Author | Message |
---|---|
batcat
Posts: 11
|
Posted 18:56 Aug 25, 2009 |
See attachment |
p0941
Posts: 95
|
Posted 12:26 Aug 26, 2009 |
Only one comment. The naive bayesian classification fails where the training data pool is too small to add one to zero as numerator and to denominator. |
xieguahu
Posts: 50
|
Posted 15:40 Aug 26, 2009 |
For the test record 2, The final calculation should be
P(fish | X2) = (a) x (1/12) x ( .5) = .042a |
batcat
Posts: 11
|
Posted 22:32 Aug 26, 2009 |
I don't think I agree. The probability of class fish is 2/12 not 1/12 |
HelloWorld
Posts: 88
|
Posted 18:17 Aug 29, 2009 |
Confirming this answer is right.. |
alomo
Posts: 70
|
Posted 16:39 Aug 30, 2009 |
Could you explain why in P(X2|reptile) = ... = 2/3 x 2/3 x 2/3 x 0/3 = 0 I am getting 3/3 x 3/3 x 3/3 x 0/3. It will not change the result, but why 2/3 and not 3/3? |
HelloWorld
Posts: 88
|
Posted 16:51 Aug 30, 2009 |
I got the same with you
P(X2 | Reptile) = P(Body Temperature = cold-blooded | Reptile) x P(Skin Cover = scales | Reptile) x P(Gives Birth = no | Reptile) x P(Aquatic Creature = yes | Reptile) x P(Aerial Creature = no | Reptile) x P(Has Legs = no | Reptile) x P(Hibernates = no | Reptile) = 3/3 x 3/3 x 3/3 x 0/3 = 0 |
batcat
Posts: 11
|
Posted 17:20 Aug 30, 2009 |
Yes sorry those were all typos Should be 3/3's not 2/3's |
cysun
Posts: 2935
|
Posted 16:36 Aug 31, 2009 |
p0941 is correct. We cannot let one zero term to zero out all others. The way to deal with this is to add 1 to all counts - if the sample size is sufficiently large, adding one would not affect the results. Sorry I forgot to mention this in the class. |
p0941
Posts: 95
|
Posted 23:44 Aug 31, 2009 |
As I can remember you did talk about this in the class |