CODEWITHSUNDEEP
pythonpcasvd
Bitswazsky
Bitswazsky

Reputation: 4698

Trouble in understanding how PCA is achieving image compression and reducing dimension

I was going through this amazing playlist for SVD by Steve Brunton in youtube. I think I got majority of the concepts but there are some gaps. Let me add a couple of screenshots so that it's easier for me to explain.

svd-1

He is considering the input matrix X to be a collection of images. So, considering an image is 28x28 pixels, we flatten it to create a 784x1 column vector. So, each column denotes an image, and the rows denote pixel indices. Let's take the dimension of X to be n x m. Now, after computing the economy SVD, if we keep only the first r (<< m) singular values, then the approximation of X is given by

X' = σ1.u1.v1(T) + σ2.u2.v2(T) + ... + σr.ur.vr(T)

svd-2

I understand that here, we're throwing away information, so the reconstructed images would be pixelated but they would still be of the same dimension (28x28). So, how are we achieving compression here? Is it because instead of storing 784m pixel values, we'll have to store r x (28 (length of each u) + 28 (length of each v)) pixels? Or is there something more to it?

My second question is, if I try to draw an analogy to numerical features, e.g. let's say a housing price dataset, that has 50 features, and 1000 data points. So, our X matrix has dimension 50 x 1000 (each column being a feature vector). In that case, if there are useless features, we'll get << 50 features (maybe 20, or 10... whatever) after applying PCA, right? I'm not able to grasp how that smaller feature vector is derived when we select only the biggest r singular values. Because X and X' have the same dimensions.

Let's have a sample code. The dimensions are reversed because of how sklearn expects it.

pca = PCA(n_components=10)
pca.fit(X)
X_pca = pca.transform(X)
print("original shape:   ", X.shape)      # original shape:    (1000, 50)
print("transformed shape:", X_pca.shape)  # transformed shape: (1000, 10)

So, how are we going from 50 to 10 here? I get that that in this case there would be 50 U basis vectors. So, even if we choose top r from these 50, the dimensions will still be the same, right? Any help is appreciated.

Upvotes: 2

Views: 290

Answers (1)

Bitswazsky
Bitswazsky

Reputation: 4698

I've been searching for the answer all over the web, and finally it clicked when I saw this video tutorial. We know X = U x ∑ x V.T. Here, columns of U give us the principal components for the colspace of X. Similarly rows of V.T give us the principal components for the rowspace of X. Since, in pca we tend to represent a feature vector by a row (unlike svd), so we'd select the first r principal components from the matrix V.T.

Let's assume the dimensions of X to be mxn. So, we have m samples each having n features. That gives us the following dimensions for the SVD:

  • U: mxm
  • ∑: mxn
  • V: nxn

Now, if we select only r (<< n) principal components then the projection of X to the r-dimensional space would be given by X.[v1 v2 ... vr]. Here each of v1, v2, ... vr is a column vector. So, the dimension of [v1 v2 ... vr] is nxr. If we now multiply X with this vector we get an nxr matrix, which is nothing but the projection of all the data points to r dimensions.

Upvotes: 1

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