因子分析中完成的步骤与 PCA 中完成的步骤相比

机器算法验证 主成分分析 因子分析 算法
2022-01-19 17:09:31

我知道如何执行 PCA(主成分分析),但我想知道应该用于因子分析的步骤。

为了执行 PCA,让我们考虑一些矩阵,例如:A

         3     1    -1
         2     4     0
         4    -2    -5
        11    22    20

我计算了它的相关矩阵B = corr(A)

        1.0000    0.9087    0.9250
        0.9087    1.0000    0.9970
        0.9250    0.9970    1.0000

然后我做了特征值分解[V,D] = eig(B),得到特征向量:

        0.5662    0.8209   -0.0740
        0.5812   -0.4613   -0.6703
        0.5844   -0.3366    0.7383

和特征值:

        2.8877         0         0
             0    0.1101         0
             0         0    0.0022

PCA背后的一般思想是选择重要的组件,形成具有列特征向量的新矩阵,然后我们需要投影原始矩阵(在PCA中它是零中心的)。但是在因子分析中,例如我们应该选择具有高于奇异值的组件,然后我们正在使用因子旋转,请告诉我它是如何完成的?例如在这种情况下。1

与 PCA 步骤相比,请帮助我理解因子分析步骤。

1个回答

这个答案是为了显示 PCA 和因子分析之间具体的计算异同。有关它们之间的一般理论差异,请参阅问题/答案12345

下面我将逐步虹膜数据(仅限“setosa”物种)进行主成分分析(PCA),然后对相同数据进行因子分析。因子分析 (FA)将通过基于 PCA 方法的迭代主轴( PAF ) 方法完成,因此可以逐步比较 PCA 和 FA。

虹膜数据(仅限 setosa):

  id  SLength   SWidth  PLength   PWidth species 

   1      5.1      3.5      1.4       .2 setosa 
   2      4.9      3.0      1.4       .2 setosa 
   3      4.7      3.2      1.3       .2 setosa 
   4      4.6      3.1      1.5       .2 setosa 
   5      5.0      3.6      1.4       .2 setosa 
   6      5.4      3.9      1.7       .4 setosa 
   7      4.6      3.4      1.4       .3 setosa 
   8      5.0      3.4      1.5       .2 setosa 
   9      4.4      2.9      1.4       .2 setosa 
  10      4.9      3.1      1.5       .1 setosa 
  11      5.4      3.7      1.5       .2 setosa 
  12      4.8      3.4      1.6       .2 setosa 
  13      4.8      3.0      1.4       .1 setosa 
  14      4.3      3.0      1.1       .1 setosa 
  15      5.8      4.0      1.2       .2 setosa 
  16      5.7      4.4      1.5       .4 setosa 
  17      5.4      3.9      1.3       .4 setosa 
  18      5.1      3.5      1.4       .3 setosa 
  19      5.7      3.8      1.7       .3 setosa 
  20      5.1      3.8      1.5       .3 setosa 
  21      5.4      3.4      1.7       .2 setosa 
  22      5.1      3.7      1.5       .4 setosa 
  23      4.6      3.6      1.0       .2 setosa 
  24      5.1      3.3      1.7       .5 setosa 
  25      4.8      3.4      1.9       .2 setosa 
  26      5.0      3.0      1.6       .2 setosa 
  27      5.0      3.4      1.6       .4 setosa 
  28      5.2      3.5      1.5       .2 setosa 
  29      5.2      3.4      1.4       .2 setosa 
  30      4.7      3.2      1.6       .2 setosa 
  31      4.8      3.1      1.6       .2 setosa 
  32      5.4      3.4      1.5       .4 setosa 
  33      5.2      4.1      1.5       .1 setosa 
  34      5.5      4.2      1.4       .2 setosa 
  35      4.9      3.1      1.5       .2 setosa 
  36      5.0      3.2      1.2       .2 setosa 
  37      5.5      3.5      1.3       .2 setosa 
  38      4.9      3.6      1.4       .1 setosa 
  39      4.4      3.0      1.3       .2 setosa 
  40      5.1      3.4      1.5       .2 setosa 
  41      5.0      3.5      1.3       .3 setosa 
  42      4.5      2.3      1.3       .3 setosa 
  43      4.4      3.2      1.3       .2 setosa 
  44      5.0      3.5      1.6       .6 setosa 
  45      5.1      3.8      1.9       .4 setosa 
  46      4.8      3.0      1.4       .3 setosa 
  47      5.1      3.8      1.6       .2 setosa 
  48      4.6      3.2      1.4       .2 setosa 
  49      5.3      3.7      1.5       .2 setosa 
  50      5.0      3.3      1.4       .2 setosa 

我们在分析中包含 4 个数值变量:SLength SWidth PLength PWidth,分析将基于协方差,这与我们分析居中变量相同。(如果我们选择分析标准化变量的相关性。基于相关性的分析产生的结果与基于协方差的分析不同。)我不会显示居中的数据。我们称这些数据矩阵X

主成分分析步骤:

Step 0. Compute centered variables X and covariance matrix S.

Covariances S (= X'*X/(n-1) matrix: see https://stats.stackexchange.com/a/22520/3277)
.12424898   .09921633   .01635510   .01033061
.09921633   .14368980   .01169796   .00929796
.01635510   .01169796   .03015918   .00606939
.01033061   .00929796   .00606939   .01110612

Step 1.1. Decompose data X or matrix S to get eigenvalues and right eigenvectors.
          You may use svd or eigen decomposition (see https://stats.stackexchange.com/q/79043/3277)

Eigenvalues L (component variances) and the proportion of overall variance explained
           L            Prop
PC1   .2364556901   .7647237023 
PC2   .0369187324   .1193992401 
PC3   .0267963986   .0866624997 
PC4   .0090332606   .0292145579    

Eigenvectors V (cosines of rotation of variables into components)
              PC1           PC2           PC3           PC4
SLength   .6690784044   .5978840102  -.4399627716  -.0360771206 
SWidth    .7341478283  -.6206734170   .2746074698  -.0195502716 
PLength   .0965438987   .4900555922   .8324494972  -.2399012853 
PWidth    .0635635941   .1309379098   .1950675055   .9699296890 

Step 1.2. Decide on the number M of first PCs you want to retain.
          You may decide it now or later on - no difference, because in PCA values of components do not depend on M.
          Let's M=2. So, leave only 2 first eigenvalues and 2 first eigenvector columns.

Step 2. Compute loadings A. May skip if you don't need to interpret PCs anyhow.
Loadings are eigenvectors normalized to respective eigenvalues: A value = V value * sqrt(L value)
Loadings are the covariances between variables and components.

Loadings A
              PC1           PC2           
SLength    .32535081     .11487892
SWidth     .35699193    -.11925773
PLength    .04694612     .09416050
PWidth     .03090888     .02515873

Sums of squares in columns of A are components' variances, the eigenvalues

Standardized (rescaled) loadings.
St. loading is Loading / sqrt(Variable's variance);
these loadings are computed if you analyse covariances, and are suitable for interpretation of PCs
(if you analyse correlations, A are already standardized).
              PC1           PC2      
SLength    .92300804     .32590717
SWidth     .94177127    -.31461076
PLength    .27032731     .54219930
PWidth     .29329327     .23873031

Step 3. Compute component scores (values of PCs).

Regression coefficients B to compute Standardized component scores are: B = A*diag(1/L) = inv(S)*A
B
              PC1           PC2  
SLength   1.375948338   3.111670112 
SWidth    1.509762499  -3.230276923 
PLength    .198540883   2.550480216 
PWidth     .130717448    .681462580 

Standardized component scores (having variances 1) = X*B
      PC1           PC2
  .219719506   -.129560000 
 -.810351411    .863244439 
 -.803442667   -.660192989 
-1.052305574   -.138236265 
  .233100923   -.763754703 
 1.322114762    .413266845 
 -.606159168  -1.294221106 
 -.048997489    .137348703 
  ...

Raw component scores (having variances = eigenvalues) can of course be computed from standardized ones.
In PCA, they are also computed directly as X*V
      PC1           PC2
  .106842367   -.024893980 
 -.394047228    .165865927 
 -.390687734   -.126851118 
 -.511701577   -.026561059 
  .113349309   -.146749722 
  .642900908    .079406116 
 -.294755259   -.248674852 
 -.023825867    .026390520 
  ...

FA(迭代主轴提取法)步骤:

Step 0.1. Compute centered variables X and covariance matrix S.

Step 0.2. Decide on the number of factors M to extract.
          (There exist several well-known methods in help to decide, let's omit mentioning them. Most of them require that you do PCA first.)
          Note that you have to select M before you proceed further because, unlike in PCA, in FA loadings and factor values depend on M.
          Let's M=2.

Step 0.3. Set initial communalities on the diagonal of S.
          Most often quantities called "images" are used as initial communalities (see https://stats.stackexchange.com/a/43224/3277).
          Images are diagonal elements of matrix S-D, where D is diagonal matrix with diagonal = 1 / diagonal of inv(S).
          (If S is correlation matrix, images are the squared multiple correlation coefficients.)

With covariance matrix, image is the squared multiple correlation multiplied by the variable variance.
S with images as initial communalities on the diagonal
.07146025  .09921633  .01635510  .01033061
.09921633  .07946595  .01169796  .00929796
.01635510  .01169796  .00437017  .00606939
.01033061  .00929796  .00606939  .00167624

Step 1. Decompose that modified S to get eigenvalues and right eigenvectors.
        Use eigen decomposition, not svd.
        (Some last eigenvalues may be negative. This is because a reduced covariance matrix can be not positive semidefinite.)

Eigenvalues L
F1   .1782099114
F2   .0062074477
    -.0030958623
    -.0243488794

Eigenvectors V
               F1            F2 
SLength   .6875564132   .0145988554   .0466389510   .7244845480
SWidth    .7122191394   .1808121121  -.0560070806  -.6759542030
PLength   .1154657746  -.7640573143   .6203992617  -.1341224497
PWidth    .0817173855  -.6191205651  -.7808922917  -.0148062006

Leave the first M=2 values in L and columns in V.

Step 2.1. Compute loadings A.
Loadings are eigenvectors normalized to respective eigenvalues: A value = V value * sqrt(L value)
               F1            F2 
SLength   .2902513607   .0011502052
SWidth    .3006627098   .0142457085
PLength   .0487437795  -.0601980567
PWidth    .0344969255  -.0487788732

Step 2.2. Compute row sums of squared loadings. These are updated communalities.
          Reset the diagonal of S to them

S with updated communalities on the diagonal
.08424718  .09921633  .01635510  .01033061
.09921633  .09060101  .01169796  .00929796
.01635510  .01169796  .00599976  .00606939
.01033061  .00929796  .00606939  .00356942

REPEAT Steps 1-2 many times (iterations, say, 25)

Extraction of factors is done.

Let us look at the final eigenvalues of the reduced covariance 
matrix after iterations:
Eigenvalues L
F1   .2026316056
F2   .0137096989
     .0005000572
    -.0005882867

The eigenvalues are the factors' (F1 and F2) variances. The overall common variance is .2026316056 + .0137096989 = .2163413036,
so F1, for example, explains .2026316056/.2163413036 = 93.7% of the common variance.
That 93.7% of the common variance amounts to .2026316056/.3092040816 = 65.5% of the total variability 
(.3092040816, the total variance, is the trace of the initial, non-reduced covariance matrix).
[Note. The .0005000572 + -.0005882867 do not count a common variance; these "dross" eigenvalues are nonzero due to the fact
the 2-factor model does not predict the covariances without any error.]

Final loadings A and communalities (row sums of squares in A).
Loadings are the covariances between variables and factors.
Communality is the degree to what the factors load a variable, it is the "common variance" in the variable.
               F1            F2                        Comm
SLength   .3125767362   .0128306509                .0978688416
SWidth    .3187577564  -.0323523347                .1026531808
PLength   .0476237419   .1034495601                .0129698323
PWidth    .0324478281   .0423861795                .0028494498

Sums of squares in columns of A are the factors' variances: .2026316056 and .0137096989.

The main goal of factor analysis is to explain correlations or covariances by means of the loadings.
A*t(A) is the restored covariances:
.0978688416   .0992211576   .0162133990   .0106862785 
.0992211576   .1026531808   .0118336023   .0089717050 
.0162133990   .0118336023   .0129698323   .0059301186 
.0106862785   .0089717050   .0059301186   .0028494498

See that off-diagonal elements above are quite close to those of the input covariance matrix:
S
.1242489796   .0992163265   .0163551020   .0103306122    
.0992163265   .1436897959   .0116979592   .0092979592    
.0163551020   .0116979592   .0301591837   .0060693878    
.0103306122   .0092979592   .0060693878   .0111061224

Standardized (rescaled) loadings and communalities.
St. loading is Loading / sqrt(Variable's variance);
these loadings are computed if you analyse covariances, and are suitable for interpretation of Fs
(if you analyse correlations, A are already standardized).
               F1            F2                        Comm
SLength   .8867684574   .0364000747                .7876832626
SWidth    .8409066701  -.0853478652                .7144082859
PLength   .2742292179   .5956880078                .4300458666
PWidth    .3078962532   .4022009053                .2565656710

Step 3. Compute factor scores (values of Fs).
        Unlike component scores in PCA, factor scores are not exact, they are reasonable approximations.
        Several methods of computation exist (https://stats.stackexchange.com/q/126885/3277).
        Here is regressional method which is the same as the one used in PCA.

Regression coefficients B to compute Standardized factor scores are: B = inv(S)*A (original S is used)
B
              F1           F2  
SLength  1.597852081   -.023604439
SWidth   1.070410719   -.637149341
PLength   .212220217   3.157497050
PWidth    .423222047   2.646300951

Standardized factor scores = X*B
These "Standardized factor scores" have variance not 1; the variance of a factor is SSregression of the factor by variables / (n-1).
      F1           F2
  .194641800   -.365588231
 -.660133976   -.042292672
 -.786844270   -.480751358
-1.011226507    .216823430
  .141897664   -.426942721
 1.250472186    .848980006
 -.669003108   -.025440982
 -.050962459    .016236852
  ...

Factors are extracted as orthogonal. And they are.
However, regressionally computed factor scores are not fully uncorrelated.
Covariance matrix between computed factor scores.
      F1      F2
F1   .864   .026
F2   .026   .459

Factor variances are their squared loadings.
You can easily recompute the above "standardized" factor scores to "raw" factor scores having those variances:
raw score = st. score * sqrt(factor variance / st. scores variance).

提取后(如上所示),可能会发生可选的旋转。在 FA 中经常进行轮换。有时它在 PCA 中以完全相同的方式完成。旋转将加载矩阵A旋转成某种形式的“简单结构”,这极大地促进了因子的解释(然后可以重新计算旋转的分数)。由于旋转并不是 FA 与 PCA 数学上的区别,而且因为它是一个单独的大主题,所以我不会触及它。