Thanks! That got me there, with only the following minor adjustments (recorded
for posterity):
First, it's nnet.predict_nnet() (not nnet.predict())
Also, I had to convert my pandas DataFrame to an R data.frame this way:
import pandas.rpy.common as com
r_df = com.convert_to_r_dataframe(pd_df)
-Alessandro
From: Laurent Gautier <lgaut...@gmail.com<mailto:lgaut...@gmail.com>>
Date: Monday, December 2, 2013 3:45 PM
To: Alessandro
<alessandro.gaglia...@glassdoor.com<mailto:alessandro.gaglia...@glassdoor.com>>
Cc: "RPy help, support and design discussion list"
<rpy-list@lists.sourceforge.net<mailto:rpy-list@lists.sourceforge.net>>
Subject: Re: [Rpy] load nnet in rpy2
Ah, yes. Good you did. The licensing had all the attention.
The following is completely untested, but should get you going (and with some
luck, just work as it is)
from rpy2.robjects.packages import importr
base = importr('base')
nnet = importr('nnet')
# load: the side effect is to populate the R global environment
# with whatever in the file. I am assuming that 'err_net' comes from it.
base.load('err_net.RData')
from rpy2.robjects import globalenv
err_net = globalenv['err_net']
y = nnet.predict(err_net, x)
On 12/02/2013 01:47 PM, Alessandro Gagliardi wrote:
I didn't get a response last week so I figured I should try again:
How do I load a nnet model from a file in rpy2?
Thanks,
-Alessandro
From: Alessandro
<alessandro.gaglia...@glassdoor.com<mailto:alessandro.gaglia...@glassdoor.com>>
Date: Friday, November 22, 2013 4:27 PM
To: "rpy-list@lists.sourceforge.net<mailto:rpy-list@lists.sourceforge.net>"
<rpy-list@lists.sourceforge.net<mailto:rpy-list@lists.sourceforge.net>>
Subject: load nnet in rpy2
Hi all,
I created a nnet model called err_net in R that I've saved in err_net.RData
which I'd like to employ programmatically in a python module. If I were to do
it in IPython Notebook using rmagic I might have a cell that look something
like:
%%R -i x -o y
library(nnet)
load('err_net.RData')
y <- predict(err_net, x)
What would be the best way to do this without relying on magics?
Thanks in advance,
-Alessandro
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