numpy.allclose

numpy.allclose(a, b, rtol=1e-05, atol=1e-08, equal_nan=False)[源代码]

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.

If either array contains one or more NaNs, False is returned. Infs are treated as equal if they are in the same place and of the same sign in both arrays.

Parameters:
a, b : array_like

Input arrays to compare.

rtol : float

The relative tolerance parameter (see Notes).

atol : float

The absolute tolerance parameter (see Notes).

equal_nan : bool

Whether to compare NaN’s as equal. If True, NaN’s in a will be considered equal to NaN’s in b in the output array.

1.10.0 新版功能.

Returns:
allclose : bool

Returns True if the two arrays are equal within the given tolerance; False otherwise.

参见

isclose, all, any, equal

Notes

If the following equation is element-wise True, then allclose returns True.

absolute(a - b) <= (atol + rtol * absolute(b))

The above equation is not symmetric in a and b, so that allclose(a, b) might be different from allclose(b, a) in some rare cases.

The comparison of a and b uses standard broadcasting, which means that a and b need not have the same shape in order for allclose(a, b) to evaluate to True. The same is true for equal but not array_equal.

Examples

>>> np.allclose([1e10,1e-7], [1.00001e10,1e-8])
False
>>> np.allclose([1e10,1e-8], [1.00001e10,1e-9])
True
>>> np.allclose([1e10,1e-8], [1.0001e10,1e-9])
False
>>> np.allclose([1.0, np.nan], [1.0, np.nan])
False
>>> np.allclose([1.0, np.nan], [1.0, np.nan], equal_nan=True)
True