Closed reynoldscem closed 7 years ago
This was calculated using research-grade pre-release code, which perhaps got cut out of the public release. But this is easy enough to do with a Python script:
import fileinput
import sys
p = 0.859
s = 0.
n = 0.
for line in fileinput.input():
try:
s += abs(p - float(line))
n += 1
except:
continue
print s/n
You can run it using the following command:
gzcat learning_traces.13m.csv.gz | cut -d ',' -f 1 | python fixedmae.py
The result is slightly different (0.199606186716
) because this is making predictions over the whole data set and the published result is just for the test set (last 10%). But you get the idea. 😄
Good luck!
I have tried to reproduce your figure of
0.175
MAE for the fixed p-bar of 0.859. On neither the test set, nor the training set can I get this value. Could you detail for me how you have calculated it?For context I am seeing if I can reproduce some of your results, or try various different modules on the same set of data. Firstly, I would like to reproduce your results as a baseline, but this one has eluded me.
Thanks.