enter image description hereI am trying to use randomforest to generate a spatial prediction map.
I developed my model by using random forest regression, but I met a little difficulty in the last step to use the best predictors for building the predictive map. I want to create a map prediction map.
My code:
library(raster)
library(randomForest)
set.seed(12)
s <- stack("Density.tif", "Aqui.tif", "Rech.tif", "Rainfall.tif","Land Use.tif", "Cond.tif", "Nitrogen.tif", "Regions.tif","Soil.tif","Topo.tif", "Climatclass.tif", "Depth.tif")
points <- read.table("Coordonnées3.txt",header=TRUE, sep="\t", dec=",",strip.white=TRUE)
d <- extract(s, points)
rf <-randomForest(nitrate~ . , data=d, importance=TRUE, ntree=500, na.action = na.roughfix)
p <- predict(s, rf)
plot(p)
Sample Data:
> head(points)
LAT LONG
1 -13.057007 27.549580
2 -4.255000 15.233745
3 5.300000 -1.983610
4 7.245675 -4.233336
5 12.096330 15.036016
6 -4.255000 15.233745
The error when I run my short code is:
Error in eval(expr, envir, enclos) : object 'nitrate' not found.
I am guessing the error happens when you fit the model.
Why would there be a variable called nitrate. Given how you create your RasterStack, perhaps there is one called Nitrogen. Either way you can find out by looking at names(s) and colnames(d).
NOTE that your points are not good! They are in reverse order. The order should be (longitude, latitude).
Based on your comments (please edit your question instead), you should
add nitrate the points file (the third column) or something like that. Then do
xy <- points[, 2:1]
nitrate <- points[,3]
Extract points and combine with your observed data
d <- extract(s, xy)
d <- cbind(nitrate=nitrate, d)
Build model and predict
rf <-randomForest(nitrate~ . , data=d, importance=TRUE, ntree=500, na.action = na.roughfix)
p <- predict(s, rf)
It sounds like the error is coming when you are trying to build the forest. It may be most helpful to not use the formula interface. Also, if d is large, then using the formula interface is not advisable. From the help file on randomForest: "For large data sets, especially those with large number of variables, calling randomForest via the formula interface is not advised: There may be too much overhead in handling the formula."
Assuming d$nitrate exists then the solution is randomForest(y = d$nitrate, x = subset(d, select = -nitrate), importance=TRUE, ntree=500, na.action = na.roughfix)
I'm learning to create a forecasting model for time series that has multiple seasonalities. Following is the subset of dataset that I'm refering to. This dataset includes hourly data points and I wish to include daily as well as weekly seasonalities in my arima model. Following is the subset of dataset:
data= c(4,4,1,2,6,21,105,257,291,172,72,10,35,42,77,72,133,192,122,59,29,25,24,5,7,3,3,0,7,15,91,230,284,147,67,53,54,55,63,73,114,154,137,57,27,31,25,11,4,4,4,2,7,18,68,218,251,131,71,43,55,62,63,80,120,144,107,42,27,11,10,16,8,10,7,1,4,3,12,17,58,59,68,76,91,95,89,115,107,107,41,40,25,18,14,15,6,12,2,4,1,6,9,14,43,67,67,94,100,129,126,122,132,118,68,26,19,12,9,5,4,2,5,1,3,16,89,233,304,174,53,55,53,52,59,92,117,214,139,73,37,28,15,11,8,1,2,5,4,22,103,258,317,163,58,29,37,46,54,62,95,197,152,58,32,30,17,9,8,1,3,1,3,16,109,245,302,156,53,34,47,46,54,65,102,155,116,51,30,24,17,10,7,4,8,0,11,0,2,225,282,141,4,87,44,60,52,74,135,157,113,57,44,26,29,17,8,7,4,4,2,10,57,125,182,100,33,27,41,39,35,50,69,92,66,30,11,10,11,9,6,5,10,4,1,7,9,17,24,21,29,28,48,38,30,21,26,25,35,10,9,4,4,4,3,5,4,4,4,3,5,10,16,28,47,63,40,49,28,22,18,27,18,10,5,8,7,3,2,2,4,1,4,19,59,167,235,130,57,45,46,42,40,49,64,96,54,27,17,18,15,7,6,2,3,1,2,21,88,187,253,130,77,47,49,48,53,77,109,147,109,45,41,35,16,13)
The code I'm trying to use is following:
tsdata = ts (data, frequency = 24)
aicvalstemp = NULL
aicvals= NULL
for (i in 1:5) {
for (j in 1:5) {
xreg1 = fourier(tsdata,i,24)
xreg2 = fourier(tsdata,j,168)
xregs = cbind(xreg1,xreg2)
armodel = auto.arima(bike_TS_west, xreg = xregs)
aicvalstemp = cbind(i,j,armodel$aic)
aicvals = rbind(aicvals,aicvalstemp)
}
}
The cbind command in the above command fails because the number of rows in xreg1 and xreg2 are different. I even tried using 1:length(data) argument in the fourier function but that also gave me an error. If someone can rectify the mistakes in the above code to produce a forecast of next 24 hours using an arima model with minimum AIC values, it would be really helpful. Also if you can include datasplitting in your code by creating training and testing data sets, it would be totally awesome. Thanks for your help.
I don't understand the desire to fit a weekly "season" to these data as there is no evidence for one in the data subset you provided. Also, you should really log-transform the data because they do not reflect a Gaussian process as is.
So, here's how you could fit models with a some form of hourly signals.
## the data are not normal, so log transform to meet assumption of Gaussian errors
ln_dat <- log(tsdata)
## number of hours to forecast
hrs_out <- 24
## max number of Fourier terms
max_F <- 5
## empty list for model fits
mod_res <- vector("list", max_F)
## fit models with increasing Fourier terms
for (i in 1:max_F) {
xreg <- fourier(ln_dat,i)
mod_res[[i]] <- auto.arima(tsdata, xreg = xreg)
}
## table of AIC results
aic_tbl <- data.frame(F=seq(max_F), AIC=sapply(mod_res, AIC))
## number of Fourier terms in best model
F_best <- which(aic_tbl$AIC==min(aic_tbl$AIC))
## forecast from best model
fore <- forecast(mod_res[[F_best]], xreg=fourierf(ln_dat,F_best,hrs_out))
I am trying to using ordinary kriging to spatially predict data where an animal will occur based on predictor variables using the gstat or automap package in R. I have many (over 100) duplicate coordinate points, which I cannot throw out since those stations were sampled multiple times over many years. Every time that I run the code below for ordinary kriging, I get an LDL error, which is due to the duplicate points. Does anyone know how to fix this problem without throwing out data? I have tried the code from the automap package that is supposed to correct for duplicates but I can't get that to work. Thank you for the help!
coordinates(fish) <- ~ LONGITUDE+LATITUDE
x.range <- range(fish#coords[,1])
y.range <- range(fish#coords[,2])
grd <- expand.grid(x=seq(from=x.range[1], to=x.range[2], by=3), y=seq(from=y.range[1], to=y.range[2], by=3))
coordinates(grd) <- ~ x+y
plot(grd, pch=16, cex=.5)
gridded(grd) <- TRUE
library(gstat)
zerodist(fish) ###146 duplicate points
v <- variogram(log(WATER_TEMP) ~1, fish, na.rm=TRUE)
plot(v)
vgm()
f <- vgm(1, "Sph", 300, 0.5)
print(f)
v.fit <- fit.variogram(v,f)
plot(v, model=v.fit) ####In fit.variogram(v, d) : Warning: singular model in variogram fit
krg <- krige(log(WATER_TEMP) ~ 1, fish, grd, v.fit)
## [using ordinary kriging]
##"chfactor.c", line 131: singular matrix in function LDLfactor()Error in predict.gstat(g, newdata = newdata, block = block, nsim = nsim,: LDLfactor
##automap code for correcting for duplicates
fish.dup = rbind(fish, fish[1,]) # Create duplicate
coordinates(fish.dup) = ~LONGITUDE + LATITUDE
kr = autoKrige(WATER_TEMP, fish.dup, grd)
###Error in inherits(formula, "SpatialPointsDataFrame"):object 'WATER_TEMP' not found
###somehow my predictor variables are no longer available when in a Spatial Points Data Frame??
automap::autoKrige expects a formula as first argument, try
kr = autoKrige(WATER_TEMP~1, fish.dup, grd)
automaphas a very simple fix for duplicate observations, and that is to discard them. So, automapdoes not really solves the issue you have. I see some options:
Discard the duplicates.
Slightly perturb the coordinates of the duplicates so that they are not on exactly the same location anymore.
Perform space-time kriging using gstat.
In regard to your specific issue, please make your example reproducible. What I can guess is that rbind of your fish object is not doing what you expect...
Alternatively you can use the function jitterDupCoords of geoR package.
https://cran.r-project.org/web/packages/geoR/geoR.pdf
In the past few days I have developed multiple PLS models in R for spectral data (wavebands as explanatory variables) and various vegetation parameters (as individual response variables). In total, the dataset comprises of 56. The first 28 (training set) have been used for model calibration, now all I want to do is to predict the response values for the remaining 28 observations in the tesset. For some reason, however, R keeps on the returning the fitted values of the calibration set for a given number of components rather than predictions for the independent test set. Here is what the model looks like in short.
# first simulate some data
set.seed(123)
bands=101
data <- data.frame(matrix(runif(56*bands),ncol=bands))
colnames(data) <- paste0(1:bands)
data$height <- rpois(56,10)
data$fbm <- rpois(56,10)
data$nitrogen <- rpois(56,10)
data$carbon <- rpois(56,10)
data$chl <- rpois(56,10)
data$ID <- 1:56
data <- as.data.frame(data)
caldata <- data[1:28,] # define model training set
valdata <- data[29:56,] # define model testing set
# define explanatory variables (x)
spectra <- caldata[,1:101]
# build PLS model using training data only
library(pls)
refl.pls <- plsr(height ~ spectra, data = caldata, ncomp = 10, validation =
"LOO", jackknife = TRUE)
It was then identified that a model comprising of 3 components yielded the best performance without over-fitting. Hence, the following command was used to predict the values of the 28 observations in the testing set using the above calibrated PLS model with 3 components:
predict(refl.pls, ncomp = 3, newdata = valdata)
Sensible as the output may seem, I soon discovered that all this piece of code generates are the fitted values of the PLS model for the calibration/training data, rather than predictions. I discovered this because the below code, in which newdata = is omitted, yields identical results.
predict(refl.pls, ncomp = 3)
Surely something must be going wrong, although I cannot seem to find out what specifically is. Is there someone out there who can, and is willing to help me move in the right direction?
I think the problem is with the nature of the input data. Looking at ?plsr and str(yarn) that goes with the example, plsr requires a very specific data frame that I find tricky to work with. The input data frame should have a matrix as one of its elements (in your case, the spectral data). I think the following works correctly (note I changed the size of the training set so that it wasn't half the original data, for troubleshooting):
library("pls")
set.seed(123)
bands=101
spectra = matrix(runif(56*bands),ncol=bands)
DF <- data.frame(spectra = I(spectra),
height = rpois(56,10),
fbm = rpois(56,10),
nitrogen = rpois(56,10),
carbon = rpois(56,10),
chl = rpois(56,10),
ID = 1:56)
class(DF$spectra) <- "matrix" # just to be certain, it was "AsIs"
str(DF)
DF$train <- rep(FALSE, 56)
DF$train[1:20] <- TRUE
refl.pls <- plsr(height ~ spectra, data = DF, ncomp = 10, validation =
"LOO", jackknife = TRUE, subset = train)
res <- predict(refl.pls, ncomp = 3, newdata = DF[!DF$train,])
Note that I got the spectral data into the data frame as a matrix by protecting it with I which equates to AsIs. There might be a more standard way to do this, but it works. As I said, to me a matrix inside of a data frame is not completely intuitive or easy to grok.
As to why your version didn't work quite right, I think the best explanation is that everything needs to be in the one data frame you pass to plsr for the data sources to be completely unambiguous.