3

Suppose I want to normalize a matrix A. I came across this code:

A_norm = (A / np.sqrt((A ** 2).sum(-1))[..., np.newaxis]).astype(np.float32)

We're subtracting a mean of 0 here, I assume. My problem is the denominator. We're taking the square root of something we've squared and summed, but I don't understand what.

Specifically, what does this do:

np.sqrt((A ** 2).sum(-1))[..., np.newaxis]
StatsSorceress
  • 3,019
  • 7
  • 41
  • 82
  • 1
    Square-root of summed up values along the last axis. That `[..., np.newaxis]` basically keeps the number of dims same that was reduced by the summing. Alternative is using [keepdims` with summing](https://docs.scipy.org/doc/numpy-1.10.0/reference/generated/numpy.sum.html). On the ellipsis `...` to extend the last dim, see [`here`](http://stackoverflow.com/a/40383002/3293881). More info on ellipsis : http://stackoverflow.com/questions/772124/what-does-the-python-ellipsis-object-do – Divakar Dec 02 '16 at 21:04
  • Recent question on `keepdims`, with example: http://stackoverflow.com/questions/40927156/what-the-role-of-keepdims-in-python – hpaulj Dec 02 '16 at 21:39
  • [`np.newaxis`](https://docs.scipy.org/doc/numpy/reference/arrays.indexing.html#numpy.newaxis) is something worth figuring out even though it’s confusing. It’s basically Numpy’s answer to the question, “How do we convert rank-1 arrays (vectors) to rank-2 and rank-3 and rank-N arrays (matrix-like arrays, volumes, N-dimensional volumes, respectively)”. Experiment with the following in Python (IPython, Jupyter Notebook, etc.): `np.random.rand(3)[:, np.newaxis]` and `np.random.rand(3,3)[:, :, np.newaxis]`, replacing the colors with `...` and replacing `np.newaxis` with `None`. – Ahmed Fasih Dec 03 '16 at 00:11

1 Answers1

3

We have

np.sqrt((A ** 2).sum(-1))

which can be decomposed as 

B = A**2
C = B.sum(-1)
D = np.sqrt(C)

where C is the row-sum of B, which has been flatten by the operation (the column sum would have been B.sum(0)) and D is the entrywise squareroot of C.

For A to be correctly normalized, we need the denominator to be correctly shaped, so as to divide the kth row of A by the kth term of D. Thus we must explicitly reshape the flatten result of np.sqrt(C) as being a column vector as follows

D = np.sqrt(C)[..., np.newaxis]
keepAlive
  • 6,369
  • 5
  • 24
  • 39