Klasifikasi
Data Mining
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Klasifikasi 1 Decision Tree Induction
2 Bayesian Classification 3 Neural Network 4 Model Evaluation and Selection
5 Techniques to Improve Classification Accuracy: Ensemble Methods 2
1 Decision Tree
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Tahapan Algoritma Decision Tree 1. Siapkan data training 2. Pilih atribut sebagai akar n
Entropy( S ) pi * log 2 pi i 1
n
| Si | * Entropy( S i ) i 1 | S |
Gain( S , A) Entropy( S )
3. Buat cabang untuk tiap-tiap nilai 4. Ulangi proses untuk setiap cabang sampai semua kasus pada cabang memiliki kelas yg sama 4
1. Siapkan data training
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2. Pilih atribut sebagai akar • Untuk memilih atribut akar, didasarkan pada nilai Gain tertinggi dari atribut-atribut yang ada. Untuk mendapatkan nilai Gain, harus ditentukan terlebih dahulu nilai Entropy • Rumus Entropy:
n
Entropy( S ) pi * log 2 pi i 1
• S = Himpunan Kasus • n = Jumlah Partisi S • pi = Proporsi dari Si terhadap S
• Rumus Gain: • • • • •
n
| Si | Gain( S , A) Entropy( S ) * Entropy( S i ) i 1 | S |
S = Himpunan Kasus A = Atribut n = Jumlah Partisi Atribut A | Si | = Jumlah Kasus pada partisi ke-i | S | = Jumlah Kasus dalam S 6
Perhitungan Entropy dan Gain Akar
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Penghitungan Entropy Akar • Entropy Total • Entropy (Outlook)
• Entropy (Temperature)
• Entropy (Humidity)
• Entropy (Windy) 8
Penghitungan Entropy Akar NODE 1
JML KASUS TIDAK YA (Si) ENTROPY (S) (Si) 14 10 4 0,86312
ATRIBUT TOTAL OUTLOOK
4 5 5
4 4 2
0 1 3
0 0,72193 0,97095
4 4 6
0 2 2
4 2 4
0 1 0,91830
NORMAL
7 7
4 7
3 0
0,98523 0
FALSE TRUE
8 6
2 4
6 2
0,81128 0,91830
CLOUDY RAINY SUNNY
TEMPERATURE COOL HOT MILD
HUMADITY HIGH
WINDY
9
GAIN
Penghitungan Gain Akar
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Penghitungan Gain Akar NODE 1
ATRIBUT TOTAL OUTLOOK
JML KASUS (S) 14
YA (Si) 10
TIDAK (Si) 4
ENTROPY
GAIN
0,86312
0,25852 CLOUDY RAINY
SUNNY
4 5 5
4 4 2
0 1 3
0 0,72193 0,97095
TEMPERATURE
0,18385 COOL
HOT MILD
4 4 6
0 2 2
4 2 4
0 1 0,91830
HUMIDITY
0,37051 HIGH NORMAL
7 7
4 7
3 0
0,98523 0
WINDY
0,00598 FALSE TRUE
8 6
2 4
11
6 2
0,81128 0,91830
Gain Tertinggi Sebagai Akar • Dari hasil pada Node 1, dapat diketahui bahwa atribut dengan Gain tertinggi adalah HUMIDITY yaitu sebesar 0.37051 • Dengan demikian HUMIDITY dapat menjadi node akar
• Ada 2 nilai atribut dari HUMIDITY yaitu HIGH dan NORMAL. Dari kedua nilai atribut tersebut, nilai atribut NORMAL sudah mengklasifikasikan kasus menjadi 1 yaitu keputusan-nya Yes, sehingga tidak perlu dilakukan perhitungan lebih lanjut • Tetapi untuk nilai atribut HIGH masih perlu dilakukan perhitungan lagi
1. HUMIDITY
High
1.1 ????? 12
Normal
Yes
2. Buat cabang untuk tiap-tiap nilai • Untuk memudahkan, dataset di-filter dengan mengambil data yang memiliki kelembaban HUMIDITY=HIGH untuk membuat Tabel Node 1.1
OUTLOOK Sunny Sunny Cloudy Rainy Sunny Cloudy Rainy
TEMPERATURE Hot Hot Hot Mild Mild Mild Mild
HUMIDITY High High High High High High High
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WINDY FALSE TRUE FALSE FALSE FALSE TRUE TRUE
PLAY No No Yes Yes No Yes No
Perhitungan Entropi Dan Gain Cabang NODE 1.1
JML KASUS TIDAK YA (Si) (S) (Si) 7 3 4
ATRIBUT HUMIDITY OUTLOOK
ENTROPY
GAIN
0,98523
0,69951 CLOUDY RAINY SUNNY
2 2 3
2 1 0
0 1 3
0 1 0
TEMPERATURE
0,02024 COOL HOT MILD
0 3 4
0 1 2
0 2 2
0 0,91830 1
WINDY
0,02024 FALSE TRUE
4 3
2 1
14
2 2
1 0,91830
Gain Tertinggi Sebagai Node 1.1 • Dari hasil pada Tabel Node 1.1, dapat diketahui bahwa atribut dengan Gain tertinggi adalah OUTLOOK yaitu sebesar 0.69951 •
Dengan demikian OUTLOOK dapat menjadi node kedua
• Artibut CLOUDY = YES dan SUNNY= NO sudah mengklasifikasikan kasus menjadi 1 keputusan, sehingga tidak perlu dilakukan perhitungan lebih lanjut •
Tetapi untuk nilai atribut RAINY masih perlu dilakukan perhitungan lagi
1. HUMIDITY High
1.1 OUTLOOK
Cloudy Rainy
1.1.2 ?????
Yes
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Normal
Yes
Sunny
No
3. Ulangi proses untuk setiap cabang sampai semua kasus pada cabang memiliki kelas yg sama OUTLOOK Rainy Rainy
TEMPERATURE HUMIDITY Mild High Mild High
NODE 1.1.2
WINDY FALSE TRUE
PLAY Yes No
JML KASUS YA (Si) TIDAK (Si) ENTROPY (S)
ATRIBUT HUMADITY HIGH & OUTLOOK RAINY TEMPERATURE
2
1
1
GAIN
1 0
COOL HOT MILD
0 0 2
0 0 1
0 0 1
0 0 1
WINDY
1 FALSE TRUE
1 1
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1 0
0 1
0 0
Gain Tertinggi Sebagai Node 1.1.2 1. HUMIDIT Y
• Dari tabel, Gain Tertinggi adalah WINDY dan menjadi node cabang dari atribut RAINY
High
1.1 OUTLOOK
• Karena semua kasus sudah masuk dalam kelas • Jadi, pohon keputusan pada Gambar merupakan pohon keputusan terakhir yang terbentuk
Normal
Cloudy
Yes
Sunny
Rainy
1.1.2 WINDY
Yes False
Yes
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No True
No
Decision Tree Induction: An Example • Training data set: Buys_computer
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age
income
student
credit_rating
buys_computer
<=30 <=30 31…40 >40 >40 >40 31…40 <=30 <=30 >40 <=30 31…40 31…40 >40
high high high medium low low low medium low medium medium medium high medium
no no no no yes yes yes no yes yes yes no yes no
fair excellent fair fair fair excellent excellent fair fair fair excellent excellent fair excellent
no no yes yes yes no yes no yes yes yes yes yes no
Overfitting and Tree Pruning • Overfitting: An induced tree may overfit the training data • Too many branches, some may reflect anomalies due to noise or outliers • Poor accuracy for unseen samples • Two approaches to avoid overfitting 1. Prepruning: Halt tree construction early ̵ do not split a node if this would result in the goodness measure falling below a threshold • Difficult to choose an appropriate threshold
2. Postpruning: Remove branches from a “fully grown” tree -get a sequence of progressively pruned trees • Use a set of data different from the training data to decide which is the “best pruned tree” 19
Pruning
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Why is decision tree induction popular? • Relatively faster learning speed (than other classification methods) • Convertible to simple and easy to understand classification rules • Can use SQL queries for accessing databases • Comparable classification accuracy with other methods
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Latihan • Lakukan eksperimen untuk mengumpulkan dataset yang memiliki 4-5 atribut dan analisis dengan decision tree pada dataset tersebut.
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2 Bayesian Classification
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Bayesian Classification: Why? • A statistical classifier: performs probabilistic prediction, i.e., predicts class membership probabilities • Foundation: Based on Bayes’ Theorem. • Performance: A simple Bayesian classifier, naïve Bayesian classifier, has comparable performance with decision tree and selected neural network classifiers • Incremental: Each training example can incrementally increase/decrease the probability that a hypothesis is correct — prior knowledge can be combined with observed data • Standard: Even when Bayesian methods are computationally intractable, they can provide a standard of optimal decision making against which other methods can be measured 24
Tahapan Algoritma Naïve Bayes 1. Baca Data Training 2. Hitung jumlah class 3. Hitung jumlah kasus yang sama dengan class yang sama 4. Kalikan semua nilai hasil sesuai dengan data X yang dicari class-nya
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1. Baca Data Training
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Teorema Bayes P ( H | X)
P( X | H ) P( H )
P ( X | H ) P ( H ) / P ( X)
P ( X) • X • H • • • •
Data dengan class yang belum diketahui Hipotesis data X yang merupakan suatu class yang lebih spesifik P (H|X) Probabilitas hipotesis H berdasarkan kondisi X (posteriori probability) P (H) Probabilitas hipotesis H (prior probability) P (X|H) Probabilitas X berdasarkan kondisi pada hipotesis H P (X) Probabilitas X 27
2. Hitung jumlah class/label • Terdapat 2 class dari data training tersebut, yaitu: • C1 (Class 1) Play = yes 9 record • C2 (Class 2) Play = no 5 record • Total = 14 record
• Maka: • P (C1) = 9/14 = 0.642857143 • P (C2) = 5/14 = 0.357142857
• Pertanyaan: • Data X = (outlook=rainy, temperature=cool, humidity=high, windy=true) • Naik gunung atau tidak? 28
3. Hitung jumlah kasus yang sama dengan class yang sama • Untuk P(Ci) yaitu P(C1) dan P(C2) sudah diketahui hasilnya di langkah sebelumnya. • Selanjutnya Hitung P(X|Ci) untuk i = 1 dan 2 • P(outlook=“sunny”|play=“yes”)=2/9=0.222222222 • P(outlook=“sunny”|play=“no”)=3/5=0.6 • P(outlook=“overcast”|play=“yes”)=4/9=0.444444444 • P(outlook=“overcast”|play=“no”)=0/5=0 • P(outlook=“rainy”|play=“yes”)=3/9=0.333333333 • P(outlook=“rainy”|play=“no”)=2/5=0.4 29
3. Hitung jumlah kasus yang sama dengan class yang sama • Jika semua atribut dihitung, maka didapat hasil akhirnya seperti berikut ini: Atribute Outlook Outlook Outlook Temperature Temperature Temperature Humidity Humidity Windy Windy
Parameter value=sunny value=overcast value=rainy value=hot value=mild value=cool value=high value=normal value=false value=true
No 0.6 0.0 0.4 0.4 0.4 0.2 0.8 0.2 0.4 0.6
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Yes 0.2222222222222222 0.4444444444444444 0.3333333333333333 0.2222222222222222 0.4444444444444444 0.3333333333333333 0.3333333333333333 0.6666666666666666 0.6666666666666666 0.3333333333333333
4. Kalikan semua nilai hasil sesuai dengan data X yang dicari class-nya • Pertanyaan:
• Data X = (outlook=rainy, temperature=cool, humidity=high, windy=true) • Naik gunung atau tidak?
• Kalikan semua nilai hasil dari data X
• P(X|play=“yes”) = 0.333333333* 0.333333333* 0.333333333*0.333333333 = 0.012345679 • P(X|play=“no”) = 0.4*0.2*0.8*0.6=0.0384 • P(X|play=“yes”)*P(C1) = 0.012345679*0.642857143 = 0.007936508 • P(X|play=“no”)*P(C2) = 0.0384*0.357142857 = 0.013714286
• Nilai “no” lebih besar dari nilai “yes” maka class dari data X tersebut adalah “No” 31
Avoiding the Zero-Probability Problem • Naïve Bayesian prediction requires each conditional prob. be non-zero. Otherwise, the predicted prob. will be zero P( X | C i )
n P( x k | C i ) k 1
• Ex. Suppose a dataset with 1000 tuples, income=low (0), income= medium (990), and income = high (10) • Use Laplacian correction (or Laplacian estimator) • Adding 1 to each case Prob(income = low) = 1/1003 Prob(income = medium) = 991/1003 Prob(income = high) = 11/1003
• The “corrected” prob. estimates are close to their “uncorrected” counterparts 32
Naïve Bayes Classifier: Comments • Advantages • Easy to implement • Good results obtained in most of the cases
• Disadvantages • Assumption: class conditional independence, therefore loss of accuracy • Practically, dependencies exist among variables, e.g.: • Hospitals Patients Profile: age, family history, etc. • Symptoms: fever, cough etc., • Disease: lung cancer, diabetes, etc.
• Dependencies among these cannot be modeled by Naïve Bayes Classifier • How to deal with these dependencies? Bayesian Belief Networks 33
3 Neural Network
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Neural Network • Neural Network adalah suatu model yang dibuat untuk meniru fungsi belajar yang dimiliki otak manusia atau jaringan dari sekelompok unit pemroses kecil yang dimodelkan berdasarkan jaringan saraf manusia
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Neural Network • Model Perceptron adalah model jaringan yang terdiri dari beberapa unit masukan (ditambah dengan sebuah bias), dan memiliki sebuah unit keluaran • Fungsi aktivasi bukan hanya merupakan fungsi biner (0,1) melainkan bipolar (1,0,-1) • Untuk suatu harga threshold ѳ yang ditentukan:
F (net) =
1 0 -1
Jika net > ѳ Jika – ѳ ≤ net ≤ ѳ Jika net < - ѳ
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Fungsi Aktivasi Macam fungsi aktivasi yang dipakai untuk mengaktifkan net di berbagai jenis neural network: 1. Aktivasi linear, Rumus:
y = sign(v) = v
2. Aktivasi step, Rumus: 3. Aktivasi sigmoid biner, Rumus: 4. Aktivasi sigmoid bipolar, Rumus:
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Tahapan Algoritma Perceptron 1. Inisialisasi semua bobot dan bias (umumnya wi = b = 0) 2. Selama ada elemen vektor masukan yang respon unit keluarannya tidak sama dengan target, lakukan: 2.1 Set aktivasi unit masukan xi = Si (i = 1,...,n) 2.2 Hitung respon unit keluaran: net = +b 1 Jika net > ѳ F (net) = 0 Jika – ѳ ≤ net ≤ ѳ -1 Jika net < - ѳ 2.3 Perbaiki bobot pola yang mengadung kesalahan menurut persamaan: wi (baru) = wi (lama) + ∆w (i = 1,...,n) dengan ∆w = α t xi b (baru) = b(lama) + ∆ b dengan ∆b = α t Dimana: α = Laju pembelajaran (Learning rate) yang ditentukan ѳ = Threshold yang ditentukan t = Target 2.4 Ulangi iterasi sampai perubahan bobot (∆wn = 0) tidak ada
Semakin besar α, semakin sedikit iterasi. Namun, α terlalu besar akan merusak pola yang sudah benar sehingga pemahaman menjadi lambat. 38
Studi Kasus • Diketahui sebuah dataset kelulusan berdasarkan IPK untuk program S1: Status Lulus Tidak Lulus Tidak Lulus Tidak lulus
IPK 2.9 2.8 2.3 2.7
Semester 1 3 5 6
• Jika ada mahasiswa IPK 2.85 dan masih semester 1, maka masuk ke dalam manakah status tersebut ?
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1: Inisialisasi Bobot • Inisialisasi Bobot dan bias awal: b = 0 dan bias = 1
t 1 -1 -1 -1
X1 2,9 2.8 2.3 2,7
X2 1 3 5 6
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2.1: Set aktivasi unit masukan • Treshold (batasan), θ = 0 , yang artinya : 1 Jika net > 0 F (net) = 0 Jika net = 0 -1 Jika net < 0
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2.2 - 2.3 Hitung Respon dan Perbaiki Bobot • Hitung Response Keluaran iterasi 1 • Perbaiki bobot pola yang mengandung kesalahan
MASUKAN X1
X2
TARGET 1
t
y= NET
f(NET)
PERUBAHAN BOBOT ∆W1
∆W2
∆b
INISIALISASI
BOBOT BARU W1
W2
b
0
0
0
2,9
1
1
1
0
0
2.9
1
1
2.9
1
1
2,8
3
1
-1
12.12
1
-2.8
-3
-1
0.1
-2
0
2,3
5
1
-1
-9.77
-1
0
0
0
0.1
-2
0
2,7
6
1
-1
-11.7
-1
0
0
0
0.1
-2
0
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2.4 Ulangi iterasi sampai perubahan bobot (∆wn = 0) tidak ada (Iterasi 2) • Hitung Response Keluaran iterasi 2 • Perbaiki bobot pola yang mengandung kesalahan
MASUKAN X1
X2
TARGET 1
t
y= NET
f(NET)
PERUBAHAN BOBOT ∆W1
∆W2
∆b
INISIALISASI
BOBOT BARU W1
W2
b
0.1
-2
0
2,9
1
1
1
-1.71
-1
2.9
1
1
3
-1
1
2,8
3
1
-1
6.4
1
-2.8
-3
-1
0.2
-4
0
2,3
5
1
-1
-19.5
-1
0
0
0
0.2
-4
0
2,7
6
1
-1
-23.5
-1
0
0
0
0.2
-4
0
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2.4 Ulangi iterasi sampai perubahan bobot (∆wn = 0) tidak ada .... (Iterasi 5) • Hitung Response Keluaran iterasi 3
• Perbaiki bobot pola yang mengandung kesalahan MASUKAN X1
X2
TARGET 1
t
y= NET
f(NET)
PERUBAHAN BOBOT ∆W1
∆W2
∆b
INISIALISASI
BOBOT BARU W1
W2
b
3.2
-5
1
2,9
1
1
1
5.28
1
0
0
0
3.2
-5
1
2,8
3
1
-1
-5.04
-1
0
0
0
3.2
-5
1
2,3
5
1
-1
-16.6
-1
0
0
0
3.2
-5
1
2,7
6
1
-1
-20.4
-1
0
0
0
3.2
-5
1
• Semua pola f(net) = target, maka jaringan sudah mengenal semua pola dan iterasi dihentikan. • Untuk data IPK memiliki pola 3.2 x - 5 y + 1 = 0 dapat dihitung prediksinya menggunakan bobot yang terakhir didapat: net = X1*W1 + X2*W2 + b = 3,2 * 2,85 - 5*1 +1 = 5,12 f(net)=1 44 (Lulus)
Latihan • Lakukan eksperimen untuk mengumpulkan dataset yang memiliki 4-5 atribut dan analisis dengan Naïve Bayes dan neural network pada dataset tersebut
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4 Model Evaluation and Selection
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Model Evaluation and Selection • Evaluation metrics: How can we measure accuracy? Other metrics to consider? • Use validation test set of class-labeled tuples instead of training set when assessing accuracy • Methods for estimating a classifier’s accuracy: • Holdout method, random subsampling • Cross-validation • Bootstrap
• Comparing classifiers: • Confidence intervals • Cost-benefit analysis and ROC Curves
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Evaluating Classifier Accuracy: Holdout & Cross-Validation Methods • Holdout method • Given data is randomly partitioned into two independent sets • Training set (e.g., 2/3) for model construction • Test set (e.g., 1/3) for accuracy estimation
• Random sampling: a variation of holdout • Repeat holdout k times, accuracy = avg. of the accuracies obtained
• Cross-validation (k-fold, where k = 10 is most popular) • Randomly partition the data into k mutually exclusive subsets, each approximately equal size • At i-th iteration, use Di as test set and others as training set • Leave-one-out: k folds where k = # of tuples, for small sized data • *Stratified cross-validation*: folds are stratified so that class dist. in each fold is approx. the same as that in the initial data 48
Evaluating Classifier Accuracy: Bootstrap • Bootstrap • Works well with small data sets
• Samples the given training tuples uniformly with replacement, i.e., each time a tuple is selected, it is equally likely to be selected again and readded to the training set
• Several bootstrap methods, and a common one is .632 boostrap 1.
A data set with d tuples is sampled d times, with replacement, resulting in a training set of d samples
2.
The data tuples that did not make it into the training set end up forming the test set. About 63.2% of the original data end up in the bootstrap, and the remaining 36.8% form the test set (since (1 – 1/d)d ≈ e-1 = 0.368)
3.
Repeat the sampling procedure k times, overall accuracy of the model:
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Estimating Confidence Intervals: Classifier Models M1 vs. M2 • Suppose we have two classifiers, M1 and M2, which one is better? • Use 10-fold cross-validation to obtain and • These mean error rates are just estimates of error on the true population of future data cases
• What if the difference between the two error rates is just attributed to chance? • Use a test of statistical significance • Obtain confidence limits for our error estimates
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Estimating Confidence Intervals: Null Hypothesis 1. Perform 10-fold cross-validation 2. Assume samples follow a t distribution with k–1 degrees of freedom (here, k=10)
3. Use t-test (or Student’s t-test) 4. Null Hypothesis: M1 & M2 are the same
5. If we can reject null hypothesis, then 1. we conclude that the difference between M1 & M2 is statistically significant 2. Chose model with lower error rate
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Estimating Confidence Intervals: t-test • If only 1 test set available: pairwise comparison • For ith round of 10-fold cross-validation, the same cross partitioning is used to obtain err(M1)i and err(M2)i • Average over 10 rounds to get • t-test computes t-statistic with k-1 degrees of freedom: where
• If two test sets available: use non-paired t-test where
where k1 & k2 are # of cross-validation samples used for M1 & M2, resp. 52
Estimating Confidence Intervals: Table for t-distribution
• Symmetric • Significance level, e.g., sig = 0.05 or 5% means M1 & M2 are significantly different for 95% of population • Confidence limit, z = sig/2 53
Estimating Confidence Intervals: Statistical Significance Are M1 & M2 significantly different? 1. Compute t. Select significance level (e.g. sig = 5%) 2. Consult table for t-distribution: Find t value corresponding to k-1 degrees of freedom (here, 9) 3. t-distribution is symmetric: typically upper % points of distribution shown → look up value for confidence limit z=sig/2 (here, 0.025) 4. If t > z or t < -z, then t value lies in rejection region: 1. 2.
Reject null hypothesis that mean error rates of M1 & M2 are same Conclude: statistically significant difference between M1 & M2
5. Otherwise, conclude that any difference is chance
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Model Selection: ROC Curves • ROC (Receiver Operating Characteristics) curves: for visual comparison of classification models • Originated from signal detection theory • Shows the trade-off between the true positive rate and the false positive rate
• The area under the ROC curve is a measure of the accuracy of the model • Rank the test tuples in decreasing order: the one that is most likely to belong to the positive class appears at the top of the list
• Vertical axis represents the true positive rate
• Horizontal axis rep. the false positive rate • The closer to the diagonal line (i.e., the closer the area is to 0.5), the less • The plot also shows a diagonal line accurate is the model • A model with perfect accuracy will have an area of 1.0 55
Issues Affecting Model Selection • Accuracy • classifier accuracy: predicting class label • Speed • time to construct the model (training time) • time to use the model (classification/prediction time) • Robustness: handling noise and missing values • Scalability: efficiency in disk-resident databases • Interpretability • understanding and insight provided by the model • Other measures, e.g., goodness of rules, such as decision tree size or compactness of classification rules 56
5 Techniques to Improve Classification Accuracy: Ensemble Methods
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Ensemble Methods: Increasing the Accuracy
• Ensemble methods • Use a combination of models to increase accuracy • Combine a series of k learned models, M1, M2, …, Mk, with the aim of creating an improved model M*
• Popular ensemble methods • Bagging: averaging the prediction over a collection of classifiers • Boosting: weighted vote with a collection of classifiers • Ensemble: combining a set of heterogeneous classifiers 58
Bagging: Boostrap Aggregation • Analogy: Diagnosis based on multiple doctors’ majority vote • Training • Given a set D of d tuples, at each iteration i, a training set Di of d tuples is sampled with replacement from D (i.e., bootstrap) • A classifier model Mi is learned for each training set Di • Classification: classify an unknown sample X • Each classifier Mi returns its class prediction • The bagged classifier M* counts the votes and assigns the class with the most votes to X • Prediction: can be applied to the prediction of continuous values by taking the average value of each prediction for a given test tuple • Accuracy • Often significantly better than a single classifier derived from D • For noise data: not considerably worse, more robust • Proved improved accuracy in prediction 59
Boosting • Analogy: Consult several doctors, based on a combination of weighted diagnoses—weight assigned based on the previous diagnosis accuracy • How boosting works? 1. Weights are assigned to each training tuple 2. A series of k classifiers is iteratively learned 3. After a classifier Mi is learned, the weights are updated to allow the subsequent classifier, Mi+1, to pay more attention to the training tuples that were misclassified by Mi 4. The final M* combines the votes of each individual classifier, where the weight of each classifier's vote is a function of its accuracy • Boosting algorithm can be extended for numeric prediction • Comparing with bagging: Boosting tends to have greater accuracy, but it also risks overfitting the model to misclassified data 60
Adaboost (Freund and Schapire, 1997) 1.
Given a set of d class-labeled tuples, (X1, y1), …, (Xd, yd)
2.
Initially, all the weights of tuples are set the same (1/d)
3.
Generate k classifiers in k rounds. At round i, 1. Tuples from D are sampled (with replacement) to form a training set Di of the same size 2. Each tuple’s chance of being selected is based on its weight 3. A classification model Mi is derived from Di 4. Its error rate is calculated using Di as a test set 5. If a tuple is misclassified, its weight is increased, o.w. it is decreased
4.
Error rate: err(Xj) is the misclassification error of tuple Xj. Classifier Mi error rate is the sum of the weights of the misclassified tuples: d
error ( M i ) w j err ( X j ) j
5.
The weight of classifier Mi’s vote is 61
log
1 error ( M i ) error ( M i )
Random Forest (Breiman 2001) • Random Forest: • Each classifier in the ensemble is a decision tree classifier and is generated using a random selection of attributes at each node to determine the split • During classification, each tree votes and the most popular class is returned
• Two Methods to construct Random Forest: 1. 2.
Forest-RI (random input selection): Randomly select, at each node, F attributes as candidates for the split at the node. The CART methodology is used to grow the trees to maximum size Forest-RC (random linear combinations): Creates new attributes (or features) that are a linear combination of the existing attributes (reduces the correlation between individual classifiers)
• Comparable in accuracy to Adaboost, but more robust to errors and outliers • Insensitive to the number of attributes selected for consideration at each split, and faster than bagging or boosting 62
Classification of Class-Imbalanced Data Sets • Class-imbalance problem: Rare positive example but numerous negative ones, e.g., medical diagnosis, fraud, oilspill, fault, etc. • Traditional methods assume a balanced distribution of classes and equal error costs: not suitable for classimbalanced data • Typical methods for imbalance data in 2-class classification: 1. Oversampling: re-sampling of data from positive class 2. Under-sampling: randomly eliminate tuples from negative class 3. Threshold-moving: moves the decision threshold, t, so that the rare class tuples are easier to classify, and hence, less chance of costly false negative errors 4. Ensemble techniques: Ensemble multiple classifiers introduced above • Still difficult for class imbalance problem on multiclass tasks 63
Rangkuman • Classification is a form of data analysis that extracts models describing important data classes • Effective and scalable methods have been developed for decision tree induction, Naive Bayesian classification, rule-based classification, and many other classification methods • Evaluation metrics include: accuracy, sensitivity, specificity, precision, recall, F measure, and Fß measure • Stratified k-fold cross-validation is recommended for accuracy estimation. Bagging and boosting can be used to increase overall accuracy by learning and combining a series of individual models 64
Rangkuman • Significance tests and ROC curves are useful for model selection. • There have been numerous comparisons of the different classification methods; the matter remains a research topic • No single method has been found to be superior over all others for all data sets • Issues such as accuracy, training time, robustness, scalability, and interpretability must be considered and can involve trade-offs, further complicating the quest for an overall superior method 65
TERIMA KASIH
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Credit • Romi Satria Wahono • http://romisatriawahono.net/lecture/dm
• Jong Jek Siang • Jaringan syaraf tiruan dan pemrogramannya menggunakan matlab, ANDI Offset.
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