有没有办法让 Python all() 函数处理多维数组?

Is there a way to get Python all() function to work with multi-dimensional arrays?

我正在尝试为基础 class 实现通用且灵活的 __eq__ 方法,该方法将适用于尽可能多的对象类型,包括可迭代对象和 numpy 数组。

这是我目前的情况:

class Environment:

    def __init__(self, state):
        self.state = state

    def __eq__(self, other):
        """Compare two environments based on their states.
        """
        if isinstance(other, self.__class__):
            try:
                return all(self.state == other.state)
            except TypeError:
                return self.state == other.state
        return False

这适用于大多数对象类型,包括一维数组:

s = 'abcdef'
e1 = Environment(s)
e2 = Environment(s)

e1 == e2  # True

s = [[1, 2, 3], [4, 5, 6]]
e1 = Environment(s)
e2 = Environment(s)

e1 == e2  # True

s = np.array(range(6))
e1 = Environment(s)
e2 = Environment(s)

e1 == e2  # True

问题是,当 self.state 是一个多维的 numpy 数组时,它 returns 一个 ValueError。

s = np.array(range(6)).reshape((2, 3))
e1 = Environment(s)
e2 = Environment(s)

e1 == e2

产生:

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

显然,我可以检查 isinstance(other, np.ndarray),然后执行 (return self.state == other.state).all(),但只是认为可能有一种更通用的方法可以用一个语句处理所有类型的所有迭代、集合和数组。

我也有点困惑,为什么 all() 不像 array.all() 那样遍历数组的所有元素。有没有办法触发 np.nditer 并可能这样做?

Signature: all(iterable, /)
Docstring:
Return True if bool(x) is True for all values x in the iterable.

对于一维数组:

In [200]: x=np.ones(3)                                                               
In [201]: x                                                                          
Out[201]: array([1., 1., 1.])
In [202]: y = x==x                                                                   
In [203]: y          # 1d array of booleans                                                                      
Out[203]: array([ True,  True,  True])
In [204]: bool(y[0])                                                                 
Out[204]: True
In [205]: all(y)                                                                     
Out[205]: True

对于二维数组:

In [206]: x=np.ones((2,3))                                                           
In [207]: x                                                                          
Out[207]: 
array([[1., 1., 1.],
       [1., 1., 1.]])
In [208]: y = x==x                                                                   
In [209]: y                                                                          
Out[209]: 
array([[ True,  True,  True],
       [ True,  True,  True]])
In [210]: y[0]                                                                       
Out[210]: array([ True,  True,  True])
In [211]: bool(y[0])                                                                 
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-211-d0ce0868392c> in <module>
----> 1 bool(y[0])

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

但对于不同的二维数组:

In [212]: x=np.ones((3,1))                                                           
In [213]: y = x==x                                                                   
In [214]: y                                                                          
Out[214]: 
array([[ True],
       [ True],
       [ True]])
In [215]: y[0]                                                                       
Out[215]: array([ True])
In [216]: bool(y[0])                                                                 
Out[216]: True
In [217]: all(y)                                                                     
Out[217]: True

numpy 数组的迭代发生在第一个维度上。 [i for i in x]

只要在需要标量布尔值的上下文中使用多值布尔数组,就会引发这种歧义 ValueError。 ifor/and 表达式是常见的。

In [223]: x=np.ones((2,3))                                                           
In [224]: y = x==x                                                                   
In [225]: np.all(y)                                                                  
Out[225]: True

np.all 与 Python all 的不同之处在于它 'knows' 关于维度。在这种情况下,它执行 ravel 将数组视为 1d:

The default (axis = None) is to perform a logical AND over all the dimensions of the input array.

这不是我希望的简洁解决方案,可能效率低下,但我认为它适用于任何 n-dimensional 可迭代对象:

def nd_true(nd_object):
    try:
        iterator = iter(nd_object)
    except TypeError:
        return nd_object
    else:
        return all([nd_true(x) for x in iterator])

class Environment:

    def __init__(self, state):
        self.state = state

    def __eq__(self, other):
        """Compare two environments based on their states.
        """
        if isinstance(other, self.__class__):
            return nd_true(self.state == other.state)
        return False

# Tests    
s = 'abcdef'
e1 = Environment(s)
e2 = Environment(s)

e1 == e2  # True

s = [[1, 2, 3], [4, 5, 6]]
e1 = Environment(s)
e2 = Environment(s)

e1 == e2  # True

s = np.array(range(6))
e1 = Environment(s)
e2 = Environment(s)

e1 == e2  # True

s = np.array(range(6)).reshape((2, 3))
e1 = Environment(s)
e2 = Environment(s)

e1 == e2  # True

s = np.array(range(27)).reshape((3, 3, 3))
e1 = Environment(s)
e2 = Environment(s)

e1 == e2  # True