8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 1/51 PCA for the uninitiated Intuitive motivation via maximum variance interpretation Ben Mabey benmabey.com github.com/bmabey @bmabey D ow nload

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 2/51 For PDF viewers... This deck can be found in its original (and better) HTML5 form at benmabey.com/presentations/pca-tutorial/ .N.B.: The deck isn't completely standalone since I don't explain every step made as I did when actually presenting it. That said I think the deck should be useful for anyone who wants to get a quick idea of what PCA is and the math behind it (I only take into account conventional PCA, not probabilistic interpretations). I am inconsistent with some of my equations to make some of the algebra easier (all legal though!) which I explained during the actual presentation. For people who want to go deeper and follow the math more closely I highly recommend the tutorial by Jonathan Shlens which is where I got most of my derivations. See the last slide of the deck for additional resources. 2/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 3/51 The ubiquitous & versatile PCA Dimensionality Reduction Noise Reduction Exploration Feature Extraction Regression (Orthogonal) · Data Visualization Learn faster Lossy Data Compression - - - · · · · Unsupervised Learning Algorithm K-Means Computer Graphics (e.g. Bounded Volumes) and many more across various domains... · Anomaly Detection (not the best) Matching/Distance (e.g. Eigenfaces, LSI) - - · · · 3/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 4/51 Majority of PCA tutorials... 1. Organize dataset as matrix. 2. Subtract off the mean for each measurement. 3. Calculate the covariance matrix and perform eigendecomposition. 4. Profit! 4/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 5/51 Majority of PCA tutorials... 1. Organize dataset as matrix. 2. Subtract off the mean for each measurement. 3. Calculate the covariance correlation matrix and perform eigendecomposition. 4. Profit! 5/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 6/51 Majority of PCA tutorials... 1. Organize dataset as matrix. 2. Subtract off the mean for each measurement. 3. Calculate the covariance correlation matrix and perform eigendecomposition. 4. Perform SVD. 5. Profit! 6/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 7/51 7/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 8/51 The intuitive Magic Math behind PCA Maximize the variance. Minimize the projection error. · · 8/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 9/51 = P m×m X m×n Y m×n 9/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 10/51 http://www.squidoo.com/noise-sources-signal-noise-ratio-snr-and-a-look-at-them-in-the-frequency-domain

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 11/51 SNR = 2 2 signal 2 2 noise 11/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 12/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 13/51 Rotate to maximize variance 13/51

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8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 15/51 l i b r a r y ( P e r f o r m a n c e A n a l y t i c s ) c h a r t . C o r r e l a t i o n ( i r i s [ - 5 ] , b g = i r i s $ S p e c i e s , p c h = 2 1 ) 15/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 16/51 c h a r t . C o r r e l a t i o n ( d e c o r r e l a t e d . i r i s , b g = i r i s $ S p e c i e s , p c h = 2 1 ) 16/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 17/51 Variance and Covariance MATHEMATICALLY USEFUL INTUITIVE Dispersion Relationship unitless measure or is if and only if and are uncorrelated. = var(A) 2 2 A = = E[(A J ] + A ) 2 ( J 1 n ∑ i=1 n a i + A )2 = stddev(A) = 2 A var(A) _ _ _ _ _ _ R = cov(A, B) 2 AB = = E[(A J )(B J )] + A + B ( J )( J ) 1 n ∑ i=1 n a i + A b i + B = = 0 AB 2 AB 2 A 2 B cov(AB) stddev(A) stddev(B) ( J 1.0..1.0) cov(A, A) = var(A) 2 AB 0 AB 0 A B 17/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 18/51 Covariance Matrix Preprocess so that it has zero mean. Now ¤ = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 2 1,1 2 2,1 ' 2 n,1 2 1,2 2 2,2 ' 2 n,2 ( ( * ( 2 1,n 2 2,n ' 2 n,n ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ X = 2 AB 1 n I n i=1 a i b i = X ¤ X 1 n X T 18/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 19/51 c e n t e r < - f u n c t i o n ( x ) x - m e a n ( x ) i r i s . c e n t e r e d < - a p p l y ( a s . m a t r i x ( i r i s [ - 5 ] ) , 2 , c e n t e r ) ( t ( i r i s . c e n t e r e d ) % * % i r i s . c e n t e r e d ) / ( n r o w ( i r i s ) - 1 ) # # S e p a l . L e n g t h S e p a l . W i d t h P e t a l . L e n g t h P e t a l . W i d t h # # S e p a l . L e n g t h 0 . 6 8 5 6 9 - 0 . 0 4 2 4 3 1 . 2 7 4 3 0 . 5 1 6 3 # # S e p a l . W i d t h - 0 . 0 4 2 4 3 0 . 1 8 9 9 8 - 0 . 3 2 9 7 - 0 . 1 2 1 6 # # P e t a l . L e n g t h 1 . 2 7 4 3 2 - 0 . 3 2 9 6 6 3 . 1 1 6 3 1 . 2 9 5 6 # # P e t a l . W i d t h 0 . 5 1 6 2 7 - 0 . 1 2 1 6 4 1 . 2 9 5 6 0 . 5 8 1 0 19/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 20/51 c e n t e r < - f u n c t i o n ( x ) x - m e a n ( x ) m . c e n t e r e d < - a p p l y ( a s . m a t r i x ( i r i s [ - 5 ] ) , 2 , c e n t e r ) ( t ( m . c e n t e r e d ) % * % m . c e n t e r e d ) / ( n r o w ( i r i s ) - 1 ) # # S e p a l . L e n g t h S e p a l . W i d t h P e t a l . L e n g t h P e t a l . W i d t h # # S e p a l . L e n g t h 0 . 6 8 5 6 9 - 0 . 0 4 2 4 3 1 . 2 7 4 3 0 . 5 1 6 3 # # S e p a l . W i d t h - 0 . 0 4 2 4 3 0 . 1 8 9 9 8 - 0 . 3 2 9 7 - 0 . 1 2 1 6 # # P e t a l . L e n g t h 1 . 2 7 4 3 2 - 0 . 3 2 9 6 6 3 . 1 1 6 3 1 . 2 9 5 6 # # P e t a l . W i d t h 0 . 5 1 6 2 7 - 0 . 1 2 1 6 4 1 . 2 9 5 6 0 . 5 8 1 0 c o v ( i r i s [ - 5 ] ) # # S e p a l . L e n g t h S e p a l . W i d t h P e t a l . L e n g t h P e t a l . W i d t h # # S e p a l . L e n g t h 0 . 6 8 5 6 9 - 0 . 0 4 2 4 3 1 . 2 7 4 3 0 . 5 1 6 3 # # S e p a l . W i d t h - 0 . 0 4 2 4 3 0 . 1 8 9 9 8 - 0 . 3 2 9 7 - 0 . 1 2 1 6 # # P e t a l . L e n g t h 1 . 2 7 4 3 2 - 0 . 3 2 9 6 6 3 . 1 1 6 3 1 . 2 9 5 6 # # P e t a l . W i d t h 0 . 5 1 6 2 7 - 0 . 1 2 1 6 4 1 . 2 9 5 6 0 . 5 8 1 0 20/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 21/51 What would our ideal look like? i.e. is decorrelated. ¤ Y PX = Y = ¤ Y ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 2 2 1 2 2 2 0 0 * 2 2 n ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ Y 21/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 22/51 Our goal... Find some orthonormal matrix in such that is a diagonal matrix. The rows of are the principal components of . Note, that I transposed the design matrix (the data) so that covariance calculation is also reversed. This will make our life easier... P PX = Y = Y ¤ Y Y T Y n P X 22/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 23/51 Rewrite in terms of the unknown... ¤ Y ¤ Y ¤ Y = = = = = Y 1 n Y T (PX)(PX 1 n ) T PX 1 n X T P T P( X ) 1 n X T P T P ¤ X P T 23/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 24/51 Spectral Theorem / Principal Axis Theorem Every symmetric matrix has the eigendecomposition (i.e. can be diagnolized) of: A = Q | = Q | Q J 1 Q T

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 25/51 Remember, we are choosing what is... P PX = Y 25/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 26/51 Remember, we are choosing what is... Let every row, , be an eigenvector of . What this means is that where comes from the eigendecomposition of . P p i ¤ X P = Q T Q ¤ X = Q | ¤ X Q T 26/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 27/51 Turn the Algebra crank... ¤ Y ¤ Y = = = = = = P ¤ X P T P(Q | ) Q T P T P( | P) P T P T (P ) | (P ) P T P T I | I | ¤ X The principal components are linear combinations of original features of . The principal components of are the eigenvectors of . The corresponding eigenvaules lie in and represent the variance. · X · X ¤ X · ¤ Y 27/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 28/51 Manual PCA in R i r i s . e i g e n = e i g e n ( c o v ( i r i s . c e n t e r e d ) ) r o w n a m e s ( i r i s . e i g e n $ v e c t o r s ) = c o l n a m e s ( i r i s . c e n t e r e d ) c o l n a m e s ( i r i s . e i g e n $ v e c t o r s ) = c ( " P C 1 " , " P C 2 " , " P C 3 " , " P C 4 " ) i r i s . e i g e n # # $ v a l u e s # # [ 1 ] 4 . 2 2 8 2 4 0 . 2 4 2 6 7 0 . 0 7 8 2 1 0 . 0 2 3 8 4 # # # # $ v e c t o r s # # P C 1 P C 2 P C 3 P C 4 # # S e p a l . L e n g t h 0 . 3 6 1 3 9 - 0 . 6 5 6 5 9 - 0 . 5 8 2 0 3 0 . 3 1 5 5 # # S e p a l . W i d t h - 0 . 0 8 4 5 2 - 0 . 7 3 0 1 6 0 . 5 9 7 9 1 - 0 . 3 1 9 7 # # P e t a l . L e n g t h 0 . 8 5 6 6 7 0 . 1 7 3 3 7 0 . 0 7 6 2 4 - 0 . 4 7 9 8 # # P e t a l . W i d t h 0 . 3 5 8 2 9 0 . 0 7 5 4 8 0 . 5 4 5 8 3 0 . 7 5 3 7 28/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 29/51 Make the contributions intuitive... i r i s . e i g e n $ v e c t o r s ^ 2 # # P C 1 P C 2 P C 3 P C 4 # # S e p a l . L e n g t h 0 . 1 3 0 6 0 0 0 . 4 3 1 1 0 9 0 . 3 3 8 7 5 9 0 . 0 9 9 5 3 # # S e p a l . W i d t h 0 . 0 0 7 1 4 4 0 . 5 3 3 1 3 6 0 . 3 5 7 4 9 7 0 . 1 0 2 2 2 # # P e t a l . L e n g t h 0 . 7 3 3 8 8 5 0 . 0 3 0 0 5 8 0 . 0 0 5 8 1 2 0 . 2 3 0 2 5 # # P e t a l . W i d t h 0 . 1 2 8 3 7 1 0 . 0 0 5 6 9 7 0 . 2 9 7 9 3 2 0 . 5 6 8 0 0 29/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 30/51 s q u a r e d < - i r i s . e i g e n $ v e c t o r s ^ 2 s o r t e d . s q u a r e s < - s q u a r e d [ o r d e r ( s q u a r e d [ , 1 ] ) , 1 ] d o t p l o t ( s o r t e d . s q u a r e s , m a i n = " V a r i a b l e C o n t r i b u t i o n s t o P C 1 " , c e x = 1 . 5 , c o l = " r e d " ) 30/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 31/51 # l i b r a r y ( F a c t o M i n e R ) ; i r i s . p c a < - P C A ( i r i s , q u a l i . s u p = 5 ) p l o t ( i r i s . p c a , c h o i x = " v a r " , t i t l e = " C o r r e l a t i o n C i r c l e " ) 31/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 32/51 # r e s . p c a < - P C A ( d e c a t h l o n , q u a n t i . s u p = 1 1 : 1 2 , q u a l i . s u p = 1 3 ) p l o t ( r e s . p c a , c h o i x = " v a r " , t i t l e = " C o r r e l a t i o n C i r c l e " ) 32/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 33/51 What does the variance (eigenvaules) tell us? i r i s . e i g e n $ v a l u e s # T h e v a r i a n c e f o r e a c h c o r r e s p o n d i n g P C # # [ 1 ] 4 . 2 2 8 2 4 0 . 2 4 2 6 7 0 . 0 7 8 2 1 0 . 0 2 3 8 4 33/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 34/51 # l i b r a r y ( F a c t o M i n e R ) ; i r i s . p c a < - P C A ( i r i s , q u a l i . s u p = 5 ) p l o t ( i r i s . p c a , h a b i l l a g e = 5 , c o l . h a b = c ( " g r e e n " , " b l u e " , " r e d " ) , t i t l e = " D a t a s e t p r o j e c t e d o n t o P C 1 - 2 S u 34/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 35/51 How many components should you keep? Ratio of variance retained (e.g. 99% is common): Ik i=1 2 i In i=1 2 i c u m s u m ( i r i s . e i g e n $ v a l u e s / s u m ( i r i s . e i g e n $ v a l u e s ) ) # # [ 1 ] 0 . 9 2 4 6 0 . 9 7 7 7 0 . 9 9 4 8 1 . 0 0 0 0 35/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 36/51 The Elbow Test i r i s . p r c o m p < - p r c o m p ( i r i s [ - 5 ] , c e n t e r = T R U E , s c a l e = F A L S E ) s c r e e p l o t ( i r i s . p r c o m p , t y p e = " l i n e " , m a i n = " S c r e e P l o t " ) 36/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 37/51 Kaiser Criterion Keep only the components whose eigenvalue is larger than the average eigenvalue. For a correlation PCA, this rule boils down to the standard advice to "keep only the eigenvalues larger than 1". e i g e n ( c o r ( i r i s . c e n t e r e d ) ) $ v a l u e s # # [ 1 ] 2 . 9 1 8 5 0 0 . 9 1 4 0 3 0 . 1 4 6 7 6 0 . 0 2 0 7 1 37/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 38/51 Remeber, always... CROSS VALIDATE! PCA is overused and commonly misused, so always verify it is helping by cross validating. 38/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 39/51 Lots of other ways to aid interpretation... i r i s . p r c o m p < - p r c o m p ( i r i s [ - 5 ] , c e n t e r = T R U E , s c a l e = F A L S E ) b i p l o t ( i r i s . p r c o m p ) 39/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 40/51 Learn more... 40/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 41/51

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8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 43/51 How will PCA perform? s c a l e d . i r i s < - i r i s s c a l e d . i r i s $ P e t a l . L e n g t h < - i r i s $ P e t a l . L e n g t h / 1 0 0 0 s c a l e d . i r i s $ P e t a l . W i d t h < - i r i s $ P e t a l . W i d t h / 1 0 0 0 s c a l e d . i r i s $ S e p a l . W i d t h < - i r i s $ S e p a l . W i d t h * 1 0 43/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 44/51 Scale Matters 44/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 45/51 Correlation Matrix - Standardize the data # ( I n p r a c t i c e j u s t u s e t h e b u i l t - i n c o r f u n c t i o n ) s t a n d a r d i z e < - f u n c t i o n ( x ) { c e n t e r e d < - x - m e a n ( x ) c e n t e r e d / s d ( c e n t e r e d ) } s c a l e d . i r i s . s t a n d a r d i z e d < - a p p l y ( a s . m a t r i x ( s c a l e d . i r i s [ - 5 ] ) , 2 , s t a n d a r d i z e ) ( t ( s c a l e d . i r i s . s t a n d a r d i z e d ) % * % s c a l e d . i r i s . s t a n d a r d i z e d ) / ( n r o w ( i r i s ) - 1 ) # # S e p a l . L e n g t h S e p a l . W i d t h P e t a l . L e n g t h P e t a l . W i d t h # # S e p a l . L e n g t h 1 . 0 0 0 0 - 0 . 1 1 7 6 0 . 8 7 1 8 0 . 8 1 7 9 # # S e p a l . W i d t h - 0 . 1 1 7 6 1 . 0 0 0 0 - 0 . 4 2 8 4 - 0 . 3 6 6 1 # # P e t a l . L e n g t h 0 . 8 7 1 8 - 0 . 4 2 8 4 1 . 0 0 0 0 0 . 9 6 2 9 # # P e t a l . W i d t h 0 . 8 1 7 9 - 0 . 3 6 6 1 0 . 9 6 2 9 1 . 0 0 0 0 45/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 46/51 Ok, so why SVD? And how is it equivalent? Short answer on why: SVD is more numerically stable More efficient Especially when operating on a wide matrix.. you skip the step of calculating the covariance matrix There are a lot of SVD algoritms and implementations to choose from · · · 46/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 47/51 "absolutely a high point of linear algebra" Every matrix has the singular value decomposition (SVD) of: A = UDV T

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 48/51 Hey, and look familar... Recall that eigendecomposition for an orthonormal matrix is . Therefore are the eigenvectors of and are the eigenvalues. Likewise are the eigenvectors of and are the eigenvalues. AA T A A T A AA T AA T = = = = = UDV T UD (UD V T V T ) T UD V V T D T U T UD ( V = I since V, and U, are orthonormal) D T U T V T U (since D is a diagnol matrix) D 2 U T A = Q | Q T U AA T D 2 V A A T D 2 48/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 49/51 Turn the crank once more... Let a new matrix where each column of is mean centered. So, if we run SVD on our then will contain the eigenvectors of ... 's principal components! Our eigenvalues, the variances, will be . Y = 1 n √ X T Y Y Y T Y Y T = = = ( ( ) 1 n _ _ √ X T ) T 1 n _ _ √ X T X 1 n X T ¤ X Y V ¤ X X D 2 49/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 50/51 Tada! y < - i r i s . c e n t e r e d / s q r t ( n r o w ( i r i s ) - 1 ) y . s v d < - s v d ( y ) p c s < - y . s v d $ v r o w n a m e s ( p c s ) = c o l n a m e s ( i r i s . c e n t e r e d ) c o l n a m e s ( p c s ) = c ( " P C 1 " , " P C 2 " , " P C 3 " , " P C 4 " ) p c s # # P C 1 P C 2 P C 3 P C 4 # # S e p a l . L e n g t h 0 . 3 6 1 3 9 - 0 . 6 5 6 5 9 0 . 5 8 2 0 3 0 . 3 1 5 5 # # S e p a l . W i d t h - 0 . 0 8 4 5 2 - 0 . 7 3 0 1 6 - 0 . 5 9 7 9 1 - 0 . 3 1 9 7 # # P e t a l . L e n g t h 0 . 8 5 6 6 7 0 . 1 7 3 3 7 - 0 . 0 7 6 2 4 - 0 . 4 7 9 8 # # P e t a l . W i d t h 0 . 3 5 8 2 9 0 . 0 7 5 4 8 - 0 . 5 4 5 8 3 0 . 7 5 3 7 y . s v d $ d ^ 2 # v a r i a n c e s # # [ 1 ] 4 . 2 2 8 2 4 0 . 2 4 2 6 7 0 . 0 7 8 2 1 0 . 0 2 3 8 4 50/51

8/22/13 PCA for the uninitiated benmabey.com/presentations/pca-tutorial/#2 51/51 References and Resources 1. Jon Shlens (versions 2.0 and 3.1), Tutorial on Principal Component Analysis 2. H Abdi and L J Williams (2010), Principal component analysis 3. Andrew Ng (2009), cs229 Lecture Notes 10 4. Andrew Ng (2009), cs229 Lectures 14 & 15 5. Christopher Bishop (2006), Pattern Recognition and Machine Learning, section 12.1 6. Steve Pittard (2012), Principal Components Analysis Using R 7. Quick-R, Principal Components and Factor Analysis (good pointers to additional R packages) 8. C Ding, X He (2004), K-means Clustering via Principal Component Analysis