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Question: How do I format the 'const' argument for the predict function in the Terra package?

Background and details:

I am using Terra's predict function to create spatial maps from a model. The function worked as expected when I created full rasters for my three constant values. When I saw predict has an argument 'const' for constants, I decided to use that instead. Unfortunately, the help document doesn't specify how to structure the data beyond that they should be in a dataframe.

Argument description for 'const' from ?terra::predict:

data.frame. Can be used to add a constant value as a predictor variable so that you do not need to make a SpatRaster layer for it

The argument is not used in any of the examples listed at the bottom of the document either.

What I tried:

I tried formating the dataframe as a single row with named columns for each constant value that is required in the model. The function accepted it with the warning:

Warning messages: 1: In data.frame(..., check.names = FALSE) : row names were found from a short variable and have been discarded

But this produced clearly incorrect values in the resulting outputs. Possibly it was just using the first value for all three constants, but I don't know.

I also tried formating the dataframe with two columns, one for the variable name and one for the variable values. This failed with the following error:

Error in h(simpleError(msg, call)) : error in evaluating the argument 'x' in selecting a method for function 'writeRaster': arguments imply differing number of rows: 10000, 3

10000 may be referring to the dimenions of the raster data (10k x 10k) and the 3 is likely referring to the three rows in the constant dataframe I supplied.

I don't want to go back to writing rasters for each constant because I am producing many maps and each one would require 3 10k x 10k rasters to be generated with the constant values, which is time consuming and may lead to memory issues.

Any suggestions would be greatly appreciated!

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1 Answer 1

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Here is an illustration

library(terra)
logo <- rast(system.file("ex/logo.tif", package="terra"))   
names(logo) <- c("red", "green", "blue")
p <- matrix(c(48, 48, 48, 53, 50, 46, 54, 70, 84, 85, 74, 84, 95, 85, 66, 42, 26, 4, 19, 17, 7, 14, 26, 29, 39, 45, 51, 56, 46, 38, 31, 22, 34, 60, 70, 73, 63, 46, 43, 28), ncol=2)
a <- matrix(c(22, 33, 64, 85, 92, 94, 59, 27, 30, 64, 60, 33, 31, 9,  99, 67, 15, 5, 4, 30, 8, 37, 42, 27, 19, 69, 60, 73, 3, 5, 21,37, 52, 70, 74, 9, 13, 4, 17, 47), ncol=2)
xy <- rbind(cbind(1, p), cbind(0, a))
e <- extract(logo, xy[,2:3])
v <- data.frame(cbind(pa=xy[,1], e))

model <- glm(formula=pa~., data=v)

names(model$coefficients)[-1]
#[1] "red"   "green" "blue" 
names(logo)
#[1] "red"   "green" "blue" 

# standard approach
r <- predict(logo, model)

# one or more constants 
r <- predict(logo[[2:3]], model, const=data.frame(red=0))
r <- predict(logo$blue, model, const=data.frame(red=100, green=0))

# you could even use constants only 
x <- logo[[1]]
names(x) <- "A"
r <- predict(x, model, const=data.frame(blue=50, red=100, green=0))

The output is as expected:

predict(model, data.frame(blue=50, red=100, green=0))
#        1 
#-4.538653 

r[1]
#          A
#1 -4.538653
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3 Comments

I'm still having an issue with this. I'm no longer getting the warning message, but I am getting clearly incorrect results in the output. Using rasters for the constants provids good results but using the const argument does not. It appears some data layers are being ignored or misassigned when I use the const argument. I verified that the values are numeric and the column names are identical to the predictor names.
I ran the predict function several times with different const arguments. I am getting very similar (though not identical) results when I entered the actual values (27.67, 180, 1093.15), all zeros (0, 0, 0) and all strings ("A", "A", "A"). This is especially weird because the three constant variables make up about 95% of the model weights so any changes should be obvious.
All I can say is that the example above works. That does not mean that it works in all cases. If something similar does not work for you than you should make an example like the above that shows that something does not work, and ask a new question. Using strings is suspect.

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