# scipy.linalg.clarkson_woodruff_transform#

scipy.linalg.clarkson_woodruff_transform(input_matrix, sketch_size, seed=None)[source]#

Applies a Clarkson-Woodruff Transform/sketch to the input matrix.

Given an input_matrix A of size (n, d), compute a matrix A' of size (sketch_size, d) so that

$\|Ax\| \approx \|A'x\|$

with high probability via the Clarkson-Woodruff Transform, otherwise known as the CountSketch matrix.

Parameters:
input_matrixarray_like

Input matrix, of shape (n, d).

sketch_sizeint

Number of rows for the sketch.

seed{None, int, numpy.random.Generator, numpy.random.RandomState}, optional

If seed is None (or np.random), the numpy.random.RandomState singleton is used. If seed is an int, a new RandomState instance is used, seeded with seed. If seed is already a Generator or RandomState instance then that instance is used.

Returns:
A’array_like

Sketch of the input matrix A, of size (sketch_size, d).

Notes

To make the statement

$\|Ax\| \approx \|A'x\|$

precise, observe the following result which is adapted from the proof of Theorem 14 of  via Markov’s Inequality. If we have a sketch size sketch_size=k which is at least

$k \geq \frac{2}{\epsilon^2\delta}$

Then for any fixed vector x,

$\|Ax\| = (1\pm\epsilon)\|A'x\|$

with probability at least one minus delta.

This implementation takes advantage of sparsity: computing a sketch takes time proportional to A.nnz. Data A which is in scipy.sparse.csc_matrix format gives the quickest computation time for sparse input.

>>> import numpy as np
>>> from scipy import linalg
>>> from scipy import sparse
>>> rng = np.random.default_rng()
>>> n_rows, n_columns, density, sketch_n_rows = 15000, 100, 0.01, 200
>>> A = sparse.rand(n_rows, n_columns, density=density, format='csc')
>>> B = sparse.rand(n_rows, n_columns, density=density, format='csr')
>>> C = sparse.rand(n_rows, n_columns, density=density, format='coo')
>>> D = rng.standard_normal((n_rows, n_columns))
>>> SA = linalg.clarkson_woodruff_transform(A, sketch_n_rows) # fastest
>>> SB = linalg.clarkson_woodruff_transform(B, sketch_n_rows) # fast
>>> SC = linalg.clarkson_woodruff_transform(C, sketch_n_rows) # slower
>>> SD = linalg.clarkson_woodruff_transform(D, sketch_n_rows) # slowest


That said, this method does perform well on dense inputs, just slower on a relative scale.

References



Kenneth L. Clarkson and David P. Woodruff. Low rank approximation and regression in input sparsity time. In STOC, 2013.



David P. Woodruff. Sketching as a tool for numerical linear algebra. In Foundations and Trends in Theoretical Computer Science, 2014.

Examples

Create a big dense matrix A for the example:

>>> import numpy as np
>>> from scipy import linalg
>>> n_rows, n_columns  = 15000, 100
>>> rng = np.random.default_rng()
>>> A = rng.standard_normal((n_rows, n_columns))


Apply the transform to create a new matrix with 200 rows:

>>> sketch_n_rows = 200
>>> sketch = linalg.clarkson_woodruff_transform(A, sketch_n_rows, seed=rng)
>>> sketch.shape
(200, 100)


Now with high probability, the true norm is close to the sketched norm in absolute value.

>>> linalg.norm(A)
1224.2812927123198
>>> linalg.norm(sketch)
1226.518328407333


Similarly, applying our sketch preserves the solution to a linear regression of $$\min \|Ax - b\|$$.

>>> b = rng.standard_normal(n_rows)
>>> x = linalg.lstsq(A, b)
>>> Ab = np.hstack((A, b.reshape(-1, 1)))
>>> SAb = linalg.clarkson_woodruff_transform(Ab, sketch_n_rows, seed=rng)
>>> SA, Sb = SAb[:, :-1], SAb[:, -1]
>>> x_sketched = linalg.lstsq(SA, Sb)


As with the matrix norm example, linalg.norm(A @ x - b) is close to linalg.norm(A @ x_sketched - b) with high probability.

>>> linalg.norm(A @ x - b)
122.83242365433877
>>> linalg.norm(A @ x_sketched - b)
166.58473879945151