比较矩阵和NAN时，numpy.isclose中可能存在一个错误

考虑下一段代码：

In [90]: m1 = np.matrix([1,2,3], dtype=np.float32)

In [91]: m2 = np.matrix([1,2,3], dtype=np.float32)

In [92]: m3 = np.matrix([1,2,'nan'], dtype=np.float32)

In [93]: np.isclose(m1, m2, equal_nan=True)
Out[93]: matrix([[ True,  True,  True]], dtype=bool)

In [94]: np.isclose(m1, m3, equal_nan=True)
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-94-5d2b979bc263> in <module>()
----> 1 np.isclose(m1, m3, equal_nan=True)

/usr/local/lib/python2.7/dist-packages/numpy/core/numeric.pyc in isclose(a, b, rtol, atol, equal_nan)
   2571         # Ideally, we'd just do x, y = broadcast_arrays(x, y). It's in
   2572         # lib.stride_tricks, though, so we can't import it here.
-> 2573         x = x * ones_like(cond)
   2574         y = y * ones_like(cond)
   2575         # Avoid subtraction with infinite/nan values...

/usr/local/lib/python2.7/dist-packages/numpy/matrixlib/defmatrix.pyc in __mul__(self, other)
    341         if isinstance(other, (N.ndarray, list, tuple)) :
    342             # This promotes 1-D vectors to row vectors
--> 343             return N.dot(self, asmatrix(other))
    344         if isscalar(other) or not hasattr(other, '__rmul__') :
    345             return N.dot(self, other)

ValueError: shapes (1,3) and (1,3) not aligned: 3 (dim 1) != 1 (dim 0)

为什么np.isclose会失败？从文件上看，它似乎应该起作用

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谢谢

这个问题来自于

np.nan==np.nan

，它在浮点逻辑中是

False

In [39]: np.nan == np.nan
Out[39]: False

The `equal_nan` parameter is to force two `nan` values to be considered as equal , not to consider any value to be equal to `nan`.

In [37]: np.isclose(m3,m3)
Out[37]: array([ True,  True, False], dtype=bool)

In [38]: np.isclose(m3,m3,equal_nan=True)
Out[38]: array([ True,  True,  True], dtype=bool)

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矩阵类型是罪魁祸首（适用于数组）。不知道如何将该类型的文档解释为输入！您使用的是什么np版本？对我来说，行np.isclose（m3，m3）导致ValueError:形状（1,3）和（1,3）未对齐：3（尺寸1）！=1（0）对不起。我看不出你在处理矩阵。正如在评论中所说的，问题是isclose只接受Ndarays。矩阵是一种准弃用结构。

In [39]: np.nan == np.nan
Out[39]: False

The `equal_nan` parameter is to force two `nan` values to be considered as equal , not to consider any value to be equal to `nan`.

In [37]: np.isclose(m3,m3)
Out[37]: array([ True,  True, False], dtype=bool)

In [38]: np.isclose(m3,m3,equal_nan=True)
Out[38]: array([ True,  True,  True], dtype=bool)