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Classification.. continued. Prediction and Classification Last week we discussed the classification problem.. – Used the Naïve Bayes Method Today..we.

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Presentation on theme: "Classification.. continued. Prediction and Classification Last week we discussed the classification problem.. – Used the Naïve Bayes Method Today..we."— Presentation transcript:

1 Classification.. continued

2 Prediction and Classification Last week we discussed the classification problem.. – Used the Naïve Bayes Method Today..we will dive into more details.. But first how do we evaluate classifier

3 Abstract Binary Classification Problem Given n data samples where x i is a data vector and y i is label {-1,1}. Aim is to learn a function Such that f is “accurate” on unseen data. [ill-specified as defined]

4 Algorithms to Learn Classifier We can use an algorithm A to learn the function f: X  Y Then we write f as f A One example of A is Naïve Bayes. Other examples {Logistic Regression, Neural Networks, Support Vector Machines, Decision Trees, Random Forests,….}

5 Training vs. Test Data In practice to take care of the “unseen” part…we split the data into training and test sets We learn f A on the training set using an algorithm A The learned function f A is then evaluated on the test set.

6 Example Suppose we learn a function F on training set. Our test set consists of four data points (z1,1),(z2,- 1),(z3,1),(z4,-1). We apply F on the four data points (without labels) and we get F(z1)=1, F(z2)=1,F(z3)=-1 and F(z4) = -1. Then F correctly classified z1 and z4 but incorrectly classified z2 and z3.

7 Confusion Matrix Actual Label (1)Actual Label (-1) Predicted Label (1)True Positive (N1)False Positive (N2) Predicted Label (-1)False Negatives (N3)True Negatives (N4) Label 1 is called Positive, Label -1 is called Negative Let the number of test samples be N N = N1 + N2 + N3 + N4. True Positive Rate (TPR) = N1/(N1+N3) True Negative Rate (TNR) = N4/(N4+N2) False Positive Rate (FPR) = N2/(N2+N4) False Negative Rate (FNR) = N3/(N1+N3) Accuracy = (N1+N4)/(N1+N2+N3+N4) Precision = N1/(N1+N2)Recall = N1/(N1+N3)

8 Example Actual Label (1)Actual Label (-1) Predicted Label (1)103 Predicted Label (-1)220 TPR = 5/6; TNR = 20/23; FPR = 3/23; FNR = 2/12; Accuracy = 30/35 Precision = 10/13 and Recall = 10/12

9 ROC (Receiver Operating Characteristic) Curves Generally a learning algorithm A will return a real number…but what we want is a label {1 or -1} We can apply a threshold..T A T= True Label A T= True Label TPR = 3/4 FPR = 2/5 TPR = 2/4 FPR = 2/5

10 ROC Curve An ROC Curve is the plot where the x-axis is FPR, the y-axis is the TPR and for each threshold t, the point on the plot represents the pair (FPR(t), TPR(t)) Lets Look at the Wikipedia ROC EntryWikipedia ROC Entry

11 Discussion.. If F: Symptoms  {Disease, No-Disease} – Higher Recall or Precision ? – What is the relative cost of a mis-diagnosis (and which way) If F: Banner Ad  {Click, No-Click} – Higher Precision means more revenue?

12 Random Variables A r.v. is a numerical quantity associated with events in an experiment. Suppose we roll two dice. Let X = k be the sum of the two faces. X can take values ranging from {2….12}. P(X=12) = 1/36. Why ? – Event associated with X=12 is {(6,6)} P(X=7) = 6/36 = 1/6 – Associated Event: {(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)}

13 Random Variable A random variable X can take values in a set which is: – discrete and finite. Lets toss a coin and X = 1 if it’s a head and X=0 if it’s a tail. X is random variable – discrete and infinite (countable) Let X be the number of accidents in Sydney in a day.. Then X = 0,1,2,….. – Infinite (uncountable) Let X be the height of a Sydney-sider. – X = 150, , ,……

14 Random Variable Properties Let X be a discrete valued random variable taking values in a set S. The Expected (average) Value of X, E(X) is The Variance is

15 Examples Let X be a random variable which takes values 1 with probability p and 0 with probability 1-p. Then

16 Examples Let X be a random variable which denotes the number of “spam s” in a batch of n s. Assuming the probability of spam is p. X={0,1,2,3,4,5} X is a r.v. which follows a binomial distribution with parameters (n,p)… X ~ Binomial(n,p) – E(X) = np ; Var(X) = np(1-p)

17 Examples Let X be a random variable which denotes the number of tcp packets that arrive in a unit time. Then X can be modeled to follow a Poisson distribution.. E(X) = Var(X) = λ

18 Continuous Distribution Ofcourse the most common continuous distribution is the Normal/Gaussian distribution… denoted

19 How to use r.v. for classification To use r.v. in classification…we have to make an assumption. – For example..Sepal Length follows a Normal Distribution. – Is this a good/reasonable assumption. Then we use data to estimate the parameters of the distribution.. – The parameters of a Normal distribution are the mean and the variance (square of standard deviation). – For the moment we can just use Matlab/program to do that… – Once we have the parameters we can use the distribution to estimate the “probability” of Sepal Length taking a new value..

20 Fitting Distributions..Examples 0,1,0,1,0,0 – Assume data from a Binomial distribution with 6 trials and 2 successes In Matlab:>> binofit(2,6) = ,20,5,3,3,100 – Assume data is from a Poisson distribution – X=[ ]; – Poissfit(X); – Ans: What is happening ? We are just taking sample averages. The more data we have the more reliable these estimates become.. Suppose we take Sepal Length…data vector x >> [mean,std] = normfit(x); >> ans: mean = 5.8, std=0.81

21 Return to the Iris Example We will redo the Iris Classification Example..but now will use “continuous” values for the attributes…


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