如何找到原始PCA和旋转的PCA负载矩阵之间的角度? [英] how do I find the angles between an original and a rotated PCA loadings matrix?

问题描述

假设我有两个矩阵，分别是loa.orig和loa.rot，并且知道loa.rot是loa.orig的旋转(手动或其他方式).

Suppose I have two matrices of PCA loadings loa.orig, and loa.rot, and I know that loa.rot is a rotation (by-hand or otherwise) of loa.orig.

(loa.orig可能已经通过varimax或其他东西进行了正交旋转，但我认为这并不重要).

(loa.orig might also have been already orthogonally rotated by varimax or something, but I don't think that matters).

我知道想知道>旋转到loa.rot所经过的角度.

I know want to know the angles by which loa.orig has been rotated to arrive at loa.rot.

我从成对(平面方向)旋转的顺序也很重要.

I understand from this comment on another question that "rotations don't commute", and therefore, the order of pair-wise (plane-wise) rotations also matters.

因此要从loa.orig再现 ，我需要知道一系列必要的旋转，理想情况下是按照以下rots给出的顺序进行.

So to reproduce loa.rot from loa.orig I would need to know a series of necessary rotations, ideally in the order given by rots in the below.

这是MWE:

library(psych) # this allows manual rotation

# suppose I have some ORIGINAL loadings matrix, from a principal components analysis, with three retained components
loa.orig <- cbind(c(0.6101496, 0.7114088, 0.3356003, 0.7318809, 0.5980133, 0.4102817, 0.7059148, 0.6080662, 0.5089014, 0.587025, 0.6166816, 0.6728603, 0.7482675, 0.5409658, 0.6415472, 0.3655053, 0.6313868), c(-0.205317, 0.3273207, 0.7551585, -0.1981179, -0.423377, -0.07281187, -0.04180098, 0.5003459, -0.504371, 0.1942334, -0.3285095, 0.5221494, 0.1850734, -0.2993066, -0.08715662, -0.02191772, -0.2002428), c(-0.4692407, 0.1581682, -0.04574932, -0.1189175, 0.2449018, -0.5283772, 0.02826476, 0.1703277, 0.2305158, 0.2135566, -0.2783354, -0.05187637, -0.104919, 0.5054129, -0.2403471, 0.5380329, -0.07999642))

# I then rotate 1-2 by 90°, and 1-3 by 45°
loa.rot <- factor.rotate(f = loa.orig, angle = 90, col1 = 1, col2 = 2)
loa.rot <- factor.rotate(f = loa.rot, angle = 45, col1 = 1, col2 = 3)

# predictably, loa.rot and loa.orig are now different
any(loa.rot == loa.orig) # are any of them the same?

很明显，在这种情况下，我知道角度和顺序，但让我们假设我不知道. 另外，我们假设在实际使用情况下，可能会保留并旋转了许多个组件，而不仅仅是三个.

Obviously, in this case, I know the angles and order, but let's assume I don't. Also, let's assume that in the real use case there may be many components retained and rotated, not just three.

我有点不确定报告组件对(平面方向)旋转角度的顺序的常规方法，但是我想像一下可能的组合(~~不是排列~~)应该做.

I am a bit unsure what would be a conventional way to report the order of component-pair-wise (plane-wise) rotation angles, but I'd imagine the list of possible combinations (~~not permutations~~) should do.

combs <- combn(x = ncol(loa.orig), m = 2, simplify = TRUE)  # find all possible combinations of factors
rots <- data.frame(t(combs), stringsAsFactors = FALSE)  # transpose
rots  # these rows give the *order* in which the rotations should be done

rots给出了这些排列.

很高兴知道如何从loa.orig到达loa.rot，并按rots中的行所给定的方式旋转组件对.

Would be great to know how to arrive at loa.rot from loa.orig, rotating component pairs as given by the rows in rots.

更新:尝试根据以下答案

基于以下答案，我尝试将一个函数组合在一起，并使用varimax旋转和 real 数据集对其进行测试. (对于varimax，没有特别的理由–我只想旋转一些我们实际上不知道的角度.)

Based on the below answer, I've tried to put together a function and test it with a varimax rotation and a real dataset. (No reason in particular for varimax – I just wanted some rotation where we actually do not know the angles.).

然后我使用提取的角度测试是否可以从香草载荷中重新创建varimax旋转.

I then test whether I can actually re-create the varimax rotation from the vanilla loadings, using the extracted angles.

library(psych)
data("Harman74.cor")  # notice the correlation matrix is called "cov", but doc says its a cor matrix
vanilla <- principal(r = Harman74.cor$cov, nfactors = 4, rotate = "none", )$loadings  # this is unrotated
class(vanilla) <- NULL  # print methods only causes confusion
varimax <- principal(r = Harman74.cor$cov, nfactors = 4, rotate = "varimax")$loadings  # this is rotated
class(varimax) <- NULL  # print methods only causes confusion

find.rot.instr <- function(original, rotated) {
  # original <- vanilla$loadings  # testing
  # rotated <- varimax$loadings  # testing
  getAngle <- function(A, B) acos(sum(A*B) / (norm(A, "F") * norm(B, "F"))) * 180/pi

  rots <- combn(x = ncol(original), m = 2, simplify = FALSE)  # find all possible combinations of factor pairs

  tmp <- original
  angles <- sapply(rots, function(cols) {
    angle <- getAngle(tmp[, cols], rotated[, cols])
    tmp <<- factor.rotate(tmp, angle = angle, col1 = cols[1], col2 = cols[2])
    return(angle)
  })
  return(angles)
}

vanilla.to.varimax.instr <- find.rot.instr(original = vanilla, rotated = varimax)  # these are the angles we would need to transform in this order

rots <- combn(x = ncol(vanilla), m = 2, simplify = FALSE)  # find all possible combinations of factor pairs
# this is again, because above is in function

# now let's implement the extracted "recipe"
varimax.recreated <- vanilla # start with original loadings
varimax.recreated == vanilla  # confirm that it IS the same
for (i in 1:length(rots)) {  # loop over all combinations, starting from the top
  varimax.recreated <- factor.rotate(f = varimax.recreated, angle = vanilla.to.varimax.instr[i], col1 = rots[[i]][1], col2 = rots[[i]][2])
}
varimax == varimax.recreated  # test whether they are the same
varimax - varimax.recreated  # are the close?

不幸的是，它们不一样，甚至不一样:(

Unfortunately, they are not the same, not even similar :(

> varimax == varimax.recreated  # test whether they are the same
PC1   PC3   PC2   PC4
VisualPerception       FALSE FALSE FALSE FALSE
Cubes                  FALSE FALSE FALSE FALSE
PaperFormBoard         FALSE FALSE FALSE FALSE
Flags                  FALSE FALSE FALSE FALSE
GeneralInformation     FALSE FALSE FALSE FALSE
PargraphComprehension  FALSE FALSE FALSE FALSE
SentenceCompletion     FALSE FALSE FALSE FALSE
WordClassification     FALSE FALSE FALSE FALSE
WordMeaning            FALSE FALSE FALSE FALSE
Addition               FALSE FALSE FALSE FALSE
Code                   FALSE FALSE FALSE FALSE
CountingDots           FALSE FALSE FALSE FALSE
StraightCurvedCapitals FALSE FALSE FALSE FALSE
WordRecognition        FALSE FALSE FALSE FALSE
NumberRecognition      FALSE FALSE FALSE FALSE
FigureRecognition      FALSE FALSE FALSE FALSE
ObjectNumber           FALSE FALSE FALSE FALSE
NumberFigure           FALSE FALSE FALSE FALSE
FigureWord             FALSE FALSE FALSE FALSE
Deduction              FALSE FALSE FALSE FALSE
NumericalPuzzles       FALSE FALSE FALSE FALSE
ProblemReasoning       FALSE FALSE FALSE FALSE
SeriesCompletion       FALSE FALSE FALSE FALSE
ArithmeticProblems     FALSE FALSE FALSE FALSE
> varimax - varimax.recreated  # are the close?
PC1         PC3          PC2       PC4
VisualPerception        0.2975463  1.06789735  0.467850675 0.7740766
Cubes                   0.2317711  0.91086618  0.361004861 0.4366521
PaperFormBoard          0.1840995  0.98694002  0.369663215 0.5496151
Flags                   0.4158185  0.82820078  0.439876777 0.5312143
GeneralInformation      0.8807097 -0.33385999  0.428455899 0.7537385
PargraphComprehension   0.7604679 -0.30162120  0.389727192 0.8329341
SentenceCompletion      0.9682664 -0.39302764  0.445263121 0.6673116
WordClassification      0.7714312  0.03747430  0.460461099 0.7643221
WordMeaning             0.8010876 -0.35125832  0.396077591 0.8201986
Addition                0.4236932 -0.32573100  0.204307400 0.6380764
Code                    0.1654224 -0.01757153  0.194533996 0.9777764
CountingDots            0.3585004  0.28032822  0.301148474 0.5929926
StraightCurvedCapitals  0.5313385  0.55251701  0.452293566 0.6859854
WordRecognition        -0.3157408 -0.13019630 -0.034647588 1.1235253
NumberRecognition      -0.4221889  0.10729098 -0.035324356 1.0963785
FigureRecognition      -0.3213392  0.76012989  0.158748259 1.1327322
ObjectNumber           -0.3234966 -0.02363732 -0.007830001 1.1804147
NumberFigure           -0.2033601  0.59238705  0.170467459 1.0831672
FigureWord             -0.0788080  0.35303097  0.154132395 0.9097971
Deduction               0.3423495  0.41210812  0.363022937 0.9181519
NumericalPuzzles        0.3573858  0.57718626  0.393958036 0.8206304
ProblemReasoning        0.3430690  0.39082641  0.358095577 0.9133117
SeriesCompletion        0.4933886  0.56821932  0.465602192 0.9062039
ArithmeticProblems      0.4835965 -0.03474482  0.332889805 0.9364874

很明显，我在犯错.

推荐答案

我认为，如果您进行矩阵代数运算，问题会更容易(我这样做是为了进行正交旋转，而不是倾斜变换，但是逻辑是相同的.)

I think the problem is easier if you do the matrix algebra (I am doing this for orthogonal rotations, not oblique transformations, but the logic is the same.)

首先，请注意，任何旋转t都会产生旋转分量矩阵c，使得c = Ct，其中C是未旋转的PCA解.

First, notice that any rotation, t, produces a rotated components matrix c such that c = Ct where C is the unrotated PCA solution.

然后，由于C'C是原始的相关矩阵R，因此我们可以通过将C'C的两边乘以R，然后乘以R的逆来求解t.这导致

Then, since C'C is the original correlation matrix, R, we can solve for t by premultiplying both sides by C' and then by the inverse of R. This leads to

t = C'R ^ -1 c

t = C' R^-1 c

以您的示例为例，

R <- Harman74.cor$cov
P <- principal(Harman74.cor$cov,nfactors=4,rotate="none")
p <- principal(Harman74.cor$cov,nfactors=4)  #the default is to do varimax
rotation <- t(P$loadings) %*% solve(R) %*% p$loadings

然后，看看这是否正确

P$loadings %*% rotation - p$loadings

此外，即将发布的psych现在报告了旋转矩阵，因此我们可以将旋转与p $ rot.mat进行比较

In addition, the coming release of psych now reports the rotation matrix, so we can compare rotation to p$rot.mat

round(p$rot.mat,2)
round(rotation,2)

这些成分的顺序有所不同，但这是因为Psych在旋转后对成分进行了重新排序，以反映旋转后的成分平方和的大小.

These differ in the order of the components, but that is because psych reorders the components after rotation to reflect the size of the rotated component sum of squares.

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