I am using h2o version 3.10.4.8.
library(h2o)
h2o.init(nthreads = -1)
df <- as.h2o(data.frame(x = 1:5, y = 11:15))
I'm trying to understand how to use the apply() function in H2O.
The following works as expected:
h2o::apply(df, 2, mean)
h2o::apply(df, 2, sum)
h2o::apply(df, 2, function(x) {2*x + 1})
But this does not:
h2o::apply(df, 2, sd)
The error returned is:
[1] "Lookup failed to find is.H2OFrame" Error in .process.stmnt(stmnt, formalz, envs) : Don't know what to do with statement: is.H2OFrame x
I also thought that H2O is actually using its own functions to do the computation, so the following should work:
h2o::apply(df, 2, h2o.mean)
h2o::apply(df, 2, h2o.sum)
h2o::apply(df, 2, h2o.sd)
But it does not. The first two lines give the following error:
[1] "Lookup failed to find .newExpr" Error in .process.stmnt(stmnt, formalz, envs) : Don't know what to do with statement: .newExpr sd x na.rm
While the third line gives the following error:
[1] "Lookup failed to find .newExpr" Error in .process.stmnt(stmnt, formalz, envs) : Don't know what to do with statement: .newExpr sd x na.rm
What's going on and what are the things I should be aware of when passing stuff into FUN parameter in the apply() function? The documentation simply describes FUN as "the function to be applied".