As semi-variograms gives semi-variance and lag distance for DEM. I want to know at what DEM elevation the sill value is flattening using penta spherical model.
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I am currently trying to calculate "weighted" spatial daily values for pressure using era5 data. This is due to the size of the area being represented differently towards the poles relative to lower-latitude regions. I am a little confused though. Should I multiply each value with the cosine of its latitude? So pressure * (cos(latitude)). The idea is then to apply PCA to the field. Thanks in advance!
I am trying to account for spatial autocorrelation in a model in R. Each observation is a country for which I have the average latitude and longitude. Here's some sample data:
country <- c("IQ", "MX", "IN", "PY")
long <- c(43.94511, -94.87018, 78.10349, -59.15377)
lat <- c(33.9415073, 18.2283975, 23.8462264, -23.3900255)
Pathogen <- c(10.937891, 13.326284, 12.472374, 12.541716)
Answer.values <- c(0, 0, 1, 0)
data <- data.frame(country, long, lat, Pathogen, Answer.values)
I know spatial autocorrelation is an issue (Moran's i is significant in the whole dataset). This is the model I am testing (Answer Values (a 0/1 variable) ~ Pathogen Prevalence (a continuous variable)).
model <- glm(Answer.values ~ Pathogen,
na.action = na.omit,
data = data,
family = "binomial")
How would I account for spatial autocorrelation with a data structure like that?
There are a lot of potential answers to this. One easy(ish) way is to use mgcv::gam() to add a spatial smoother. Most of your model would stay the same:
library(mgcv)
gam(Answer.values ~ Pathogen +s([something]),
family="binomial",
data=data)
where s([something]) is some form of smooth spatial term. Three possible/reasonable choices would be:
a spherical spline (?mgcv::smooth.construct.sos.smooth.spec), which takes lat/long as input; this would be useful if (1) you have data over a significant fraction of the earth's surface (so that a smoother that constructs a 2D planar spatial smooth is less reasonable); (2) you want to account for distance between locations in a continuous way
a Markov random field (?mgcv::smooth.construct.mrf.smooth.spec). This is essentially the spatial analogue of a discrete order-1 autoregressive structure (i.e. countries are directly correlated only with their direct neighbours, however you choose to define that). In order to do this you have to come up somehow with a neighbourhood list (i.e. a list of countries, where the elements are lists of countries that are neighbours of the original countries). You could do this however you like, e.g. by finding nearest neighbours geographically. (Check out some introductions to spatial statistics/spatial data analysis in R.) (On the other hand, if you're testing Moran's I then you've presumably already come up with some way to identify first-order neighbours ...)
if you're comfortable treating lat/long as coordinates in a 2D plane, then you have lot of choices of smoothing basis, e.g. ?mgcv::smooth.construct.gp.smooth.spec (Gaussian process smoothers, which include most of the standard spatial autocorrelation models as special cases)
A helpful link for getting up to speed with GAMs in R ...
I'm doing a spatial analysis on my data and I have their localization in terms of longitude and latitude. I calculated the sample variogram in R with geoR and gstat but what I get is the sample variogram with the distance in grades (longitude and latitude). I would like to have the same information but with the distance in km, like what happens when I calculate the spatio-temporal variogram with variogramST (gstat).
Thank you very much for those who are going to answer, it would help me a lot!
You get that with gstat if you specify for your dataset that the coordinates are degrees latitude longitude, e.g. in case you use sp, with proj4string(x) = "+proj=longlat". "distances in degrees" do not exist, they are plane wrong (except on the equator).
I was trying to use the dbscan package in R to try to cluster some spatial data. The dbscan::dbscan function takes eps and minpts as input. I have a dataframe with two columns longitude and latitude expressed in degree decimals like in the following:
df <- data.frame(lon = c(seq(1,5,1), seq(1,5,1)),
lat = c(1.1,3.1,1.2,4.1,2.1,2.2,3.2,2.4,1.4,5.1))
and I apply the algorithm:
db <- fpc::dbscan(df, eps = 1, MinPts = 2)
will eps here be defined in degrees or in some other unit ? I'm really trying to understand in which unit this maximum distance eps value is expressed so any help is appreciated
Never use the fpc package, always use dbscan::dbscan instead.
If you have latitude and longitude, you need to choose an appropriate distance function such as Haversine.
The default distance function, Euclidean, ignores the spherical nature of earth. The eps value then is a mixture of degrees latitude and longitude, but these do not correspond to uniform distances! One degree east at the equator is much farther than one degree east in Vancouver.
Even then, you need to pay attention to units. One implementation of Haversine may yield radians, another one meters, and of course someone crazy will work in miles.
Unfortunately, as far as I can tell, none of the R implementations can accelerate Haversine distance. So it may be much faster to cluster the data in ELKI instead (you need to add an index yourself though).
If your data is small enough, you can however use a precomputed distance matrix (dist object) in R. But that will take O(n²) time and memory, so it is not very scalable.
I use the haversine formula to calculate the distance among the points. The result of this formula is in meter or in kmeter?
http://en.wikipedia.org/wiki/Haversine_formula
Anyone can help me?
Haversine formula is used for finding distance between to points, if latitude and longitude for both points are known.
Formula:
ACOS(SIN(Lat1)*SIN(Lat2) +COS(Lat1)*COS(Lat2)*COS(Lon2-Lon1)) *6371
Excel formula:
=ACOS(COS(RADIANS(90-Lat1)) *COS(RADIANS(90-Lat2)) +SIN(RADIANS(90-Lat1)) *SIN(RADIANS(90-Lat2)) *COS(RADIANS(Long1-Long2))) *6371
Note:
Replace 6371 with 3958.756 if you want the answer in miles.
For further detail:
http://bluemm.blogspot.in/2007/01/excel-formula-to-calculate-distance.html