标：我无法使用参数中的模拟数据运行这段代码，特别是对于经度(X)值(请参见行: 24)。我想用模拟的数据(下面)绘制家庭的范围。

Error：“getverticeshr.estUD中的错误(x[i]，百分比，ida =name(X)i，unin )：网格太小，无法估计归属范围。您应该用更大的范围参数重新运行kernelUD。”

# 1. Packages
library(adehabitatHR)         # Package for spatal analysis

# 2. Empty Dataframe
points <- data.frame(ID = double())
XY_cor <- data.frame(X = double(),
                 Y = double())
# 3. Assigning values (this will be our spatial coordinates)
set.seed(17)
for(i in c(1:100)){
  if(i >= 50){points[i, 1] <- 1}
  else {points[i, 1] <- 2}

  XY_cor[i, 1] <- runif(1, -78.86887, -78.86440) ## error is here!
  XY_cor[i, 2] <- runif(1, 0.958533, 0.960777)} 

# 4. Transform to SpatialDataframe
coordinates(points) <- XY_cor[, c("X", "Y")]
class(points)

# 5. Domain
x <- seq(-80.0, -77.0, by=1.) # resolution is the pixel size you desire 
y <- seq(-200, 200, by=1.)
xy <- expand.grid(x=x,y=y)
coordinates(xy) <- ~x+y
gridded(xy) <- TRUE
class(xy)

# 6. Kernel Density
kud_points <- kernelUD(points, h = "href", grid = xy)
image(kud_points)

# 7. Get the Volum
 vud_points <- getvolumeUD(kud_points)

# 8. Get contour
levels <- c(50, 75, 95)
list <- vector(mode="list", length = 2)

list[[1]] <- as.image.SpatialGridDataFrame(vud_points[[1]])
list[[2]] <- as.image.SpatialGridDataFrame(vud_points[[2]])

# 9. Plot
par(mfrow = c(2, 1))
image(vud_points[[1]])
contour(list[[1]], add=TRUE, levels=levels)
image(vud_points[[2]])
contour(list[[2]], add=TRUE, levels=levels)


# 10. Get vertices 
vkde_points <- getverticeshr(kud_points, percent = 50,
                         unin = 'm', unout='m2')
plot(vkde_points)

Answer 1

我能够用下面的代码来解决这个问题。希望这能帮助别人在同样的问题上！我不得不手动改变我的包围盒的分辨率。

5.领域

x <- seq(737326.0, 737639.6, by=5)        
y <-seq(106071.4, 106259.9, by=5)
xy <- expand.grid(x=x,y=y)
coordinates(xy) <- ~x+y    
gridded(xy) <- TRUE