Timeline for Which attribute to choose first (the root) when making a decision tree?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 22, 2017 at 14:28 | vote | accept | Siyah | ||
| Mar 22, 2017 at 14:13 | comment | added | carlo | higher entropy. I see you don't understand how does it all work, so I suggest you to read some more complete explanations, like: stackoverflow.com/questions/1859554/…, or math.unipd.it/~aiolli/corsi/0708/IR/Lez12.pdf (first pages only) | |
| Mar 22, 2017 at 12:12 | comment | added | Siyah | You say no, but you also state that a node with 2 yes and 2 no is higher... higher what? Higher information gain? If so, why do you say no? Can you be more concise / precise please? | |
| Mar 21, 2017 at 23:43 | comment | added | carlo | no, entropy of a node wtih 2 yes and 2 no is higher than one with 3 yes and 5 no, so there's an information loss. however, this is averaged with the information gain from the other node, which has a lower entropy, and the overall information gain is positive | |
| Mar 21, 2017 at 19:18 | comment | added | Siyah | Okay, but does this mean that the information gain of an element with 2 yes and 2 no is higher than 3 yes 5 no? | |
| Mar 21, 2017 at 19:07 | comment | added | carlo | you choose the variable for performing the split for the information gain that you will get choosing it, and that's calculeted on nodes created by the split, so, yes. | |
| Mar 21, 2017 at 19:01 | comment | added | Siyah | But does this mean that you first need to check what comes AFTER you would hypothetically choose some root? I mean, do you choose the root here because it's 2 YES and 2 NO or because of the possibilities after you would choose this? | |
| Mar 21, 2017 at 17:58 | history | answered | carlo | CC BY-SA 3.0 |