所以这就是我所假设的两个问题，就像我一样，基本的潜在混淆。我希望没问题。
这里有一些代码：
import numpy as np
class new_array(np.ndarray):
def __new__(cls, array, foo):
obj = array.view(cls)
obj.foo = foo
return obj
def __array_finalize__(self, obj):
print "__array_finalize"
if obj is None: return
self.foo = getattr(obj, 'foo', None)
def __getitem__(self, key):
print "__getitem__"
print "key is %s"%repr(key)
print "self.foo is %d, self.view(np.ndarray) is %s"%(
self.foo,
repr(self.view(np.ndarray))
)
self.foo += 1
return super(new_array, self).__getitem__(key)
print "Block 1"
print "Object construction calls"
base_array = np.arange(20).reshape(4,5)
print "base_array is %s"%repr(base_array)
p = new_array(base_array, 0)
print "\n\n"
print "Block 2"
print "Call sequence for p[-1:] is:"
p[-1:]
print "p[-1].foo is %d\n\n"%p.foo
print "Block 3"
print "Call sequence for s = p[-1:] is:"
s = p[-1:]
print "p[-1].foo is now %d"%p.foo
print "s.foo is now %d"%s.foo
print "s.foo + p.foo = %d\n\n"%(s.foo + p.foo)
print "Block 4"
print "Doing q = s + s"
q = s + s
print "q.foo = %d\n\n"%q.foo
print "Block 5"
print "Printing s"
print repr(s)
print "p.foo is now %d"%p.foo
print "s.foo is now %d\n\n"%s.foo
print "Block 6"
print "Printing q"
print repr(q)
print "p.foo is now %d"%p.foo
print "s.foo is now %d"%s.foo
print "q.foo is now %d\n\n"%q.foo
print "Block 7"
print "Call sequence for p[-1]"
a = p[-1]
print "p[-1].foo is %d\n\n"%a.foo
print "Block 8"
print "Call sequence for p[slice(-1, None, None)] is:"
a = p[slice(-1, None, None)]
print "p[slice(None, -1, None)].foo is %d"%a.foo
print "p.foo is %d"%p.foo
print "s.foo + p.foo = %d\n\n"%(s.foo + p.foo)
此代码的输出是
Block 1
Object construction calls
base_array is array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]])
__array_finalize
Block 2
Call sequence for p[-1:] is:
__array_finalize
p[-1].foo is 0
Block 3
Call sequence for s = p[-1:] is:
__array_finalize
p[-1].foo is now 0
s.foo is now 0
s.foo + p.foo = 0
Block 4
Doing q = s + s
__array_finalize
q.foo = 0
Block 5
Printing s
__getitem__
key is -1
self.foo is 0, self.view(np.ndarray) is array([[15, 16, 17, 18, 19]])
__array_finalize
__getitem__
key is -5
self.foo is 1, self.view(np.ndarray) is array([15, 16, 17, 18, 19])
__getitem__
key is -4
self.foo is 2, self.view(np.ndarray) is array([15, 16, 17, 18, 19])
__getitem__
key is -3
self.foo is 3, self.view(np.ndarray) is array([15, 16, 17, 18, 19])
__getitem__
key is -2
self.foo is 4, self.view(np.ndarray) is array([15, 16, 17, 18, 19])
__getitem__
key is -1
self.foo is 5, self.view(np.ndarray) is array([15, 16, 17, 18, 19])
new_array([[15, 16, 17, 18, 19]])
p.foo is now 0
s.foo is now 1
Block 6
Printing q
__getitem__
key is -1
self.foo is 0, self.view(np.ndarray) is array([[30, 32, 34, 36, 38]])
__array_finalize
__getitem__
key is -5
self.foo is 1, self.view(np.ndarray) is array([30, 32, 34, 36, 38])
__getitem__
key is -4
self.foo is 2, self.view(np.ndarray) is array([30, 32, 34, 36, 38])
__getitem__
key is -3
self.foo is 3, self.view(np.ndarray) is array([30, 32, 34, 36, 38])
__getitem__
key is -2
self.foo is 4, self.view(np.ndarray) is array([30, 32, 34, 36, 38])
__getitem__
key is -1
self.foo is 5, self.view(np.ndarray) is array([30, 32, 34, 36, 38])
new_array([[30, 32, 34, 36, 38]])
p.foo is now 0
s.foo is now 1
q.foo is now 1
Block 7
Call sequence for p[-1]
__getitem__
key is -1
self.foo is 0, self.view(np.ndarray) is array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]])
__array_finalize
p[-1].foo is 1
Block 8
Call sequence for p[slice(-1, None, None)] is:
__getitem__
key is slice(-1, None, None)
self.foo is 1, self.view(np.ndarray) is array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]])
__array_finalize
p[slice(None, -1, None)].foo is 2
p.foo is 2
s.foo + p.foo = 3
请注意两件事：
对p[-1:]的调用不会导致对new_array.__getitem__的调用。如果将p[-1:]替换为p[0:]，p[0:-1]等内容，则属实......但p[-1]和p[slice(-1, None, None)]之类的语句会导致调用new_array.__getitem__。对于p[-1:] + p[-1:]或s = p[-1]等语句也是如此，但对于print s这样的语句则不然。您可以通过查看上面给出的“块”来看到这一点。
变量foo在调用new_array.__getitem__期间已正确更新（参见第5和第6块），但在new_array.__getitem__的评估完成后不正确（请再次参阅，第5和第6栏）。我还应该补充说，用return super(new_array, self).__getitem__(key)替换行return new_array(np.array(self.view(np.ndarray)[key]), self.foo)也不起作用。以下块是输出中唯一的差异。
Block 5
Printing s
__getitem__
key is -1
self.foo is 0, self.view(np.ndarray) is array([[15, 16, 17, 18, 19]])
__array_finalize__
__getitem__
key is -5
self.foo is 1, self.view(np.ndarray) is array([15, 16, 17, 18, 19])
__array_finalize__
__array_finalize__
__array_finalize__
__getitem__
key is -4
self.foo is 2, self.view(np.ndarray) is array([15, 16, 17, 18, 19])
__array_finalize__
__array_finalize__
__array_finalize__
__getitem__
key is -3
self.foo is 3, self.view(np.ndarray) is array([15, 16, 17, 18, 19])
__array_finalize__
__array_finalize__
__array_finalize__
__getitem__
key is -2
self.foo is 4, self.view(np.ndarray) is array([15, 16, 17, 18, 19])
__array_finalize__
__array_finalize__
__array_finalize__
__getitem__
key is -1
self.foo is 5, self.view(np.ndarray) is array([15, 16, 17, 18, 19])
__array_finalize__
__array_finalize__
__array_finalize__
new_array([[15, 16, 17, 18, 19]])
p.foo is now 0
s.foo is now 1
Block 6
Printing q
__getitem__
key is -1
self.foo is 0, self.view(np.ndarray) is array([[30, 32, 34, 36, 38]])
__array_finalize__
__getitem__
key is -5
self.foo is 1, self.view(np.ndarray) is array([30, 32, 34, 36, 38])
__array_finalize__
__array_finalize__
__array_finalize__
__getitem__
key is -4
self.foo is 2, self.view(np.ndarray) is array([30, 32, 34, 36, 38])
__array_finalize__
__array_finalize__
__array_finalize__
__getitem__
key is -3
self.foo is 3, self.view(np.ndarray) is array([30, 32, 34, 36, 38])
__array_finalize__
__array_finalize__
__array_finalize__
__getitem__
key is -2
self.foo is 4, self.view(np.ndarray) is array([30, 32, 34, 36, 38])
__array_finalize__
__array_finalize__
__array_finalize__
__getitem__
key is -1
self.foo is 5, self.view(np.ndarray) is array([30, 32, 34, 36, 38])
__array_finalize__
__array_finalize__
__array_finalize__
new_array([[30, 32, 34, 36, 38]])
p.foo is now 0
s.foo is now 1
q.foo is now 1
现在包含对new_array.__array_finalize__的过多调用，但没有
使用变量foo更改“问题”。
我希望像p[-1:]这样的new_array对p.foo = 0对象的调用会导致此语句p.foo == 1返回True。显然情况并非如此，即使在foo调用期间__getitem__被正确更新，因为p[-1:]之类的语句会导致对__getitem__的大量调用（一旦考虑延迟评估）。此外，调用p[-1:]和p[slice(-1, None, None)]会导致foo的不同值（如果计数的内容正常工作）。在前一种情况下，foo会添加5，而在后一种情况下，foo会添加1。
虽然对numpy数组切片的延迟评估在评估我的代码时不会引起问题，但使用pdb调试我的一些代码却是一个巨大的痛苦。基本上，语句在运行时和pdb中的评估方式不同。我认为这不好。这就是我偶然发现这种行为的原因。
我的代码使用__getitem__的输入来评估应返回的对象类型。在某些情况下，它返回一个相同类型的新实例，在其他情况下，它返回一个其他类型的新实例，在其他情况下，它返回一个numpy数组，标量或浮点数（取决于底层numpy数组认为是正确的） ）。我使用传递给__getitem__的密钥来确定要返回的正确对象是什么。但是如果用户通过切片，我不能这样做，例如像p[-1:]这样的东西，因为该方法只获得单独的索引，例如好像用户写了p[4]。 那么，如果我的numpy子类的key中的__getitem__没有反映用户是否正在请求切片，p[-1:]给出，或者只是一个条目，我该如何做，由p[4]提供？
作为一个侧面点，numpy indexing文档暗示了切片对象，例如slice(start, stop, step)将与start:stop:step等语句一样对待。这让我觉得我错过了一些非常基本的东西。暗示这种情况的句子很早就出现了：
当obj是切片对象（由start：stop：step表示法构造）时，会发生基本切片 括号内），整数或切片对象和整数的元组。
我不禁觉得同样的基本错误也是我认为self.foo += 1行应该计算用户请求切片的次数或{的实例的元素的原因{1}}（而不是切片中“元素”的数量）。 这两个问题实际上是否相关，如果是这样的话？
答案 0 :(得分：8)
你确实被一个讨厌的小虫咬了一口。知道我不是唯一一个人，这是一种解脱！幸运的是，它很容易解决。只需在课堂上添加以下内容即可。这实际上是我几个月前写的一些代码的复制粘贴，文档字符串告诉我们发生了什么，但你也可能想要阅读the python docs。
def __getslice__(self, start, stop) :
"""This solves a subtle bug, where __getitem__ is not called, and all
the dimensional checking not done, when a slice of only the first
dimension is taken, e.g. a[1:3]. From the Python docs:
Deprecated since version 2.0: Support slice objects as parameters
to the __getitem__() method. (However, built-in types in CPython
currently still implement __getslice__(). Therefore, you have to
override it in derived classes when implementing slicing.)
"""
return self.__getitem__(slice(start, stop))
答案 1 :(得分：0)
使用isinstance方法检查切片类型。
from __future__ import print_function
class SliceExample(object):
def __getitem__(self, key):
if isinstance(key, slice):
return key.start, key.stop
return key
sl = SliceExample()
print(repr(sl[1]))
print(repr(sl[1:2]))