在将栅格的值提取到点时，我发现我有几个NA，而不是使用buffer的{{1}}和fun个参数我希望将最近的非extract像素提取到与NA重叠的点。
我正在使用基本提取功能：
NA答案 0 :(得分：6)
这是一个不使用缓冲区的解决方案。但是，它会为数据集中的每个点单独计算距离图，因此如果数据集很大，它可能无效。
set.seed(2)
# create a 10x10 raster
r <- raster(ncol=10,nrow=10, xmn=0, xmx=10, ymn=0,ymx=10)
r[] <- 1:10
r[sample(1:ncell(r), size = 25)] <- NA
# plot the raster
plot(r, axes=F, box=F)
segments(x0 = 0, y0 = 0:10, x1 = 10, y1 = 0:10, lty=2)
segments(y0 = 0, x0 = 0:10, y1 = 10, x1 = 0:10, lty=2)
# create sample points and add them to the plot
xy = data.frame(x=runif(10,1,10), y=runif(10,1,10))
points(xy, pch=3)
text(x = xy$x, y = xy$y, labels = as.character(1:nrow(xy)), pos=4, cex=0.7, xpd=NA)
# use normal extract function to show that NAs are extracted for some points
extracted = extract(x = r, y = xy)
# then take the raster value with lowest distance to point AND non-NA value in the raster
sampled = apply(X = xy, MARGIN = 1, FUN = function(xy) r@data@values[which.min(replace(distanceFromPoints(r, xy), is.na(r), NA))])
# show output of both procedures
print(data.frame(xy, extracted, sampled))
# x y extracted sampled
#1 5.398959 6.644767 6 6
#2 2.343222 8.599861 NA 3
#3 4.213563 3.563835 5 5
#4 9.663796 7.005031 10 10
#5 2.191348 2.354228 NA 2
#6 1.093731 9.835551 2 2
#7 2.481780 3.673097 3 3
#8 8.291729 2.035757 9 9
#9 8.819749 2.468808 9 9
#10 5.628536 9.496376 6 6
答案 1 :(得分：1)
这是一种基于栅格的解决方案，首先使用最近的非NA像素值填充NA像素。 但请注意，这并未考虑像素内点的位置。相反，它会计算像素中心之间的距离，以确定最近的非NA像素。
首先，它为每个NA光栅像素计算距离最近的非NA像素的距离和方向。下一步是计算这个非NA单元格的坐标（假设投影CRS），提取其值并将该值存储在NA位置。
开始数据：投影栅格，其值与koekenbakker中的answer相同：
set.seed(2)
# set projected CRS
r <- raster(ncol=10,nrow=10, xmn=0, xmx=10, ymn=0,ymx=10, crs='+proj=utm +zone=1')
r[] <- 1:10
r[sample(1:ncell(r), size = 25)] <- NA
# create sample points
xy = data.frame(x=runif(10,1,10), y=runif(10,1,10))
# use normal extract function to show that NAs are extracted for some points
extracted <- raster::extract(x = r, y = xy)
计算从所有NA像素到最近的非NA像素的距离和方向：
dist <- distance(r)
# you can also set a maximum distance: dist[dist > maxdist] <- NA
direct <- direction(r, from=FALSE)
检索NA像素的坐标
# NA raster
rna <- is.na(r) # returns NA raster
# store coordinates in new raster: https://stackoverflow.com/a/35592230/3752258
na.x <- init(rna, 'x')
na.y <- init(rna, 'y')
# calculate coordinates of the nearest Non-NA pixel
# assume that we have a orthogonal, projected CRS, so we can use (Pythagorean) calculations
co.x <- na.x + dist * sin(direct)
co.y <- na.y + dist * cos(direct)
# matrix with point coordinates of nearest non-NA pixel
co <- cbind(co.x[], co.y[])
用坐标'co'
提取最近的非NA细胞的值# extract values of nearest non-NA cell with coordinates co
NAVals <- raster::extract(r, co, method='simple')
r.NAVals <- rna # initiate new raster
r.NAVals[] <- NAVals # store values in raster
使用新值填充原始栅格
# cover nearest non-NA value at NA locations of original raster
r.filled <- cover(x=r, y= r.NAVals)
sampled <- raster::extract(x = r.filled, y = xy)
# compare old and new values
print(data.frame(xy, extracted, sampled))
# x y extracted sampled
# 1 5.398959 6.644767 6 6
# 2 2.343222 8.599861 NA 3
# 3 4.213563 3.563835 5 5
# 4 9.663796 7.005031 10 10
# 5 2.191348 2.354228 NA 3
# 6 1.093731 9.835551 2 2
# 7 2.481780 3.673097 3 3
# 8 8.291729 2.035757 9 9
# 9 8.819749 2.468808 9 9
# 10 5.628536 9.496376 6 6
请注意，第5点采用的另一个值是Koekenbakker的answer，因为此方法没有考虑像素内点的位置（如上所述）。如果这很重要，则此解决方案可能不合适。在其他情况下，例如如果栅格单元格与点精度相比较小，则这种基于栅格的方法应该会产生良好的结果。
答案 2 :(得分：1)
对于光栅堆栈，请使用上面的@ koekenbakker解决方案，并将其转换为函数。光栅堆栈的@layers广告位是一个栅格列表，因此，请将其拉平并从那里开始。
#new layer
r2 <- raster(ncol=10,nrow=10, xmn=0, xmx=10, ymn=0,ymx=10)
r2[] <- 1:10
r2[sample(1:ncell(r2), size = 25)] <- NA
#make the stack
r_stack <- stack(r, r2)
#a function for sampling
sample_raster_NA <- function(r, xy){
apply(X = xy, MARGIN = 1,
FUN = function(xy) r@data@values[which.min(replace(distanceFromPoints(r, xy), is.na(r), NA))])
}
#lapply to get answers
lapply(r_stack@layers, function(a_layer) sample_raster_NA(a_layer, xy))
还是想要（速度提升？）
purrr::map(r_stack@layers, sample_raster_NA, xy=xy)
这让我想知道是否可以使用dplyr加速整个事情......