DISTÀNCIES, LEVERAGE I OUTLIERS EN L ANÀLISI MULTIVARIANT
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1 DISTÀNCIES, LEVERAGE I OUTLIERS EN L ANÀLISI MULTIVARIANT Anàlisi Multivariant, UPF, Tardor del 2012
2 1 Distància d ts entre dos punts, matriu D de distàncies 2
3 Distància entre punts distancia entre dos punts ordenades A (0.4, 0.4) C (0.6, 0.35) B (0.8, 0.7) abcisses
4 Distància Euclidiana Punts A i B de coordenades (x A 1,..., xa p ) i (x B 1,..., xb p ) p d AB = (xj A xj B)2 j=1 Donada una matriu de dades X(n p) podem considerar la matriu D(n n) amb elements d ts corresponents a la distància entre les files t i s.
5 Exemple: distància Euclídia, funció R: dist data=read.table(" satorra/dades/commaa.dat", header=t) X = as.matrix(data[, -1]) n= dim(x)[1] D = matrix(0,n,n) for (i in 1:n) { for (j in 1:n) { D[i,j] = sqrt(sum((x[i,]-x[j,])ˆ2))} } rownames(d)= data[,1] colnames(d)= data[,1] > D[1:4, 1:4] Andal Catal Madri Valen Andal Catal Madri Valen ### la funcio dist: D = as.matrix(dist(x, diag=t, upper=t)) rownames(d)= data[,1] colnames(d)= data[,1] D[1:5, 1:5] Andal Catal Madri Valen Casle Andal Catal Madri Valen Casle
6 Exemple: distància Minkowski, Manhattan (Eixample) D = dist(dadesr, method="minkowski", 1,diag=T, upper=t) D > D = dist(dadesr, method="manhattan",diag=t, upper=t) > D >
7 Tipus de Distàncies Objectes A : x A = (x A 1,..., xa p ), B : x B = (x B 1,..., xb p ) p 1 Euclídia d AB = j=1 (xa j xj B)2 = (x A x B ) (x A x B ) ( p ) 1/r 2 Minkowski d AB = j=1 xa j xj A r 3 Mahalanobis d AB = (x A x B ) S 1 (x A x B ), S la matriu de variàncies i covariàncies de les dades. És distància Euclídia de les dades transformades X = XS 1/2
8 Computing S 1/2 Since S 1/2 = (S 1/2 ) 1, the problem could be translated to computing S 1/2. Take ( ) 4 1 S = 1 4 Is S 1/2 = ( The answer is a resounding no, since ( ) ( ) ( We need the spectral representation of S as S = VΛV, V V = I 2 and Λ a diagonal matrix. It is easy to see that S α = VVΛ α V, the power α applying only to the diagonal elements of Λ. ) )
9 Computing S 1/2 (cont.) > S =matrix(c(4,1,1,4),2,2) > S [,1] [,2] [1,] 4 1 [2,] 1 4 > lambda= eigen(s)$values > V= eigen(s)$vectors > V%*%diag(lambda)%*%t(V) [,1] [,2] [1,] 4 1 [2,] 1 4 > Sqrt= V%*%diag(sqrt(lambda))%*%t(V) > Sqrt [,1] [,2] [1,] [2,] > Sqrt%*%Sqrt [,1] [,2] [1,] 4 1 [2,] 1 4 > > Sinsqrt= V%*%diag(1/sqrt(lambda))%*%t(V) > Sinsqrt [,1] [,2] [1,] [2,]
10 Exemple: distància Mahalanobis S = cov(x) V = eigen(s)$vectors lambda = eigen(s)$values insqrts = V%*%diag(1/sqrt(lambda)) %*%t(v) # Noteu que S%*% insqrts%*%insqrts la matriu identitad Xs = X%*%insqrtS ## Noteu que round(cov(xs),3) > round(cov(xs),3) [,1] [,2] [,3] [,4] [1,] [2,] [3,] [4,] La distancia de Mahalanobis serà: M= as.matrix(dist(xs, method="minkowski", 2,diag=T, upper=t)) round( M[1:5, 1:5],2)
11 Observacions atípiques multivariants Donada la matriu de dades X, interessa calibrar les observacions a tipiques, aquelles que s allunyen del centre del núvol de punts. Centrem les dades, considerem X c, la norma de Mahalanobis de cada fila és una mesura de la distància al centre de cada punt fila; d 2 i = x is 1 x i = (n 1)x i(x cx c ) 1 x i Els di 2 són els elements de la diagonal de la matriu (n 1)X c (X cx c ) 1 X c {di 2 } n i=1 = diag ( (n 1)X c (X cx c ) 1 X c ) S anomenen els leverage ( apalancaments ) de cada punt fila de X c.
12 funció R: leverage dades = read.table(" satorra/dades/amd1.txt", header=t) X = as.matrix(dades[,2:6]) n = dim(x)[1] Xc = scale(x, scale=f) lev = (n-1)*diag(xc%*%solve(t(xc)%*%xc)%*%t(xc)) plot(1:n, lev, type ="h", axes=t, lty = 2) axis(2) # points(1:n, lev,pch=16, col="red") abline(h=0, col="blue") text(1:n, lev+0.03,1:n, pch=2, col="red") Veure la funció leverage a la web de l assignatura.
13 7 lev :n Figure: Els leverages de les dades de CCAA
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