在R中使用randomForest包，如何映射随机森林预测？

Question

enter image description here I 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.

Answer 1

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)

Answer 2

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)