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gdancik avatar gdancik commented on July 29, 2024

This should not be the case. Can you provide a reproducible example?

For me the code below with 100 observations and 2 covariates takes about 1.7 seconds to run.

t1 <- Sys.time()
df <- data.frame(x1 = 1:100, x2 = 1:2)
y <- df$x1 + 3*df$x2 + rnorm(nrow(df))

fit = mlegp(df, y)

predict(fit)
predict(fit, data.frame(x1 = 3, x2 = 2))

t2 <- Sys.time()

print(t2-t1)

Here is my sessionInfo():

sessionInfo()
R version 4.0.4 (2021-02-15)
Platform: x86_64-apple-darwin17.0 (64-bit)
Running under: macOS Mojave 10.14.6

Matrix products: default
BLAS: /System/Library/Frameworks/Accelerate.framework/Versions/A/Frameworks/vecLib.framework/Versions/A/libBLAS.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/4.0/Resources/lib/libRlapack.dylib

locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8

attached base packages:
[1] stats graphics grDevices utils datasets
[6] methods base

other attached packages:
[1] mlegp_3.1.8

loaded via a namespace (and not attached):
[1] compiler_4.0.4 tools_4.0.4 shinyjs_2.0.0

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phit0 avatar phit0 commented on July 29, 2024

Hi, sorry I think my initial statement was not correct.
I meant, for predicting on a 1000 x 2 matrix it takes around 2 minutes.

t1 <- Sys.time()
df <- data.frame(x1 = 1:100, x2 = 1:2)
y <- df$x1 + 3*df$x2 + rnorm(nrow(df))

fit = mlegp(df, y)

predict(fit)
predict(fit, data.frame(x1 = seq(1:1000), x2 = runif(1000, min = 1, max = 2)))

t2 <- Sys.time()

print(t2-t1)
> print(t2-t1)
Time difference of 2.209696 mins
> sessionInfo()
R version 3.6.3 (2020-02-29)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Ubuntu 20.04.2 LTS

Matrix products: default
BLAS:   /usr/lib/x86_64-linux-gnu/blas/libblas.so.3.9.0
LAPACK: /usr/lib/x86_64-linux-gnu/lapack/liblapack.so.3.9.0

locale:
 [1] LC_CTYPE=en_US.UTF-8       LC_NUMERIC=C              
 [3] LC_TIME=de_DE.UTF-8        LC_COLLATE=en_US.UTF-8    
 [5] LC_MONETARY=de_DE.UTF-8    LC_MESSAGES=en_US.UTF-8   
 [7] LC_PAPER=de_DE.UTF-8       LC_NAME=C                 
 [9] LC_ADDRESS=C               LC_TELEPHONE=C            
[11] LC_MEASUREMENT=de_DE.UTF-8 LC_IDENTIFICATION=C       

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] mlegp_3.1.8   

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phit0 avatar phit0 commented on July 29, 2024

This was the original code I used:

x <- matrix(runif(60), ncol = 2)
y <- sin(x %*% rep(10, 2))
library(microbenchmark)
gp <- mlegp::mlegp(x, y)
microbenchmark(fitgp <- mlegp::predict.gp(gp, newData = data.frame(X1 = runif(1000), X2 = runif(1000))),
               times = 10, 
               unit = "s")

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gdancik avatar gdancik commented on July 29, 2024

This, unfortunately, is expected behavior for a large number of inputs.

For each predicted value, we need to calculate the correlation between it and each of the original inputs (100 calculations in the above example). This is then repeated for each input in the prediction matrix (repeated 1000 times in the above example).

The original code for this was written over 10 years ago and can possibly be optimized (or written in C). I will take a closer look when time allows, but if you want to make any changes, pull requests are welcome. The relevant code is predict.gp() which calls predictNewZ() which calls calcCorOneObs().

from mlegp.

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