[R-sig-Geo] help using SVC_mle of library varycoef
Hi, this is emanuele barca and I would like to learn the library
varycoef.
I created a simulation of an ideal spatial dataset (gaussian) and
applied SVC_mle function:
set.seed(1234)
library(gstat)
# create structure
nx <- 100
ny <- 100
xy <- expand.grid(1:nx, 1:ny)
names(xy) <- c("x","y")
g.dummy <- gstat(formula = z ~ 1 + x + I(y^0.5), locations = ~ x + y,
dummy = T, beta = c(1, 0.01, 0.005),
model = vgm(psill = 0.025, range = 20, model = 'Ste',
kappa = 10), nmax = 20)
yy <- predict(g.dummy, newdata = xy, nsim = 1)
xy.reduced <- as.data.frame(matrix(ncol = 3, nrow = 0))
for (i in 1:5000){
xy.reduced[i, ] <- yy[i*2, 1:3]
}
Fact<- 20/100 #% of reduction about 80%
training <- sample(nrow(xy.reduced), trunc(Fact*nrow(xy.reduced)))
Xtraining <- xy.reduced[training, ]
Xtest <- xy.reduced[-training, ]
df_train <- Xtraining
colnames(df_train) <- c("X", "Y", "sim")
fit_svc <- SVC_mle(sim ~ X + Y, data = df_train, locs = df_train[,
1:2])#
coef(fit_svc)
summary(fit_svc)
but it returns an error of the covariance matrix.
Could someone help me to overcome the error?
thanks in advance
emanuele
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[R-sig-Geo] automatic assessment of linear coregionalization model
Dear friend,
I am trying to garner a result similar to that of autofitvariogram()
function with fit.lmc() function.
But results are not good.
The following is my code:
g <- gstat(id = "Balance", formula = Balance ~ 1, data = mydata.sp, set
= list(nocheck = 1))
g <- gstat(g, id = "Ancillary", formula = Ancillary ~ 1, data = r.sp,
set = list(nocheck = 1))
v.g <- variogram(g)
g.fit <- fit.lmc(v.g, g, vgm("Exp",
model.df$range[2]), fit.method = 1, correct.diagonal = 1.01) #
fit.lmc = TRUE,
g.fit
the outcome is
data:
Balance : formula = Balance`~`1 ; data dim = 74 x 1
Ancillary : formula = Ancillary`~`1 ; data dim = 18683 x 1
variograms:
model psillrange
Balance[1] Nug 10.864143 0.00
Balance[2] Exp 8.128785 12946.87
Ancillary[1] Nug 43.803405 0.00
Ancillary[2] Exp 52.100889 16156.90
Balance.Ancillary[1] Nug 21.598834 0.00
Balance.Ancillary[2] Exp 20.375770 16151.70
set nocheck = 1;
the results are too weird. Is a data problem?
thanks in advance
emanuele barca
data.rar
Description: application/rar-compressed
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Re: [R-sig-Geo] regression-kriging and co-kriging (Edzer Pebesma)
------
Message: 2
Date: Thu, 15 Aug 2019 08:33:59 +0200
From: Edzer Pebesma
To: r-sig-geo@r-project.org
Subject: Re: [R-sig-Geo] regression-kriging and co-kriging
Message-ID:
Content-Type: text/plain; charset="utf-8"
On 8/12/19 8:21 PM, Emanuele Barca wrote:
Dear Edzer,
maybe I found the solution. I found this in the predict function help:
"When a non-stationary (i.e., non-constant) mean is used, both for
simulation and prediction purposes the variogram model defined should
be
that of the residual process, and not that of the raw observations"
Since my data were, actually, non-stationary, I applied the universal
co-kriging instead usual co-kriging.
now the maps of regression-kring and co-kriging are actually similar s
expected.
did I understand correctly the quoted sentence?
I think so, but hard to be sure given the information you provide.
regards
emanuele barca
--
Message: 2
Date: Sat, 10 Aug 2019 10:41:38 +0200
From: Edzer Pebesma
To: r-sig-geo@r-project.org
Subject: Re: [R-sig-Geo] regression-kriging and co-kriging
Message-ID:
Content-Type: text/plain; charset="utf-8"
Hard to tell from your script. Maybe give a reproducible example?
On 8/6/19 1:07 PM, Emanuele Barca wrote:
Dear r-sig-geo friends,
I produced two maps garnered in the following way:
# for regression-kriging
Piezo.map <-autoKrige(LivStat ~ Z, input_data = mydata.sp, new_data
= covariates, model = "Ste")
Piezork.pred <- Piezo.map$krige_output$var1.pred
Piezork.coords <- Piezo.map$krige_output@coords
Piezork.out <- as.data.frame(cbind(Piezork.coords, Piezork.pred))
colnames(Piezork.out)[1:2] <- c("X", "Y")
coordinates(Piezork.out) = ~ X + Y
gridded(Piezork.out) <- TRUE
spplot(Piezork.out, main = list(label = "R-k Hydraulic head", cex =
1.5))
#for co-kriging
g <- gstat(id = "Piezo", formula = LivStat ~ 1, data = mydata.sp,
set
= list(nocheck = 1))
g <- gstat(g, id = "Z", formula = Z ~ 1, data = mydata.sp, set =
list(nocheck = 1))
v.g <- variogram(g)
#g <- gstat(g, id = "Piezo", model = vgm(150, "Mat", 1350, 0.0,
kappa
= 1.9), fill.all = T)#
g <- gstat(g, id = "Piezo", model = vgm(0.7, "Ste", 1300, 18, kappa
=
1.9), fill.all = T)#
g.fit <- fit.lmc(v.g, g, fit.lmc = TRUE, correct.diagonal = 1.01) #
fit multivariable variogram model , fit.lmc = TRUE, correct.diagonal
= 1.01
g.fit
plot(v.g, model = g.fit, main = "Fitted Variogram Models - Raw
Data")#
#gridded(covariates) <- TRUE
g.cok <- predict(g.fit, newdata = covariates)#grid
g.cok.pred <- g.cok@data$Piezo.pred
<- na.omit(g.cok.pred)
g.cok.coords <- g.cok@coords
g.cok.out <- as.data.frame(cbind(g.cok.coords, g.cok.pred))
colnames(g.cok.out)[1:2] <- c("X", "Y")
coordinates(g.cok.out) = ~ X + Y
gridded(g.cok.out) <- TRUE
spplot(g.cok.out, main = list(label = "Hydraulic head with
Co-kriging", cex = 1.5))
###
I am unable to understand why the first map appears as a raster and
the second not, notwithstanding the fact that they are both computed
on the same "covariates" DEM???
where is the mistake???
regards
emanuele
Emanuele Barca Researcher
Water Research Institute (IRSA-CNR)
Via De Blasio, 5 70123 Bari (Italy)
Phone +39 080 5820535 Fax +39 080 5313365
Mobile +39 340 3420689
_
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Edzer Pebesma
Institute for Geoinformatics
Heisenbergstrasse 2, 48151 Muenster, Germany
Phone: +49 251 8333081
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Edzer Pebesma
Institute for Geoinformatics
[R-sig-Geo] regression-kriging and co-kriging
Dear Edzer,
maybe I found the solution. I found this in the predict function help:
"When a non-stationary (i.e., non-constant) mean is used, both for
simulation and prediction purposes the variogram model defined should be
that of the residual process, and not that of the raw observations"
Since my data were, actually, non-stationary, I applied the universal
co-kriging instead usual co-kriging.
now the maps of regression-kring and co-kriging are actually similar s
expected.
did I understand correctly the quoted sentence?
regards
emanuele barca
--
Message: 2
Date: Sat, 10 Aug 2019 10:41:38 +0200
From: Edzer Pebesma
To: r-sig-geo@r-project.org
Subject: Re: [R-sig-Geo] regression-kriging and co-kriging
Message-ID:
Content-Type: text/plain; charset="utf-8"
Hard to tell from your script. Maybe give a reproducible example?
On 8/6/19 1:07 PM, Emanuele Barca wrote:
Dear r-sig-geo friends,
I produced two maps garnered in the following way:
# for regression-kriging
Piezo.map <-autoKrige(LivStat ~ Z, input_data = mydata.sp, new_data
= covariates, model = "Ste")
Piezork.pred <- Piezo.map$krige_output$var1.pred
Piezork.coords <- Piezo.map$krige_output@coords
Piezork.out <- as.data.frame(cbind(Piezork.coords, Piezork.pred))
colnames(Piezork.out)[1:2] <- c("X", "Y")
coordinates(Piezork.out) = ~ X + Y
gridded(Piezork.out) <- TRUE
spplot(Piezork.out, main = list(label = "R-k Hydraulic head", cex =
1.5))
#for co-kriging
g <- gstat(id = "Piezo", formula = LivStat ~ 1, data = mydata.sp, set
= list(nocheck = 1))
g <- gstat(g, id = "Z", formula = Z ~ 1, data = mydata.sp, set =
list(nocheck = 1))
v.g <- variogram(g)
#g <- gstat(g, id = "Piezo", model = vgm(150, "Mat", 1350, 0.0, kappa
= 1.9), fill.all = T)#
g <- gstat(g, id = "Piezo", model = vgm(0.7, "Ste", 1300, 18, kappa =
1.9), fill.all = T)#
g.fit <- fit.lmc(v.g, g, fit.lmc = TRUE, correct.diagonal = 1.01) #
fit multivariable variogram model , fit.lmc = TRUE, correct.diagonal =
1.01
g.fit
plot(v.g, model = g.fit, main = "Fitted Variogram Models - Raw Data")#
#gridded(covariates) <- TRUE
g.cok <- predict(g.fit, newdata = covariates)#grid
g.cok.pred <- g.cok@data$Piezo.pred
<- na.omit(g.cok.pred)
g.cok.coords <- g.cok@coords
g.cok.out <- as.data.frame(cbind(g.cok.coords, g.cok.pred))
colnames(g.cok.out)[1:2] <- c("X", "Y")
coordinates(g.cok.out) = ~ X + Y
gridded(g.cok.out) <- TRUE
spplot(g.cok.out, main = list(label = "Hydraulic head with
Co-kriging", cex = 1.5))
###
I am unable to understand why the first map appears as a raster and
the second not, notwithstanding the fact that they are both computed
on the same "covariates" DEM???
where is the mistake???
regards
emanuele
Emanuele Barca Researcher
Water Research Institute (IRSA-CNR)
Via De Blasio, 5 70123 Bari (Italy)
Phone +39 080 5820535 Fax +39 080 5313365
Mobile +39 340 3420689
_
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Heisenbergstrasse 2, 48151 Muenster, Germany
Phone: +49 251 8333081
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[R-sig-Geo] regression-kriging and co-kriging
Dear r-sig-geo friends,
I produced two maps garnered in the following way:
# for regression-kriging
Piezo.map <-autoKrige(LivStat ~ Z, input_data = mydata.sp, new_data
= covariates, model = "Ste")
Piezork.pred <- Piezo.map$krige_output$var1.pred
Piezork.coords <- Piezo.map$krige_output@coords
Piezork.out <- as.data.frame(cbind(Piezork.coords, Piezork.pred))
colnames(Piezork.out)[1:2] <- c("X", "Y")
coordinates(Piezork.out) = ~ X + Y
gridded(Piezork.out) <- TRUE
spplot(Piezork.out, main = list(label = "R-k Hydraulic head", cex = 1.5))
#for co-kriging
g <- gstat(id = "Piezo", formula = LivStat ~ 1, data = mydata.sp, set
= list(nocheck = 1))
g <- gstat(g, id = "Z", formula = Z ~ 1, data = mydata.sp, set =
list(nocheck = 1))
v.g <- variogram(g)
#g <- gstat(g, id = "Piezo", model = vgm(150, "Mat", 1350, 0.0, kappa
= 1.9), fill.all = T)#
g <- gstat(g, id = "Piezo", model = vgm(0.7, "Ste", 1300, 18, kappa =
1.9), fill.all = T)#
g.fit <- fit.lmc(v.g, g, fit.lmc = TRUE, correct.diagonal = 1.01) #
fit multivariable variogram model , fit.lmc = TRUE, correct.diagonal = 1.01
g.fit
plot(v.g, model = g.fit, main = "Fitted Variogram Models - Raw Data")#
#gridded(covariates) <- TRUE
g.cok <- predict(g.fit, newdata = covariates)#grid
g.cok.pred <- g.cok@data$Piezo.pred
<- na.omit(g.cok.pred)
g.cok.coords <- g.cok@coords
g.cok.out <- as.data.frame(cbind(g.cok.coords, g.cok.pred))
colnames(g.cok.out)[1:2] <- c("X", "Y")
coordinates(g.cok.out) = ~ X + Y
gridded(g.cok.out) <- TRUE
spplot(g.cok.out, main = list(label = "Hydraulic head with
Co-kriging", cex = 1.5))
###
I am unable to understand why the first map appears as a raster and
the second not, notwithstanding the fact that they are both computed
on the same "covariates" DEM???
where is the mistake???
regards
emanuele
Emanuele Barca Researcher
Water Research Institute (IRSA-CNR)
Via De Blasio, 5 70123 Bari (Italy)
Phone +39 080 5820535 Fax +39 080 5313365
Mobile +39 340 3420689
_
---
Questa e-mail è stata controllata per individuare virus con Avast antivirus.
https://www.avast.com/antivirus
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[R-sig-Geo] problems in anisotropy detection
Dear friends, I have a dataset of hydraulic heads, mydata = and i would like to check for anisotropy. In R I found 3 functions to carry out such task: 1. likfit x_geodata <- as.geodata(mydata, coord.cols = 1:2, data.col = 3, covar.col = 4) fit_mle <- likfit(x_geodata, fix.nugget = FALSE, cov.model = "exponential", psiA = 0, psiR = 1, ini = c(var(Y), 1), fix.psiA = FALSE, fix.psiR = FALSE, hessian = TRUE) that detects no anisotropy. 2. estimateAnisotropy mydata.sp <- mydata coordinates(mydata.sp) = ~ X + Y estimateAnisotropy(mydata.sp, depVar = "LivStat", "LivStat ~ Z") that returns the following [generalized least squares trend estimation] $`ratio` [1] 1.340775 $direction [1] -35.29765 $Q Q11 Q22 Q12 [1,] 1.926136e-05 2.329241e-05 5.721893e-06 $doRotation [1] TRUE finally, 3. estiStAni vmod1 <- fit.variogram(vgm1, vgm(18, "Ste", 1300, 0.78, kappa = 1.7)) estiStAni(vgm1, c(10, 150), "vgm", vmod1) that returns an error: Error in `$<-.data.frame`(`*tmp*`, "dir.hor", value = 0) : replacement has 1 row, data has 0. I am very puzzled, can anyone help me understanding if there is anisotropy in my dataset? thanks emanuele ___ R-sig-Geo mailing list R-sig-Geo@r-project.org https://stat.ethz.ch/mailman/listinfo/r-sig-geo
Re: [R-sig-Geo] auto-correlation of nominal variables (Rich Shepard)
Thank you Rich for your suggestion, is indeed intersting! Actually, my interest is in finding a clustering tendency of crop-classes. Therefore, joincount.multi of spdep library is perfect for my aims. I would like to find a function for representing graphically such tendency. thanks once more. emanuele Il 2019-06-19 12:00 r-sig-geo-requ...@r-project.org ha scritto: Send R-sig-Geo mailing list submissions to r-sig-geo@r-project.org To subscribe or unsubscribe via the World Wide Web, visit https://stat.ethz.ch/mailman/listinfo/r-sig-geo or, via email, send a message with subject or body 'help' to r-sig-geo-requ...@r-project.org You can reach the person managing the list at r-sig-geo-ow...@r-project.org When replying, please edit your Subject line so it is more specific than "Re: Contents of R-sig-Geo digest..." Today's Topics: 1. GAM parameters: Biomod 2 (Lara Silva) 2. Re: auto-correlation of nominal variables (Rich Shepard) -- Message: 1 Date: Tue, 18 Jun 2019 13:46:08 + From: Lara Silva To: r-sig-ecol...@r-project.org, r-sig-geo@r-project.org Subject: [R-sig-Geo] GAM parameters: Biomod 2 Message-ID: Content-Type: text/plain; charset="utf-8" Hello, I am using Biomod 2 to modelling species distribution. I ran some algorithms like GAM, GLM, RF and CTA. I did not change the parameters. How can I known the quadratic terms of GAM? I used the variables with form linear (e.g. precipitation, altimetry, temperature) I did not use temperature ^2. Any suggestions? Regards, Lara <https://www.avast.com/sig-email?utm_medium=email_source=link_campaign=sig-email_content=webmail> Sem vírus. www.avast.com <https://www.avast.com/sig-email?utm_medium=email_source=link_campaign=sig-email_content=webmail> <#DAB4FAD8-2DD7-40BB-A1B8-4E2AA1F9FDF2> [[alternative HTML version deleted]] -- Message: 2 Date: Tue, 18 Jun 2019 07:51:09 -0700 (PDT) From: Rich Shepard To: r-sig-geo@r-project.org Subject: Re: [R-sig-Geo] auto-correlation of nominal variables Message-ID: Content-Type: text/plain; charset="us-ascii"; Format="flowed" On Sun, 16 Jun 2019, Emanuele Barca wrote: I received a dataset of point data, organized in the following way: a couple of coordinates and a column of "crop classes codes", three columns. I would like to compute something similar to the Moran index for each crop class code. Emanuele, What is your interest in these data? Are you interested in the pattern of crop classes (such as in epidemiology or designing a timber sale) or something else? If it is the pattern that is of interest I suggest you first test for complete spatial randomness. Look at some of the references and R packages for spatial point processes. The distribution of crop classes could be random, regular, or clumped/clustered. Peter Diggle's book, "Statistical Analysis of Spatial Point Patterns," and "Spatial Point Patterns" by Adrian Baddeley, Ege Rubak, and Rolf Turner would be excellent places to start. Hope this helps, Rich -- Subject: Digest Footer ___ R-sig-Geo mailing list R-sig-Geo@r-project.org https://stat.ethz.ch/mailman/listinfo/r-sig-geo -- End of R-sig-Geo Digest, Vol 190, Issue 13 ** ___ R-sig-Geo mailing list R-sig-Geo@r-project.org https://stat.ethz.ch/mailman/listinfo/r-sig-geo
[R-sig-Geo] auto-correlation of nominal variables
Dear friends, I received a dataset of point data, organized in the following way: a couple of coordinates and a column of "crop classes codes", three columns. I would like to compute something similar to the Moran index for each crop class code. How can I carry out such task in R? thanks in advance emanuele ___ R-sig-Geo mailing list R-sig-Geo@r-project.org https://stat.ethz.ch/mailman/listinfo/r-sig-geo
