Respected Madam, I am learning lots of things from your video. Madam, if P+ = 1, and P- = 0, then Entropy(S) = NaN, since, 0*log2(0) is NaN (not a number), but not zero. It is asymptotic. Thank You.
No. We find the limit of x->0 xlog2x . It can be found by LHospital's rule, and is equal to 0. It isn't the value of the function, it is the limit we consider at x->0.
Greetings Dr. Sudesha Sarkar, At 21:25 why is the Entropy[29+, 35-] is calculated as -29/64log2 29/64.. ... why divide by 64 when the formula does not have divide by total samples.
In Entropy formula, p+ is the probability of positive sample, which will be (no. of positive sample)/total sample. Similarly, p- is the probability of negative sample
@zinalpatel8962 According to my interpretation the 3rd point means that when only few examples are falling under a split those may be the outliers or the noisy examples. So to avoid overfitting the decision tree we should avoid that split.
This is probably one of the best tutorials I have found on net. Thanks a lot Madam, continue the good work.
Respected Madam, I am learning lots of things from your video. Madam, if P+ = 1, and P- = 0, then Entropy(S) = NaN, since, 0*log2(0) is NaN (not a number), but not zero. It is asymptotic. Thank You.
No. We find the limit of x->0 xlog2x . It can be found by LHospital's rule, and is equal to 0. It isn't the value of the function, it is the limit we consider at x->0.
Greetings Dr. Sudesha Sarkar,
At 21:25 why is the Entropy[29+, 35-] is calculated as -29/64log2 29/64.. ... why divide by 64 when the formula does not have divide by total samples.
In Entropy formula, p+ is the probability of positive sample, which will be (no. of positive sample)/total sample.
Similarly, p- is the probability of negative sample
Ohh gotcha. Thank you very much.
This is probably the best decision tree explanation I've come accross. Thank you madam.
Nice lecture,
In the topic 'when to stop ' i can't understand 3rd reason.
That means there is only one positive and negative point...you can choose from that you can choose any one of the feature
3 reasons are 1) completely dominant 2) partially dominant 3) when we run out of attributes
@zinalpatel8962
According to my interpretation the 3rd point means that when only few examples are falling under a split those may be the outliers or the noisy examples. So to avoid overfitting the decision tree we should avoid that split.
Great Professor!!!
Where can I get the content or PPT?
Just enroll to the course in nptel
( Introduction to machine learning )
thanks ma'am, have an exam tomorrow and this really helped
where is the slide ?
thank u mam
Nice series of tutorials.
thank you
Thank you mam
not good