Numpy array precision, how to overcome with search another array

You're just seeing the effects of floating point arithmetic. (The same thing is true if you used a python list instead of a numpy array.)

I'm actually rather surprised that there's not a built-in function to do floating point "close" comparisons in numpy... There's numpy.allclose which does it for comparing between two numpy arrays, but it just returns True or False rather than a boolean array.

Doing this in general is actually a bit tricky. inf will thrown in false positives and false negatives. Furthermore subtracting two arrays with inf or nan in them will raise a warning, so we generally want to avoid doing that...

import numpy as np

def close(a, b, rtol=1.e-5, atol=1.e-8, check_invalid=True):
    """Similar to numpy.allclose, but returns a boolean array.
    See numpy.allclose for an explanation of *rtol* and *atol*."""
    def within_tol(x, y, atol, rtol):
        return np.less_equal(np.abs(x-y), atol + rtol * np.abs(y))
    x = np.array(a, copy=False)
    y = np.array(b, copy=False)
    if not check_invalid:
        return within_tol(x, y, atol, rtol)
    xfin = np.isfinite(x)
    yfin = np.isfinite(y)
    if np.all(xfin) and np.all(yfin):
        return within_tol(x, y, atol, rtol)
    else:
        # Avoid subtraction with infinite/nan values...
        cond = np.zeros(np.broadcast(x, y).shape, dtype=np.bool)
        mask = xfin & yfin
        cond[mask] = within_tol(x[mask], y[mask], atol, rtol)
        # Inf and -Inf equality...
        cond[~mask] = (x[~mask] == y[~mask])
        # NaN equality...
        cond[np.isnan(x) & np.isnan(y)] = True
        return cond

# A few quick tests...
assert np.any(close(0.300001, np.array([0.1, 0.2, 0.3, 0.4])))

x = np.array([0.1, np.nan, np.inf, -np.inf])
y = np.array([0.1000001, np.nan, np.inf, -np.inf])
assert np.all(close(x, y))

x = np.array([0.1, 0.2, np.inf])
y = np.array([0.101, np.nan, 0.2])
assert not np.all(close(x, y))