I have two numpy arrays:
g1 = np.array([3118740.3553, 3520175.8121]) g2 = np.array([3118740.8553, 3520176.3121]) I want to use numpy.allclose() to test if those arrays are identical inside the floating point precision tolerance
np.allclose(g1, g2, atol=1e-7) Curiously it returns True even if the difference between those two arrays is significant. Why?
1 Answer
The call signature of np.allclose is
In [4]: np.allclose? Signature: np.allclose(a, b, rtol=1e-05, atol=1e-08, equal_nan=False) Notice that the default for rtol (relative tolerance) is 1e-05. As long as
abs(a[i] - b[i]) <= rtol * abs(b[i]) + atol for all i = 0, ..., len(a), then np.allclose returns True.
In [11]: rtol, atol = 1e-05, 1e-7 In [12]: [abs(ai - bi) < rtol * abs(bi) + atol for ai, bi in zip(g1, g2)] Out[12]: [True, True] Since the values in g2 are large, even a small rtol leads to a fairly large tolerance:
In [14]: rtol * g2.min() Out[14]: 31.187408553 If you don't want to include a relative tolerance, you must set it to zero to override the default:
In [13]: np.allclose(g1, g2, rtol=0, atol=1e-7) Out[13]: False 
