这里有一个不使用缓冲区的解决方案。但是，它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

这是基于光栅的解决方案，首先用最接近的非NA像素值填充NA像素。然而，请注意，这并不考虑像素内的点的位置。相反，它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光栅像素计算到最近的非NA像素的距离和方向。下一步是计算该非NA小区的坐标（假设投影CRS），提取其值并将该值存储在NA位置。
起始数据：投影光栅，其值与koekenbakker的answer中的值相同：

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)

NA像素的X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[])

提取坐标为“co”的最近非NA单元格的值

# 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，点5取的值不同于Koekenbakker的answer，因为该方法没有考虑点在像素内的位置（如上所述）。如果这很重要，那么这个解决方案可能不合适。在其他情况下，例如如果光栅单元与点精度相比较小，则这种基于光栅的方法应当给出给予良好的结果。

对于光栅堆栈，使用@koekenbakker上面的解决方案，并将其转换为函数。光栅堆栈的@layers插槽是光栅列表，因此，lapply它穿过并从那里开始。

#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))

或者是花式（速度改进？）

purrr::map(r_stack@layers, sample_raster_NA, xy=xy)

这让我想知道整个事情是否可以加快使用dspur.

seegSDM包中的函数nearestLand可以很好地完成这一任务：

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
extracted = extract(x = r, y = xy)

# find points that fall in NA areas on the raster
plot(r)
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)
outside_mask <- is.na(extracted)
outside_pts <- xy[outside_mask,]

points(xy, pch = 16)
text(x = xy$x, y = xy$y, labels = as.character(1:nrow(xy)), pos=4, cex=0.7, xpd=NA)

points(outside_pts, pch = 16, col = 'red')

# find the nearest cell
nearest.cells <- nearestLand(outside_pts, r, 90000)

# plot the new points (for those which were reassigned) in green
points(nearest.cells, pch = 16, col = ' green')

# plot where they moved to
arrows(outside_pts[, 1],
       outside_pts[, 2],
       nearest.cells[, 1],
       nearest.cells[, 2], length = 0.1)

#Extracting the raster values using the new coordinates:
extract(x = r, y = nearest.cells)