This vignette illustrates the process of transforming a set of variables to a new set of uncorrelated (orthogonal) variables. It carries out the Gram-Schmidt process directly by successively projecting each successive variable on the previous ones and subtracting (taking residuals). This is equivalent by replacing each successive variable by its residuals from a least squares regression on the previous variables.
When this method is used on the predictors in a regression problem, the resulting orthogonal variables have exactly the same anova() summary (based on “Type I”, sequential sums of squares) as do original variables.
We use the class data set, but convert the character factor sex to a dummy (0/1) variable male.
library(matlib)
data(class)
class$male <- as.numeric(class$sex=="M")For later use in regression, we create a variable IQ as a response variable
class <- transform(class,
IQ = round(20 + height + 3*age -.1*weight -3*male + 10*rnorm(nrow(class))))
head(class)## sex age height weight male IQ
## Alfred M 14 69.0 112.5 1 137
## Alice F 13 56.5 84.0 0 117
## Barbara F 13 65.3 98.0 0 118
## Carol F 14 62.8 102.5 0 103
## Henry M 14 63.5 102.5 1 114
## James M 12 57.3 83.0 1 123Reorder the predictors we want, forming a numeric matrix, X.
X <- as.matrix(class[,c(3,4,2,5)])
head(X)## height weight age male
## Alfred 69.0 112.5 14 1
## Alice 56.5 84.0 13 0
## Barbara 65.3 98.0 13 0
## Carol 62.8 102.5 14 0
## Henry 63.5 102.5 14 1
## James 57.3 83.0 12 1The Gram-Schmidt process treats the variables in a given order, according to the columns in X. We start with a new matrix Z consisting of X[,1]. Then, find a new variable Z[,2] orthogonal to Z[,1] by subtracting the projection of X[,2] on Z[,1].
Z <- cbind(X[,1], 0, 0, 0)
Z[,2] <- X[,2] - Proj(X[,2], Z[,1])
crossprod(Z[,1], Z[,2]) # verify orthogonality## [,1]
## [1,] 7.276e-12Continue in the same way, subtracting the projections of X[,3] on the previous columns, and so forth
Z[,3] <- X[,3] - Proj(X[,3], Z[,1]) - Proj(X[,3], Z[,2])
Z[,4] <- X[,4] - Proj(X[,4], Z[,1]) - Proj(X[,4], Z[,2]) - Proj(X[,4], Z[,3])Note that if any column of X is a linear combination of the previous columns, the corresponding column of Z will be all zeros.
These computations are similar to the following set of linear regressions:
z2 <- residuals(lm(X[,2] ~ X[,1]), type="response")
z3 <- residuals(lm(X[,3] ~ X[,1:2]), type="response")
z4 <- residuals(lm(X[,4] ~ X[,1:3]), type="response")The columns of Z are now orthogonal, but not of unit length,
zapsmall(crossprod(Z)) # check orthogonality## [,1] [,2] [,3] [,4]
## [1,] 57888 0 0 0
## [2,] 0 3249 0 0
## [3,] 0 0 7 0
## [4,] 0 0 0 2We make standardize column to unit length, giving Z as an orthonormal matrix, such that \(Z' Z = I\).
Z <- Z %*% diag(1 / len(Z)) # make each column unit length
zapsmall(crossprod(Z)) # check orthonormal## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1colnames(Z) <- colnames(X)The QR method uses essentially the same process, factoring the matrix \(\mathbf{X}\) as \(\mathbf{X = Q R}\), where \(\mathbf{Q}\) is the orthonormal matrix corresponding to Z and \(\mathbf{R}\) is an upper triangular matrix. However, the signs of the columns of \(\mathbf{Q}\) are arbitrary, and QR() returns QR(X)$Q with signs reversed, compared to Z.
# same result as QR(X)$Q, but with signs reversed
head(Z, 5)## height weight age male
## Alfred 0.2868 0.07545 -0.3687 0.12456
## Alice 0.2348 -0.08067 0.3569 -0.02177
## Barbara 0.2714 -0.07715 -0.3862 -0.45170
## Carol 0.2610 0.07058 0.1559 -0.20548
## Henry 0.2639 0.05132 0.1047 0.40538head(-QR(X)$Q, 5)## [,1] [,2] [,3] [,4]
## [1,] 0.2868 0.07545 -0.3687 0.12456
## [2,] 0.2348 -0.08067 0.3569 -0.02177
## [3,] 0.2714 -0.07715 -0.3862 -0.45170
## [4,] 0.2610 0.07058 0.1559 -0.20548
## [5,] 0.2639 0.05132 0.1047 0.40538all.equal( unname(Z), -QR(X)$Q )## [1] TRUEWe carry out two regressions of IQ on the variables in X and in Z. These are equivalent, in the sense that
anova() are the same.class2 <- data.frame(Z, IQ=class$IQ)Regression of IQ on the original variables in X
mod1 <- lm(IQ ~ height + weight + age + male, data=class)
anova(mod1)## Analysis of Variance Table
##
## Response: IQ
## Df Sum Sq Mean Sq F value Pr(>F)
## height 1 177 177 4.01 0.073 .
## weight 1 106 106 2.40 0.153
## age 1 329 329 7.46 0.021 *
## male 1 334 334 7.58 0.020 *
## Residuals 10 441 44
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Regression of IQ on the orthogonalized variables in Z
mod2 <- lm(IQ ~ height + weight + age + male, data=class2)
anova(mod2)## Analysis of Variance Table
##
## Response: IQ
## Df Sum Sq Mean Sq F value Pr(>F)
## height 1 177 177 4.01 0.073 .
## weight 1 106 106 2.40 0.153
## age 1 329 329 7.46 0.021 *
## male 1 334 334 7.58 0.020 *
## Residuals 10 441 44
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1This illustrates that anova() tests for linear models are sequential tests. They test hypotheses about the extra contribution of each variable over and above all previous ones, in a given order. These usually do not make substantive sense, except in testing ordered (“hierarchical”) models.