Hi Keith! On Mon, Jul 19, 2010 at 6:40 PM, Keith Goodman <kwgood...@gmail.com> wrote: > On Mon, Jul 19, 2010 at 6:31 PM, Ondrej Certik <ond...@certik.cz> wrote: >> Hi, >> >> I was always using something like >> >> abs(x-y) < eps >> >> or >> >> (abs(x-y) < eps).all() >> >> but today I needed to also make sure this works for larger numbers, >> where I need to compare relative errors, so I found this: >> >> http://www.cygnus-software.com/papers/comparingfloats/comparingfloats.htm >> >> and wrote this: >> >> def feq(a, b, max_relative_error=1e-12, max_absolute_error=1e-12): >> a = float(a) >> b = float(b) >> # if the numbers are close enough (absolutely), then they are equal >> if abs(a-b) < max_absolute_error: >> return True >> # if not, they can still be equal if their relative error is small >> if abs(b) > abs(a): >> relative_error = abs((a-b)/b) >> else: >> relative_error = abs((a-b)/a) >> return relative_error <= max_relative_error >> >> >> Is there any function in numpy, that implements this? Or maybe even >> the better, integer based version, as referenced in the link above? >> >> I need this in tests, where I calculate something on some mesh, then >> compare to the correct solution projected on some other mesh, so I >> have to deal with accuracy issues. > > Is allclose close enough? > > np.allclose(a, b, rtol=1.0000000000000001e-05, atol=1e-08) > > Returns True if two arrays are element-wise equal within a tolerance. > > The tolerance values are positive, typically very small numbers. The > relative difference (`rtol` * abs(`b`)) and the absolute difference > `atol` are added together to compare against the absolute difference > between `a` and `b`.
thanks for this. This should do the job. I'll give it a shot and report back. Ondrej _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
