On May 20, 10:34 pm, Robert Kern <robert.k...@gmail.com> wrote: > On Wed, May 20, 2009 at 14:24, dmitrey <dmitrey.kros...@scipy.org> wrote: > > hi all, > > > suppose I have A that is numpy ndarray of floats, with shape n x n. > > > I want to obtain dot(A, b), b is vector of length n and norm(b)=1, but > > instead of exact multiplication I want to approximate b as a vector > > [+/- 2^m0, ± 2^m1, ± 2^m2 ,,, ± 2^m_n], m_i are integers, and then > > invoke left_shift(vector_m) for rows of A. > > You don't shift floats. You only shift integers. For floats, > multiplying by an integer power of 2 should be fast because of the > floating point representation (the exponent just gets incremented or > decremented), so just do the multiplication. > > > So, what is the simplest way to do it, without cycles of course? Or it > > cannot be implemented w/o cycles with current numpy version? > > It might help if you showed us an example of an actual b vector > decomposed the way you describe. Your description is ambiguous. > > -- > Robert Kern
For the task involved (I intend to try using it for speed up ralg solver) it doesn't matter essentially (using ceil, floor or round), but for example let m_i is floor(log2(b_i)) for b_i > 1e-15, ceil(log2(-b_i)) for b_i < - 1e-15, for - 1e-15 <= b_i <= 1e-15 - don't modify the elements of A related to the b_i at all. D. _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
