GH-15991: Infogram pydocs updates

[shaunyogeshwaran]

fixes #15991
Crafted new examples to showcase the different parameters.

[tomasfryda]

Looks good to me, thank you @shaunyogeshwaran !

[valenad1]

Out of scope but I don't understand Default is -1 which gets set to 0.1.

[valenad1]

Also the first line is not correct since the default is -0.1.

A number between 0 and 1 representing a threshold for total information, defaulting to 0.1.

[shaunyogeshwaran]

cc: @hannah-tillman

[hannah-tillman]

@wendycwong any ideas about what's going on with the default values here?

(I can update the schema and fix the first line issue)

[valenad1]

@hannah-tillman I understand it now.

It comes from here:

https://github.com/h2oai/h2o-3/blob/master/h2o-admissibleml/src/main/java/hex/Infogram/Infogram.java#L185-L187

@wendycwong any reason why we not set it directly here?

Nevertheless, its out of scope of this PR. @shaunyogeshwaran.

[wendycwong]

@valenad1 @hannah-tillman

The value of -1 or -0.1 is used to denote that the user has not set any value. If the user has not set any value, we will set it to a default value of 0.1. There is a reason that the code needs to know if the user set that value. I cannot remember what it is now.

[valenad1]

Does this work the same as split_frame?

[shaunyogeshwaran]

cc: @hannah-tillman

[valenad1]

Same here. First line says number between 0-1,... Default -1 sets to 0.1..

[valenad1]

Same here. First line says number between 0-1,... Default -1 sets to 0.1..

[valenad1]

Same here. First line says number between 0-1,... Default -1 sets to 0.1..

[wendycwong]

I think @valenad1 has a good point. We could just set those values to 0.1 as default instead of setting it to -1 and then set it to 0.1.

[wendycwong]

Let me make a quick change in a new PR.

[valenad1]

All the examples works, but I would use different number than default to show how the parameter change the result, and I would rather avoid any unused stuff in the example. So the test should be used for some metric be deleted from example

[valenad1]

Thanks for the updates @shaunyogeshwaran. Can you please run the all the examples ones again. E.g this will fail for sure.

[valenad1]

And please merge it to re-3.46.0, the rel-3.44.0 is not used anymore.

[valenad1]

@wendycwong

import h2o
from h2o.estimators.infogram import H2OInfogram
h2o.init()
f = "https://erin-data.s3.amazonaws.com/admissible/data/taiwan_credit_card_uci.csv"
col_types = {'SEX': "enum", 'MARRIAGE': "enum", 'default_payment_next_month': "enum"}
df = h2o.import_file(path=f, col_types=col_types)
y = "default_payment_next_month"
x = df.columns
x.remove(y)
gbm_params = {'ntrees':3}
pcols = ["SEX", "MARRIAGE", "AGE"]
ig = H2OInfogram(protected_columns=pcols)
ig.train(y=y, x=x, training_frame=df, algorithm_params=gbm_params)
ig.algorithm_params

gives me:

TypeError                                 Traceback (most recent call last)
/var/folders/2z/9gdhqbns0djdj9crb242sgc00000gn/T/ipykernel_9099/4280293040.py in <cell line: 13>()
     11 pcols = ["SEX", "MARRIAGE", "AGE"]
     12 ig = H2OInfogram(protected_columns=pcols)
---> 13 ig.train(y=y, x=x, training_frame=df, algorithm_params=gbm_params)
     14 ig.algorithm_params

~/h2o-3/_valenad-notebooks/../h2o-py/build/main/h2o/estimators/infogram.py in train(self, x, y, training_frame, verbose, **kwargs)
   1202                                                                                    " and <= 1."
   1203 
-> 1204         parms = sup._make_parms(x,y,training_frame, extend_parms_fn = extend_parms, **kwargs)
   1205 
   1206         sup._train(parms, verbose=verbose)

TypeError: _make_parms() got an unexpected keyword argument 'algorithm_params'.

Ticket here.

The base branch was changed.