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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