Machine Learning

ðŸ‘‰ â™¡ supervised learning

â™¡ unsupervised learning

â™¡ reinforcement learning

recap:

ðŸ”– types of supervised learning

âœ” classification ðŸ“‘

âœ” regression ðŸ“ˆ

âœ” mixed âš—

- tree based :balloon:
- random forest
- neural networks
- support vector machines

ðŸŽ— enthropy

enthropy is just another word for expected value

in the past post, we decided what to use to split based on purity index. we can do the same thing mathematically (easier) by

P+ means probability of target (good day in our case)

P- means probability not target (bad day)

log2(P-) means log(P-) base 2

H = -(P+)log2(P+) - (P-)log2(P-)

the above is the formula for the purity of subset. it measures how likely you get a positive element if you select randomly from a particular subset

let us take teacher’s presence. we had

absent

good 5 bad 1

present

good 2 bad 1

H(absent) = -(5/6)log2(5/6) - (1/6)log2(1/6)

= 0.65

H(present) = -(2/3)log2(2/3) - (1/3)log2(1/3)

= 0.92

0 is extremely pure and 1 is extremely impure

so absent is purer than than present

ðŸŽ— gain

gain is used to determine what to split first by finding the feature with the highrst gain. 0 means irrelevent. 1 means very relevent

gain is defined by

gain = H(S) âˆ’ Î£ (SV/S) H(SV)

S -> total number of points in leaf

v is the possible values and SV -> subset for which we have those values

H is our enthropy as above

taking for presence, our gain is

our S is 9 that is (6+3)

gain = H(presence) - sum( SV/9 * H(SV))

gain = H(presence) - (6/9 * 0.65) - (3/9 * 0.92)

our H(presence) is -(7/9)log2(7/9) - (2/9)log2(2/9)

= 0.76

gain = 0.76 - (6/9 * 0.65) - (3/9 * 0.92)

gain = 0.02

exercise:

google up information gain related to decision trees as well as associated concepts

next:

random forest