相同的元组给出不同的泡菜

2024-02-27

这种情况很奇怪，我希望有一些关于 pickle 模块的东西我不知道。

我有两个元组s1 and s2。如果我比较它们它会返回True.

s1 == s2
True

如果我腌制它们并比较结果，它会返回False

pickle.dumps(s1) == pickle.dumps(s2)
False

让我们看看这些字节字符串。他们确实不同：

pk1 = pickle.dumps(s1)
pk2 = pickle.dumps(s2)

print(pk1)
b'\x80\x03((X!\x00\x00\x00AA1A2A5D4D4D4D4H2H2H3N0N0P4P4a2a3q\x00X!\x00\x00\x00AA1A3A5D3D4D4D4H1H2H3N0N1P3P4a2a3q\x01K\x01\x86q\x02X!\x00\x00\x00AA2A3A4D2D3D3D3H0H2H2N2N2P3P3a0a2q\x03K\x02\x86q\x04X!\x00\x00\x00AA4A5A5D1D2D2D3H2H4H4N5N5P2P2a2a4q\x05K\x05\x86q\x06X!\x00\x00\x00DA1A3A4A4D2D2D2H1H2H3N4N4P2P2a1a3q\x07K\x04\x86q\x08X!\x00\x00\x00DA2A3A3A4D1D1D2H1H2H3N3N4P0P0a1a3q\tK\x04\x86q\nX!\x00\x00\x00DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3q\x0bK\x04\x86q\x0cX!\x00\x00\x00DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3q\rK\x04\x86q\x0eX!\x00\x00\x00HA0A2A2A4D3D3D3D3H2H2N2N2P3P3a0a2q\x0fK\x02\x86q\x10X!\x00\x00\x00HA1A2A2A4D2D2D3D3H1H2N1N2P2P2a1a2q\x11K\x02\x86q\x12X!\x00\x00\x00HA2A2A3A3D1D1D2D2H1H2N3N3P1P2a0a2q\x13K\x03\x86q\x14X!\x00\x00\x00NA0A0A2A5D3D4D4D4H1H2H3N0P3P4a2a3q\x15K\x00\x86q\x16X!\x00\x00\x00NA0A1A2A5D4D4D4D4H2H2H3N0P4P4a2a3q\x17K\x00\x86q\x18X!\x00\x00\x00PA2A3A3A4D0D1D1D2H1H2H3N3N4P0a1a3q\x19K\x04\x86q\x1aX!\x00\x00\x00PA2A3A4A4D0D0D0D2H2H2H3N4N4P0a2a3q\x1bK\x04\x86q\x1cX!\x00\x00\x00aA0A2A2A4D3D3D3D3H0H2H2N2N2P3P3a2q\x1dK\x02\x86q\x1eX!\x00\x00\x00aA2A2A3A3D1D1D2D2H0H1H2N3N3P1P2a2q\x1fK\x03\x86q 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\x01\x00\x00(h\x1fh\x00K\x03\x86r!\x01\x00\x00h\x01K\x03\x86r"\x01\x00\x00h\x03K\x02\x86r#\x01\x00\x00h\x05K\x02\x86r$\x01\x00\x00h\x07K\x01\x86r%\x01\x00\x00h\tK\x01\x86r&\x01\x00\x00h\x0bK\x02\x86r\'\x01\x00\x00h\rK\x02\x86r(\x01\x00\x00h\x0fK\x02\x86r)\x01\x00\x00h\x11K\x01\x86r*\x01\x00\x00h\x13K\x00\x86r+\x01\x00\x00h\x15K\x03\x86r,\x01\x00\x00h\x17K\x03\x86r-\x01\x00\x00h\x19K\x01\x86r.\x01\x00\x00h\x1bK\x02\x86r/\x01\x00\x00h\x1dK\x02\x86r0\x01\x00\x00tr1\x01\x00\x00tr2\x01\x00\x00.'

print(pk2)
b'\x80\x03((X!\x00\x00\x00AA1A2A5D4D4D4D4H2H2H3N0N0P4P4a2a3q\x00X!\x00\x00\x00AA1A3A5D3D4D4D4H1H2H3N0N1P3P4a2a3q\x01K\x01\x86q\x02X!\x00\x00\x00AA2A3A4D2D3D3D3H0H2H2N2N2P3P3a0a2q\x03K\x02\x86q\x04X!\x00\x00\x00AA4A5A5D1D2D2D3H2H4H4N5N5P2P2a2a4q\x05K\x05\x86q\x06X!\x00\x00\x00DA1A3A4A4D2D2D2H1H2H3N4N4P2P2a1a3q\x07K\x04\x86q\x08X!\x00\x00\x00DA2A3A3A4D1D1D2H1H2H3N3N4P0P0a1a3q\tK\x04\x86q\nX!\x00\x00\x00DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3q\x0bK\x04\x86q\x0cX!\x00\x00\x00DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3q\rK\x04\x86q\x0eX!\x00\x00\x00HA0A2A2A4D3D3D3D3H2H2N2N2P3P3a0a2q\x0fK\x02\x86q\x10X!\x00\x00\x00HA1A2A2A4D2D2D3D3H1H2N1N2P2P2a1a2q\x11K\x02\x86q\x12X!\x00\x00\x00HA2A2A3A3D1D1D2D2H1H2N3N3P1P2a0a2q\x13K\x03\x86q\x14X!\x00\x00\x00NA0A0A2A5D3D4D4D4H1H2H3N0P3P4a2a3q\x15K\x00\x86q\x16X!\x00\x00\x00NA0A1A2A5D4D4D4D4H2H2H3N0P4P4a2a3q\x17K\x00\x86q\x18X!\x00\x00\x00PA2A3A3A4D0D1D1D2H1H2H3N3N4P0a1a3q\x19K\x04\x86q\x1aX!\x00\x00\x00PA2A3A4A4D0D0D0D2H2H2H3N4N4P0a2a3q\x1bK\x04\x86q\x1cX!\x00\x00\x00aA0A2A2A4D3D3D3D3H0H2H2N2N2P3P3a2q\x1dK\x02\x86q\x1eX!\x00\x00\x00aA2A2A3A3D1D1D2D2H0H1H2N3N3P1P2a2q\x1fK\x03\x86q 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\x01\x00\x00(h\x1fh\x00K\x03\x86r!\x01\x00\x00h\x01K\x03\x86r"\x01\x00\x00h\x03K\x02\x86r#\x01\x00\x00h\x05K\x02\x86r$\x01\x00\x00h\x07K\x01\x86r%\x01\x00\x00h\tK\x01\x86r&\x01\x00\x00h\x0bK\x02\x86r\'\x01\x00\x00h\rK\x02\x86r(\x01\x00\x00h\x0fK\x02\x86r)\x01\x00\x00h\x11K\x01\x86r*\x01\x00\x00h\x13K\x00\x86r+\x01\x00\x00h\x15K\x03\x86r,\x01\x00\x00h\x17K\x03\x86r-\x01\x00\x00h\x19K\x01\x86r.\x01\x00\x00h\x1bK\x02\x86r/\x01\x00\x00h\x1dK\x02\x86r0\x01\x00\x00tr1\x01\x00\x00tr2\x01\x00\x00.'

for n, (i, j) in enumerate(zip(pk1, pk2)):
    if i != j:
        print(n, i, j)

924 11 13
931 13 11
1042 13 11
1049 11 13
1359 11 13
1403 13 11
1477 13 11
1521 11 13

如果我将它们解开并比较结果，它们是相同的：

upk1 = pickle.loads(pk1)
upk2 = pickle.loads(pk2)

print(upk1 == upk2)
True

那么问题是什么导致了这个问题以及如何避免它？是否有一些元数据腌制？

嗯，我发现了一个相关问题Python pickle 不是一对一的：不同的 pickle 给出相同的对象 https://stackoverflow.com/questions/21271479/python-pickle-not-one-to-one-different-pickles-give-same-object。那么pickle对于相同的对象可以返回不同的结果吗？那么它不适合进行哈希计算吗？因为我计划根据 pickled 表示来计算 md5 哈希值。

更新：文本表示s1 and s2:

(('AA1A2A5D4D4D4D4H2H2H3N0N0P4P4a2a3',
  ('AA1A3A5D3D4D4D4H1H2H3N0N1P3P4a2a3', 1),
  ('AA2A3A4D2D3D3D3H0H2H2N2N2P3P3a0a2', 2),
  ('AA4A5A5D1D2D2D3H2H4H4N5N5P2P2a2a4', 5),
  ('DA1A3A4A4D2D2D2H1H2H3N4N4P2P2a1a3', 4),
  ('DA2A3A3A4D1D1D2H1H2H3N3N4P0P0a1a3', 4),
  ('DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3', 4),
  ('DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3', 4),
  ('HA0A2A2A4D3D3D3D3H2H2N2N2P3P3a0a2', 2),
  ('HA1A2A2A4D2D2D3D3H1H2N1N2P2P2a1a2', 2),
  ('HA2A2A3A3D1D1D2D2H1H2N3N3P1P2a0a2', 3),
  ('NA0A0A2A5D3D4D4D4H1H2H3N0P3P4a2a3', 0),
  ('NA0A1A2A5D4D4D4D4H2H2H3N0P4P4a2a3', 0),
  ('PA2A3A3A4D0D1D1D2H1H2H3N3N4P0a1a3', 4),
  ('PA2A3A4A4D0D0D0D2H2H2H3N4N4P0a2a3', 4),
  ('aA0A2A2A4D3D3D3D3H0H2H2N2N2P3P3a2', 2),
  ('aA2A2A3A3D1D1D2D2H0H1H2N3N3P1P2a2', 3)),
 ('AA1A3A5D3D4D4D4H1H2H3N0N1P3P4a2a3',
  ('AA1A2A5D4D4D4D4H2H2H3N0N0P4P4a2a3', 1),
  ('AA2A3A4D2D3D3D3H0H2H2N2N2P3P3a0a2', 3),
  ('AA4A5A5D1D2D2D3H2H4H4N5N5P2P2a2a4', 5),
  ('DA1A3A4A4D2D2D2H1H2H3N4N4P2P2a1a3', 4),
  ('DA2A3A3A4D1D1D2H1H2H3N3N4P0P0a1a3', 3),
  ('DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3', 4),
  ('DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3', 4),
  ('HA0A2A2A4D3D3D3D3H2H2N2N2P3P3a0a2', 2),
  ('HA1A2A2A4D2D2D3D3H1H2N1N2P2P2a1a2', 1),
  ('HA2A2A3A3D1D1D2D2H1H2N3N3P1P2a0a2', 3),
  ('NA0A0A2A5D3D4D4D4H1H2H3N0P3P4a2a3', 0),
  ('NA0A1A2A5D4D4D4D4H2H2H3N0P4P4a2a3', 1),
  ('PA2A3A3A4D0D1D1D2H1H2H3N3N4P0a1a3', 3),
  ('PA2A3A4A4D0D0D0D2H2H2H3N4N4P0a2a3', 4),
  ('aA0A2A2A4D3D3D3D3H0H2H2N2N2P3P3a2', 2),
  ('aA2A2A3A3D1D1D2D2H0H1H2N3N3P1P2a2', 3)),
 ('AA2A3A4D2D3D3D3H0H2H2N2N2P3P3a0a2',
  ('AA1A2A5D4D4D4D4H2H2H3N0N0P4P4a2a3', 2),
  ('AA1A3A5D3D4D4D4H1H2H3N0N1P3P4a2a3', 3),
  ('AA4A5A5D1D2D2D3H2H4H4N5N5P2P2a2a4', 4),
  ('DA1A3A4A4D2D2D2H1H2H3N4N4P2P2a1a3', 3),
  ('DA2A3A3A4D1D1D2H1H2H3N3N4P0P0a1a3', 3),
  ('DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3', 2),
  ('DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3', 3),
  ('HA0A2A2A4D3D3D3D3H2H2N2N2P3P3a0a2', 0),
  ('HA1A2A2A4D2D2D3D3H1H2N1N2P2P2a1a2', 2),
  ('HA2A2A3A3D1D1D2D2H1H2N3N3P1P2a0a2', 2),
  ('NA0A0A2A5D3D4D4D4H1H2H3N0P3P4a2a3', 2),
  ('NA0A1A2A5D4D4D4D4H2H2H3N0P4P4a2a3', 2),
  ('PA2A3A3A4D0D1D1D2H1H2H3N3N4P0a1a3', 3),
  ('PA2A3A4A4D0D0D0D2H2H2H3N4N4P0a2a3', 3),
  ('aA0A2A2A4D3D3D3D3H0H2H2N2N2P3P3a2', 0),
  ('aA2A2A3A3D1D1D2D2H0H1H2N3N3P1P2a2', 2)),
 ('AA4A5A5D1D2D2D3H2H4H4N5N5P2P2a2a4',
  ('AA1A2A5D4D4D4D4H2H2H3N0N0P4P4a2a3', 5),
  ('AA1A3A5D3D4D4D4H1H2H3N0N1P3P4a2a3', 5),
  ('AA2A3A4D2D3D3D3H0H2H2N2N2P3P3a0a2', 4),
  ('DA1A3A4A4D2D2D2H1H2H3N4N4P2P2a1a3', 1),
  ('DA2A3A3A4D1D1D2H1H2H3N3N4P0P0a1a3', 2),
  ('DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3', 2),
  ('DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3', 3),
  ('HA0A2A2A4D3D3D3D3H2H2N2N2P3P3a0a2', 4),
  ('HA1A2A2A4D2D2D3D3H1H2N1N2P2P2a1a2', 4),
  ('HA2A2A3A3D1D1D2D2H1H2N3N3P1P2a0a2', 2),
  ('NA0A0A2A5D3D4D4D4H1H2H3N0P3P4a2a3', 5),
  ('NA0A1A2A5D4D4D4D4H2H2H3N0P4P4a2a3', 5),
  ('PA2A3A3A4D0D1D1D2H1H2H3N3N4P0a1a3', 2),
  ('PA2A3A4A4D0D0D0D2H2H2H3N4N4P0a2a3', 2),
  ('aA0A2A2A4D3D3D3D3H0H2H2N2N2P3P3a2', 4),
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  ('AA1A2A5D4D4D4D4H2H2H3N0N0P4P4a2a3', 3),
  ('AA1A3A5D3D4D4D4H1H2H3N0N1P3P4a2a3', 3),
  ('AA2A3A4D2D3D3D3H0H2H2N2N2P3P3a0a2', 2),
  ('AA4A5A5D1D2D2D3H2H4H4N5N5P2P2a2a4', 2),
  ('DA1A3A4A4D2D2D2H1H2H3N4N4P2P2a1a3', 1),
  ('DA2A3A3A4D1D1D2H1H2H3N3N4P0P0a1a3', 1),
  ('DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3', 2),
  ('DA2A3A4A4D1D1D2H2H3H3N4N4P0P1a2a3', 2),
  ('HA0A2A2A4D3D3D3D3H2H2N2N2P3P3a0a2', 2),
  ('HA1A2A2A4D2D2D3D3H1H2N1N2P2P2a1a2', 1),
  ('HA2A2A3A3D1D1D2D2H1H2N3N3P1P2a0a2', 0),
  ('NA0A0A2A5D3D4D4D4H1H2H3N0P3P4a2a3', 3),
  ('NA0A1A2A5D4D4D4D4H2H2H3N0P4P4a2a3', 3),
  ('PA2A3A3A4D0D1D1D2H1H2H3N3N4P0a1a3', 1),
  ('PA2A3A4A4D0D0D0D2H2H2H3N4N4P0a2a3', 2),
  ('aA0A2A2A4D3D3D3D3H0H2H2N2N2P3P3a2', 2)))

问题是pickle 保留子对象之间的身份关系 https://docs.python.org/3/library/pickle.html#comparison-with-marshal（这是为了效率和处理递归对象）：

a1 = "a" * 30
a2 = "a" * 30
s1 = (a1, a2, a1)
s2 = (a1, a2, a2)
print(pickle.dumps(s1) == pickle.dumps(s2))
False

Here, a1 and a2是平等的，但是是不同的对象（它们有不同的id; see python什么时候选择intertern一个字符串 https://stackoverflow.com/questions/10622472/when-does-python-choose-to-intern-a-string?lq=1); s1 and s2再次相等，但在s1第三个元素是同一个物体 as the first元素，同时在s2第三个元素与第一个元素是同一个对象second元素。

我们可以通过拆解泡菜来看到这一点：

>>> pickletools.dis(pickle.dumps(s1))
    0: \x80 PROTO      3
    2: X    BINUNICODE 'aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa'
   37: q    BINPUT     0
   39: X    BINUNICODE 'aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa'
   74: q    BINPUT     1
   76: h    BINGET     0  # <=== watch this line *********************
   78: \x87 TUPLE3
   79: q    BINPUT     2
   81: .    STOP
highest protocol among opcodes = 2

>>> pickletools.dis(pickle.dumps(s2))
    0: \x80 PROTO      3
    2: X    BINUNICODE 'aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa'
   37: q    BINPUT     0
   39: X    BINUNICODE 'aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa'
   74: q    BINPUT     1
   76: h    BINGET     1  # <=== this is different *********************
   78: \x87 TUPLE3
   79: q    BINPUT     2
   81: .    STOP
highest protocol among opcodes = 2

一个快速（如果效率低下）的解决方法可能是通过使用来破坏对象标识ast.literal_eval从它们重新创建元组repr:

>>> print(pickle.dumps(ast.literal_eval(repr(s1))) == pickle.dumps(ast.literal_eval(repr(s2))))
True

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