了解QR分解

我有一个可行的示例（在R中），我试图进一步理解。我正在使用Limma创建线性模型，并试图逐步了解倍数变化计算中发生的情况。我主要是想弄清楚如何计算系数。据我所知，QR分解用于获取系数，因此我本质上是在寻找一种解释或逐步查看正在计算的方程式的方法，或者在其中寻找qr（）的源代码。 R自己追踪。

使用以下数据：

expression_data <- c(1.27135202935009, 1.41816160331787, 1.2572772420417, 1.70943398046296, 1.30290218641586, 0.632660015122616, 1.73084258791384, 0.863826352944684, 0.62481665344628, 0.356064235030147, 1.31542028558644, 0.30549909383238, 0.464963176430548, 0.132181421105667, -0.284799809563931, 0.216198538884642, -0.0841133304341238, -0.00184472290008803, -0.0924271878885008, -0.340291804468472, -0.236829711453303, 0.0529690806587626, 0.16321956624511, -0.310513510587778, -0.12970035111176, -0.126398635780533, 0.152550803185228, -0.458542514769473, 0.00243517688116406, -0.0190192219685527, 0.199329876859774, 0.0493831375210439, -0.30903829000185, -0.289604319193543, -0.110019942085281, -0.220289950537685, 0.0680403723818882, -0.210977291862137, 0.253649629045288, 0.0740109953273042, 0.115109148186167, 0.187043445057404, 0.705155251555554, 0.105479342752451, 0.344672919872447, 0.303316487542805, 0.332595721664644, 0.0512213943473417, 0.440756755046719, 0.091642538588249, 0.477236022595909, 0.109140019847968, 0.685001267317616, 0.183154080053337, 0.314190891668279, -0.123285017407119, 0.603094973500324, 1.53723917249845, 0.180518835745199, 1.5520102749957, -0.339656677699664, 0.888791974821514, 0.321402618155527, 1.31133008668306, 0.287587853884556, -0.513896569786498, 1.01400498573403, -0.145552182640197, -0.0466811491949621, 1.34418631328095, -0.188666887863983, 0.920227741574566, -0.0182196762358299, 1.18398082848213, 0.0680539755381465, 0.389472802053599, 1.14920099633956, 1.35363045061024, -0.0400907708395635, 1.14405154287124, 0.365672853509181, -0.0742688460368051, 1.60927415300638, -0.0312210890874907, -0.302097025523754, 0.214897201115632, 2.029775196118, 1.46210810601113, -0.126836819148653, -0.0799005522761045, 0.958505775644153, -0.209758749029421, 0.273568395649965, 0.488150388217536, -0.230312627718208, -0.0115780974342431, 0.351708198671371, 0.11803520077305, -0.201488605868396, 0.0814169684941098, 1.32266103732873, 1.9077004570343, 1.34748531668521, 1.37847539147601, 1.85761827653095, 1.11327229058024, 1.21377936983249, 1.167867701785, 1.3119314966728, 1.01502530573911, 1.22109375841952, 1.23026951795161, 1.30638557237133, 1.02569437924906, 0.812852833149196) 

treatment <- c('A', 'A', 'A', 'A', 'A', 'A', 'A', 'B', 'B', 'B', 'B', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'B', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'A', 'B', 'A', 'C', 'A', 'C', 'A', 'B', 'C', 'B', 'C', 'C', 'A', 'C', 'A', 'B', 'A', 'C', 'B', 'B', 'A', 'C', 'A', 'C', 'C', 'A', 'C', 'B', 'C', 'A', 'A', 'B', 'C', 'A', 'C', 'B', 'B', 'C', 'C', 'B', 'B', 'C', 'C', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A')

variation <- c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3)

...以及以下模型设计

design               <- model.matrix(~0 + factor(treatment,
                                                 levels=unique(treatment)) +
                                          factor(variation))
colnames(design)     <- c(unique(treatment),
                          paste0("b",
                                 unique(variation)[-1]))
#expression_data consists of more than the data given. The data given is just one row from the object
fit                  <- lmFit((expression_data), design)

cont_mat             <- makeContrasts(B-A,
                                      levels=design)
fit2                 <- contrasts.fit(fit,
                                      contrasts=cont_mat)
fit2                 <- eBayes(fit2)

给我倍数更改-0.8709646。

可以通过以下方式获取系数：

qr.solve(design, expression_data)

然后，这是BA获得倍数变更的一个简单案例。

现在让我感到困惑的是它的qr.solve实际工作方式，它调用了该qr函数，但是我似乎找不到该源。

有人对qr分解有很好的解释，还是我可以精确地追踪得出系数的方式？

谢谢你的帮助！

@MatthewDrury链接的帖子中已经解释了QR分解作为获取OLS估计值的过程的想法。

该函数的源代码qr是用Fortran编写的，可能很难遵循。在这里，我展示了一个最小的实现，该实现可重现OLS拟合的模型的主要结果。希望这些步骤更容易遵循。

$X$$Q$$R$$X = QR$$X'X\hat\beta = X'y$

$R^{-1}$$Q'Q$

$R$$\hat\beta$

$Q$$R$

$R$$Y$$Q'y$

QR.regression <- function(y, X)
{
  nr <- length(y)
  nc <- NCOL(X)

  # Householder transformations
  for (j in seq_len(nc))
  {
    id <- seq.int(j, nr)
    sigma <- sum(X[id,j]^2)
    s <- sqrt(sigma)
    diag_ej <- X[j,j]
    gamma <- 1.0 / (sigma + abs(s * diag_ej))
    kappa <- if (diag_ej < 0) s else -s
    X[j,j] <- X[j,j] - kappa
    if (j < nc)
    for (k in seq.int(j+1, nc))
    {
      yPrime <- sum(X[id,j] * X[id,k]) * gamma
      X[id,k] <- X[id,k] - X[id,j] * yPrime
    }

    yPrime <- sum(X[id,j] * y[id]) * gamma
    y[id] <- y[id] - X[id,j] * yPrime

    X[j,j] <- kappa

  } # end Householder

  # residual sum of squares
  rss <- sum(y[seq.int(nc+1, nr)]^2)

  # Backsolve
  beta <- rep(NA, nc)
  for (j in seq.int(nc, 1))
  {
    beta[j] <- y[j]
    if (j < nc)
    for (i in seq.int(j+1, nc))
      beta[j] <- beta[j] - X[j,i] * beta[i]
    beta[j] <- beta[j] / X[j,j]
  }

  # set zeros in the lower triangular side of X (which stores) 
  # not really necessary, this is just to return R for illustration
  for (i in seq_len(ncol(X)))
    X[seq.int(i+1, nr),i] <- 0

  list(R=X[1:nc,1:nc], y=y, beta=beta, rss=rss)
}

我们可以检查与lm获得的估计是否相同。

# benchmark results
fit <- lm(expression_data ~ 0+design)
# OLS by QR decomposition
y <- expression_data
X <- design
res <- QR.regression(y, X)
res$beta
# [1]  1.43235881  0.56139421  0.07744044 -0.15611038 -0.15021796    
all.equal(res$beta, coef(fit), check.attributes=FALSE)
# [1] TRUE
all.equal(res$rss, sum(residuals(fit)^2))
# [1] TRUE

$Q$

Q <- X %*% solve(res$R)
round(crossprod(Q), 3)
#   1 2 3 4 5
# 1 1 0 0 0 0
# 2 0 1 0 0 0
# 3 0 0 1 0 0
# 4 0 0 0 1 0
# 5 0 0 0 0 1

残差可以通过获得y - X %*% res$beta。

参考文献

DSG Pollock（1999） ：时间序列分析，信号处理和动力学手册，学术出版社。