python - 为什么 PCA 图像与原始图像完全不相似？

Question

我正在尝试在没有任何图像降维库的情况下实现 PCA。我尝试了 O'Reilly 计算机视觉书中的代码，并在示例 lenna 图片上实现了它:

    from PIL import Image
    from numpy import *

    def pca(X):
        num_data, dim = X.shape

        mean_X = X.mean(axis=0)
        X = X - mean_X

        if dim > num_data:
            # PCA compact trick
            M = np.dot(X, X.T) # covariance matrix
            e, U = np.linalg.eigh(M) # calculate eigenvalues an deigenvectors
            tmp = np.dot(X.T, U).T
            V = tmp[::-1] # reverse since the last eigenvectors are the ones we want
            S = np.sqrt(e)[::-1] #reverse since the last eigenvalues are in increasing order
            for i in range(V.shape[1]):
                V[:,i] /= S
        else:
            # normal PCA, SVD method
            U,S,V = np.linalg.svd(X)
            V = V[:num_data] # only makes sense to return the first num_data
        return V, S, mean_X
img=color.rgb2gray(io.imread('D:\lenna.png'))
x,y,z=pca(img)
plt.imshow(x)

但是 PCA 的图像图看起来根本不像原始图像。据我所知，PCA 有点减少图像尺寸，但它仍然会在某种程度上类似于原始图像，但细节较低。代码有什么问题吗？

Answer 1

好吧，您的代码本身没有任何问题，但如果我确实理解您真正想要做什么，那么您就没有显示正确的内容!

我会为您的问题写下以下内容:

def pca(X, number_of_pcs):
    num_data, dim = X.shape

    mean_X = X.mean(axis=0)
    X = X - mean_X

    if dim > num_data:
        # PCA compact trick
        M = np.dot(X, X.T) # covariance matrix
        e, U = np.linalg.eigh(M) # calculate eigenvalues an deigenvectors
        tmp = np.dot(X.T, U).T
        V = tmp[::-1] # reverse since the last eigenvectors are the ones we want
        S = np.sqrt(e)[::-1] #reverse since the last eigenvalues are in increasing order
        for i in range(V.shape[1]):
            V[:,i] /= S

        return V, S, mean_X

    else:
        # normal PCA, SVD method
        U, S, V = np.linalg.svd(X, full_matrices=False)

        # reconstruct the image using U, S and V
        # otherwise you're just outputting the eigenvectors of X*X^T
        V = V.T
        S = np.diag(S)
        X_hat = np.dot(U[:, :number_of_pcs], np.dot(S[:number_of_pcs, :number_of_pcs], V[:,:number_of_pcs].T))      

        return X_hat, S, mean_X

The change here lies in the fact that we want to reconstruct the image using a given number of eigenvectors (determined by number_of_pcs).

要记住的是，在 np.linalg.svd 中，U 的列是 X.X^T 的特征向量。

这样做时，我们获得以下结果(此处使用 1 和 10 个主成分显示):

<小时/>

X_hat, S, mean_X = pca(img, 1)
plt.imshow(X_hat)

<小时/>

X_hat, S, mean_X = pca(img, 10)
plt.imshow(X_hat)

PS:请注意，由于 matplotlib.pyplot，图片不会以灰度显示，但这是一个非常小的问题。