Time to compute correlogram from dnearneigh neighbour object

[CRyan1]

Hi, thanks for this package. I'm relatively new to R and spatial statistics. I'm wondering how long it should take to create a sp.correlogram from a dnearneigh neighbour object with an upper bound of 1.4m, and the following parameters:

Neighbour list object: Number of regions: 10735 Number of nonzero links: 42372 Percentage nonzero weights: 0.03676841 Average number of links: 3.947089

This represents a raster grid of 540 m2, with a cell size of 1m2. Input is as a data matrix for coordinates and the variable of interest. The coordinates are projected, in metres.

I've tried this several times and stopped it after an hour each time, to check to see if I've done something wrong. I have 130 grids, most 10,000 m2, so if there is a better way to go about creating correlograms for them, please let me know. Thanks

[rsbivand]

Please show your workflow in code, with details of the objects, critically their CRS. Verbal descriptions are insufficient, a reproducible examble using built-in data would be best.

[rsbivand]

Also, correlograms are not a very good idea anyway, since the underlying derivation of the measures uses the whole graph.

[CRyan1]

Thanks for getting back. Here's the code and data. The CRS is NZTM2000

library(spdep)

dat2 <-read_csv("dat2.csv") dat2.1<-read_csv("dat2.1.csv")

my.knn <- knearneigh(dat2.1, k=3) my.nb <- knn2nb(my.knn) plot(my.nb,dat2.1) summary(my.nb) dists<-unlist(nbdists(my.nb, dat2.1, longlat = FALSE)) summary(dists) # this dat2.1.csv dat2.csv shows the max dist is 2 (for use in dnearneigh upper threshold (not 1.4 as prev))

my.nb1<-dnearneigh(as.matrix(dat2.1),0,2) summary(my.nb1)

dat2.2 <- as.vector(dat2$Vegetation_cover) my.correlog<- sp.correlogram(nb1, dat2.2, order =10, method ="I", zero.policy = TRUE) print(my.correlog,p.adj.method ="holm") plot(my.correlog)

alt.1: with package pgirmess (much quicker, a few seconds)

library(pgirmess)

my.correlog_pg<- correlog(dat2.1, dat2.2, method="Moran", nbclass = NULL) plot(my.correlog_pg)

alt.2: with package ncf. Whilst this computes quickly, the relationship is unlikely (negative, straight line)??.

library("ncf")

my.correlog_ncf <-correlog( dat2.1$X_Coord_pt, dat2.1$Y_Coord_pt, dat2.2, w = NULL, increment =34, resamp = 0, latlon = FALSE, na.rm = FALSE, quiet = FALSE )

plot (my.correlog_ncf)

[rsbivand]

Thanks for the reprex. Please permit me get back to you in a day's time, travelling. Maybe compare with similar functionality in https://cran.r-project.org/package=ncf and add any code to the reprex, ncf may be better with distance measures.

[rsbivand]

Another alternative might be pgirmess https://cran.r-project.org/package=pgirmess.

[CRyan1]

Thanks, have updated the code with comments.

[rsbivand]

There are multiple problems in your workflow. For some reason you had very many more observations than 540 in the neighbour objects. So, with:

library(spdep)
dat2 <-read.csv("dat2.csv")
library(sf)
dat.2_sf <- st_as_sf(dat2, coords=c("X_Coord_pt", "Y_Coord_pt"))
dat.2_sf
my.knn <- knearneigh(dat.2_sf, k=1)
my.nb <- knn2nb(my.knn)
plot(my.nb, st_geometry(dat.2_sf))

dists <- unlist(nbdists(my.nb, dat.2_sf))
summary(dists)

The distances between the oblique grid points are all 1m.

my.nb1 <- dnearneigh(dat.2_sf, 0, 1)
my.nb1

However, before moving to your attempt to create a correlogram, I checked whether the variable you were examining was suitable. It is not, this is a categorical variable. It is possible to use a join count test (not a correlogram), but with the configuration of the data (the whole area split into two blocks), this would be inappropriate.

table(dat.2_sf$Vegetation_cover)
plot(dat.2_sf["Vegetation_cover"])

So I'm very unsure what you are trying to do, and this case does not raise any problems for sp.correlogram().

set.seed(1)
o <- sp.correlogram(my.nb1, rnorm(nrow(dat.2_sf)), order=10, method="I")
plot(o)

[CRyan1]

Ok, thank you. Makes sense.