Number of attributes for tree construction in RandomForest

Legacy User
Legacy User New Altair Community Member
edited November 5 in Community Q&A
Hi,

how can I define the number of attributes (the subset of all attributes
in the learning set) randomly chosen to grow the trees in the
forest?

Could you give me an example how to specify that n^1/2
attributes are considered.

Thanks.

Cheers,
Stephan
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Answers

  • haddock
    haddock New Altair Community Member
    I guess the answer to the first question is ...

    subset ratio: Ratio of randomly chosen attributes to test (-1: use log(m)
    + 1 features) (real; -1.0-1.0) ( Manual 5.4.53 )

    But I ponder what the answer to the second question would be if there were, say, 5 attributes  ;)
  • Legacy User
    Legacy User New Altair Community Member
    Thank you for your answer.

    But I still didn't catch what is meant with "(-1: use log(m) + 1 features) ":
    Does it mean that with a total of 'm' attributes the setting of
    subset_ratio=-1.0 yields log(m)+1 attributes used for the test?

    So, what does then the setting 1.0 mean?

    And I still don't see a way to define a specific number of attributes like my square example.
  • haddock
    haddock New Altair Community Member
    Hi Stephan,

    I think that a setting of 1.0 would mean "use all the attributes" and a setting of 0.0 would mean "use none of the attributes", neither of which are of any interest to random foresters, as far as I remember.

    A setting of 0.1 might be more like it, meaning "use a random selection of 10% of the attributes".

    So what does "-1" mean? In general this convention is used to indicate a specific setting that is calculated, rather than part of the selection range, so -0.8 would fuse the system and cause springs to fall out of the back of your machine.

    You see this convention used with quite a number of the operators.

    If I'm wrong no doubt heads more pointy than mine will intervene... :D
  • IngoRM
    IngoRM New Altair Community Member
    Hi,

    If I'm wrong no doubt heads more pointy than mine will intervene...
    no need for that  :D

    Everything was perfectly right.

    All the best,
    Ingo