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- ID3(
,
,
)
- Create a
node for the tree - If all
are positive, Return the single-node tree
, with label = + - If all
are negative, Return the single-node tree
, with label = - - If
is empty, Return the single-node tree
, with label = most common value of
in
- Otherwise Begin
-
the attribute from
that best
classifies
- The decision attribute for
- For each possible value,
, of
,
- Add a new tree branch below
, corresponding to the test
- Let
be the subset of
that have
value
for
- If
is empty
- Then below this new branch add a leaf node with label = most
common value of
in
- Else below this new branch add the subtree
- End
- Return
Patricia Riddle
Fri May 15 13:00:36 NZST 1998