# 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_matrix: array_like

Input matrix, of shape (n, d).

sketch_size: int

Number of rows for the sketch.

seedNone or int or numpy.random.mtrand.RandomState instance, optional

This parameter defines the RandomState object to use for drawing random variates. If None (or np.random), the global np.random state is used. If integer, it is used to seed the local RandomState instance. Default is None.

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 [2] 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.

>>> from scipy import linalg
>>> from scipy import sparse
>>> 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 = np.random.randn(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

1

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

2

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

Examples

Given a big dense matrix A:

>>> from scipy import linalg
>>> n_rows, n_columns, sketch_n_rows = 15000, 100, 200
>>> A = np.random.randn(n_rows, n_columns)
>>> sketch = linalg.clarkson_woodruff_transform(A, sketch_n_rows)
>>> sketch.shape
(200, 100)
>>> norm_A = np.linalg.norm(A)
>>> norm_sketch = np.linalg.norm(sketch)


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

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

>>> from scipy import linalg
>>> n_rows, n_columns, sketch_n_rows = 15000, 100, 200
>>> A = np.random.randn(n_rows, n_columns)
>>> b = np.random.randn(n_rows)
>>> x = np.linalg.lstsq(A, b, rcond=None)
>>> Ab = np.hstack((A, b.reshape(-1,1)))
>>> SAb = linalg.clarkson_woodruff_transform(Ab, sketch_n_rows)
>>> SA, Sb = SAb[:,:-1], SAb[:,-1]
>>> x_sketched = np.linalg.lstsq(SA, Sb, rcond=None)


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

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