回归与回归的视觉比较主成分分析

Question

我正在尝试完善一种比较回归和 PCA 的方法，灵感来自于博客 大脑咀嚼，这也在 所以。在我忘记之前，非常感谢 JD Long 和 Josh Ulrich 为本文提供的大部分核心内容。我将在下学期的课程中使用它。抱歉这很长！

更新：我发现了一种几乎有效的不同方法（如果可以的话请修复它！）。我把它贴在底部了。这是一个比我能想出的更聪明、更短的方法！

我基本上遵循了之前的方案：生成随机数据，找出最佳拟合线，绘制残差。这显示在下面的第二个代码块中。但我还挖掘并编写了一些函数来绘制与通过随机点（本例中的数据点）的线垂直的线。我认为这些工作得很好，并且它们在“第一个代码块”中显示以及它们工作的证明。

现在，第二个代码块使用与 @JDLong 相同的流程显示了整个操作过程，我添加了结果图的图像。黑色、红色的数据是带有残差的回归，粉色、蓝色是第一个 PC，浅蓝色应该是法线，但显然它们不是。第一个代码块中绘制这些法线的函数看起来不错，但演示中有些问题：我想我一定是误解了某些内容或传递了错误的值。我的法线是水平的，这似乎是一个有用的线索（但到目前为止，对我来说不是）。谁能看出这里出了什么问题吗？

谢谢，这个问题困扰我好久了……

第一个代码块（绘制法线的函数并证明它们有效）：

##### The functions below are based very loosely on the citation at the end

pointOnLineNearPoint <- function(Px, Py, slope, intercept) {
    # Px, Py is the point to test, can be a vector.
    # slope, intercept is the line to check distance.

    Ax <- Px-10*diff(range(Px))
    Bx <- Px+10*diff(range(Px))
    Ay <- Ax * slope + intercept
    By <- Bx * slope + intercept
    pointOnLine(Px, Py, Ax, Ay, Bx, By)
    }

pointOnLine <- function(Px, Py, Ax, Ay, Bx, By) {

    # This approach based upon comingstorm's answer on
    # stackoverflow.com/questions/3120357/get-closest-point-to-a-line
    # Vectorized by Bryan

    PB <- data.frame(x = Px - Bx, y = Py - By)
    AB <- data.frame(x = Ax - Bx, y = Ay - By)
    PB <- as.matrix(PB)
    AB <- as.matrix(AB)
    k_raw <- k <- c()
    for (n in 1:nrow(PB)) {
        k_raw[n] <- (PB[n,] %*% AB[n,])/(AB[n,] %*% AB[n,])
        if (k_raw[n] < 0)  { k[n] <- 0
            } else { if (k_raw[n] > 1) k[n] <- 1
                else k[n] <- k_raw[n] }
        }
    x = (k * Ax + (1 - k)* Bx)
    y = (k * Ay + (1 - k)* By)
    ans <- data.frame(x, y)
    ans
    }

# The following proves that pointOnLineNearPoint
# and pointOnLine work properly and accept vectors

par(mar = c(4, 4, 4, 4)) # otherwise the plot is slightly distorted
# and right angles don't appear as right angles

m <- runif(1, -5, 5)
b <- runif(1, -20, 20)
plot(-20:20, -20:20, type = "n", xlab = "x values", ylab = "y values")
abline(b, m )

Px <- rnorm(10, 0, 4)
Py <- rnorm(10, 0, 4)

res <- pointOnLineNearPoint(Px, Py, m, b)
points(Px, Py, col = "red")
segments(Px, Py, res[,1], res[,2], col = "blue")

##========================================================
##
##  Credits:
##  Theory by Paul Bourke http://local.wasp.uwa.edu.au/~pbourke/geometry/pointline/
##  Based in part on C code by Damian Coventry Tuesday, 16 July 2002
##  Based on VBA code by Brandon Crosby 9-6-05 (2 dimensions)
##  With grateful thanks for answering our needs!
##  This is an R (http://www.r-project.org) implementation by Gregoire Thomas 7/11/08
##
##========================================================

第二个代码块（绘制演示图）：

set.seed(55)
np <- 10 # number of data points
x <- 1:np
e <- rnorm(np, 0, 60)
y <- 12 + 5 * x + e

par(mar = c(4, 4, 4, 4)) # otherwise the plot is slightly distorted

plot(x, y, main = "Regression minimizes the y-residuals & PCA the normals")
yx.lm <- lm(y ~ x)
lines(x, predict(yx.lm), col = "red", lwd = 2)
segments(x, y, x, fitted(yx.lm), col = "pink")

# pca "by hand"
xyNorm <- cbind(x = x - mean(x), y = y - mean(y)) # mean centers
xyCov <- cov(xyNorm)
eigenValues <- eigen(xyCov)$values
eigenVectors <- eigen(xyCov)$vectors

# Add the first PC by denormalizing back to original coords:

new.y <- (eigenVectors[2,1]/eigenVectors[1,1] * xyNorm[x]) + mean(y)
lines(x, new.y, col = "blue", lwd = 2)

# Now add the normals

yx2.lm <- lm(new.y ~ x) # zero residuals: already a line
res <- pointOnLineNearPoint(x, y, yx2.lm$coef[2], yx2.lm$coef[1])
points(res[,1], res[,2], col = "blue", pch = 20) # segments should end here
segments(x, y, res[,1], res[,2], col = "lightblue1") # the normals

############ UPDATE

位于 Vincent Zoonekynd 的页面 我几乎找到了我想要的东西。但是，它不太有效（显然曾经有效）。以下是该网站的代码摘录，其中绘制了通过垂直轴反射的第一台 PC 的法线：

set.seed(1)
x <- rnorm(20)
y <- x + rnorm(20)
plot(y~x, asp = 1)
r <- lm(y~x)
abline(r, col='red')

r <- princomp(cbind(x,y))
b <- r$loadings[2,1] / r$loadings[1,1]
a <- r$center[2] - b * r$center[1]
abline(a, b, col = "blue")
title(main='Appears to use the reflection of PC1')

u <- r$loadings
# Projection onto the first axis
p <- matrix( c(1,0,0,0), nrow=2 )
X <- rbind(x,y)
X <- r$center + solve(u, p %*% u %*% (X - r$center))
segments( x, y, X[1,], X[2,] , col = "lightblue1")

以下是结果：

Answer 1

好吧，我得回答我自己的问题了！经过进一步阅读和比较人们在网上发布的方法，我解决了这个问题。我不确定我能否清楚地说明我“修复”的内容，因为我经历了相当多的迭代。无论如何，这是情节和代码（MWE）。为了清楚起见，辅助函数位于最后。

# Comparison of Linear Regression & PCA
# Generate sample data

set.seed(39) # gives a decent-looking example
np <- 10 # number of data points
x <- -np:np
e <- rnorm(length(x), 0, 10)
y <- rnorm(1, 0, 2) * x + 3*rnorm(1, 0, 2) + e

# Plot the main data & residuals

plot(x, y, main = "Regression minimizes the y-residuals & PCA the normals", asp = 1)
yx.lm <- lm(y ~ x)
lines(x, predict(yx.lm), col = "red", lwd = 2)
segments(x, y, x, fitted(yx.lm), col = "pink")

# Now the PCA using built-in functions
# rotation = loadings = eigenvectors

r <- prcomp(cbind(x,y), retx = TRUE)
b <- r$rotation[2,1] / r$rotation[1,1] # gets slope of loading/eigenvector 1
a <- r$center[2] - b * r$center[1]
abline(a, b, col = "blue") # Plot 1st PC

# Plot normals to 1st PC

X <- pointOnLineNearPoint(x, y, b, a)
segments( x, y, X[,1], X[,2], col = "lightblue1")

###### Needed Functions

pointOnLineNearPoint <- function(Px, Py, slope, intercept) {
    # Px, Py is the point to test, can be a vector.
    # slope, intercept is the line to check distance.

    Ax <- Px-10*diff(range(Px))
    Bx <- Px+10*diff(range(Px))
    Ay <- Ax * slope + intercept
    By <- Bx * slope + intercept
    pointOnLine(Px, Py, Ax, Ay, Bx, By)
    }

pointOnLine <- function(Px, Py, Ax, Ay, Bx, By) {

    # This approach based upon comingstorm's answer on
    # stackoverflow.com/questions/3120357/get-closest-point-to-a-line
    # Vectorized by Bryan

    PB <- data.frame(x = Px - Bx, y = Py - By)
    AB <- data.frame(x = Ax - Bx, y = Ay - By)
    PB <- as.matrix(PB)
    AB <- as.matrix(AB)
    k_raw <- k <- c()
    for (n in 1:nrow(PB)) {
        k_raw[n] <- (PB[n,] %*% AB[n,])/(AB[n,] %*% AB[n,])
        if (k_raw[n] < 0)  { k[n] <- 0
            } else { if (k_raw[n] > 1) k[n] <- 1
                else k[n] <- k_raw[n] }
        }
    x = (k * Ax + (1 - k)* Bx)
    y = (k * Ay + (1 - k)* By)
    ans <- data.frame(x, y)
    ans
    }

Answer 2

尝试将这行代码更改为：

res <- pointOnLineNearPoint(x, y, yx2.lm$coef[2], yx2.lm$coef[1])

以便

res <- pointOnLineNearPoint(x, new.y, yx2.lm$coef[2], yx2.lm$coef[1])

您调用正确的 y 值。

Answer 3

在 Vincent Zoonekynd 的代码 中，更改行 u <- r$loadings 到 u <-solve(r$loadings)。在 solve() 的第二个实例中，沿第一主轴的预测组件得分（即，第二个预测组件得分设置为零的预测得分矩阵）需要乘以 < em>载荷/特征向量的倒数。将数据乘以载荷即可得出预测分数；将预测分数除以负载即可得到数据。希望有帮助。