Hi Keith! On Mon, Jul 19, 2010 at 6:40 PM, Keith Goodman <kwgoodman@gmail.com> wrote:
On Mon, Jul 19, 2010 at 6:31 PM, Ondrej Certik <ondrej@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