I salute the hard work with which you created this video..i think you are hanging mobile on some thread 5 years ago...and putting almost full day to create this video...
1)What about the temperature in the final internal node?It wont be used as node? 2) If there is rain->temp(mild/cool)-> Humidity(high/normal)-> Wind (weak/strong); and I want to take the decision including humidity..? Anyway as per the output you got if outlook is"rain",we exclude humidity..Will that be possible in practical scenario,Can you pull me out out in this regard?
Literally I was not at all understanding anything in from text book but by watching ur video I understood very nicely that even I can explain others ...grt explanation tq so mch ..
in this example for sunny , we got gain factor as humidity and in the humidity we got complete yes for "high" and complete no for "normal". But if high falls on both yes and no , during that time which all columns i should consider to perform gain_factor, i mean should i need to consider outlook column once again
Very nice and clear explanation After watching incomplete lecture of edureka I was searching for how to select next child nodes Finally I got u Thank u bro
really good explanation, sir. but sir i have a query? when there wind' has two element weak and strong' and there when weak then all elements are yes. and other side when strong has all elements ar no thats why easily we can create tree . but when weak has mixed yes and no... then what can i do? please sir answered me.
you have really worked hard to explain such a long procedure...it was great..i would appreciate if you can explain KNN algorithm same way..keep it up..thank you
I checked online the formula of entropy and information gain and they are exactly opposite of those used by you i.e. the formula of information gain used by you is actually the formula of entropy...what is the reason behind this?
U had 2 options to start working again, sunny and Rainy, and u started with sunny. Is there any specific rule to choose which node i want to start working with first, in such cases like yours?
i m confused about temperature attribute ...where have we to put that??as descendant to humidity u directly put yes or no...either we also check here for temperature or not??
Thank a ton for this video, I got this concept very clear. The only thing I wanna ask is that at 27:44 you said that for every "strong" there is a "no" and for every "weak" there is a "yes", what if for 2 "weak" there is "yes" but for 1 "weak" there is "no"..?
The video is good but it would be great if you can summarize the steps at the end of every video. It's a long process so remembering it as an algorithm will be useful.
Good explanation. But you should have mentioned that this method is done using "ID3 algorithm". For "C4.5 algorithm" Gain Ratio is used and for "CART algorithm" 'GINI index' is used.
The formula for information gain you have used here is actually a formula for entropy. The formula for information gain is : IG = entropy of total data set - summation of p/p+n*entrop(vi) where i=ith value
Awesome, Perfect and Clear explanation. Thanks a lot to share your valuable lectures with us. If you share lectures document, then we shall be more benefited. .
i really understand clearly this. your method was out class but if you used your Native language than it will be more brilliant. English is not the purpose of this topic but the main thing is Decision Tree.
payal bajpai Thank you so much for appreciation, share it with your friends This is referred as decision tree, but on some platforms like R they are referred to by modern term CART
I think the reason that you gave for not considering "Humidity" as an attribute after going downwards from "Rain" branch is not right. "Humidity" does not come simply because its Gain is lower than "Wind" attribute with respect to T_rainy table. Same attribute can appear in multiple paths. But in a single path, same attribute will come only once.
sir is it possible to use bayesian classification with nearest neighbor clustering...if yes please explain please provide me answer as soon as possible.... thanking you
Suppose I wish to employ a one-level decision tree (i.e, a decision stump) to classify data. Recall that decision trees split data into subsets, and there are many measures that can be used to determine the best way to split data. I’m interested in two measures of impurity: Gini index and classification error. Help me decide how to classify the above data according to Feature 1 by completing the following tasks. (a) Draw and label a line to indicate the best split as measured by Gini index. (b) Draw and label a line to indicate the best split as measured by classification error
If there is a unique identifier let us suppose product_id In that case the split on product_id will give you large number of partitions as it is unique identifier each one containing one tuple, that means every partition is pure We can directly say that infogain of product_id=0, and it is useless to calculate
Amazing explanation, Thank you so much! Just having a doubt that "we are not considering temperature in our Decision Tree, is it okay to skip temperature?"
You can imagine it this way - just by knowing the roll number of a student one cannot determine whether student will pass or not. The day attribute over here is day number, a unique identifier for each day. Knowing day number doesn't contribute to your decision making. In fact, whenever an attribute column has unique values it cannot be used for decision making.
