Python:为什么每个键添加字典值取决于顺序？

Question

假设您有多个字典，每个键(在子字典中)跟踪三个浮点值。您希望能够以添加存在于多个字典中的键值的方式合并这些字典。

通过普通的字典更新，值会被覆盖，所以你将 dict() 子类化:

class StatementDict(dict):
    def add(self, statement):
        ann_id = statement[0]
        lvl_dict = statement[1]
        if ann_id in self:
            self[ann_id]['skill'] += lvl_dict['skill']
            self[ann_id]['knowledge'] += lvl_dict['knowledge']
            self[ann_id]['interest'] += lvl_dict['interest']
        else:
            self[ann_id] = lvl_dict

    def update(self, statement_dict):
        for statement in statement_dict.iteritems():
            self.add(statement)

然后将要合并/添加的字典放入普通字典的键中:

# Small example data that reproduces the error
few_statements = {}
few_statements['linkedin'] = {u'Homerun': {u'skill': 14.0,
                                           u'knowledge': 34.0,
                                           u'interest': 20.0}}
few_statements['tudelft'] = {u'Presentation': {u'skill': 14.0,
                                               u'knowledge': 34.0,
                                               u'interest': 20.0},
                             u'Future': {u'skill': 16.0,
                                         u'knowledge': 25.33,
                                         u'interest': 2.0},
                             u'Visual_perception': {u'skill': 20.46,
                                                    u'knowledge': 28.35,
                                                    u'interest': 4.0}}
few_statements['website'] = {u'Homerun': {u'skill': 1.0,
                                          u'knowledge': 3.0,
                                          u'interest': 2.0}}

few_statements['shareworks'] = {u'Presentation': {u'skill': 8.0,
                                                  u'knowledge': 20.0,
                                                  u'interest': 12.0},
                                u'Future': {u'skill': 17.0,
                                            u'knowledge': 26.33,
                                            u'interest': 3.0},
                                u'Visual_perception': {u'skill': 2.0,
                                                       u'knowledge': 3.0,
                                                       u'interest': 6.0}}

现在我们应该能够将这些键值对逐个添加到 StatementDict() 中，或者使用 StatementDict.update() 方法。源字典添加到 StatementDict 的顺序对结果没有影响。

# First we try updating in one order
small_test1a = StatementDict()
for origin in ("tudelft", "website", "linkedin", "shareworks"):
    for st in few_statements[origin].iteritems():
        small_test1a.add(st)

# And then in another order
small_test2 = StatementDict()
for origin in ("linkedin", "shareworks", "tudelft", "website"):
    for st in few_statements[origin].iteritems():
        small_test2.add(st)

print "Different order, same result?", small_test1a == small_test2
                                                # False, but why?
for key in small_test1a:
    print "Desired:", key, small_test1a[key]
    print "Unexpected:", key, small_test2[key]

唉，dict 添加的顺序确实会影响结果。但是为什么会发生意外结果呢？

Desired: Future {u'skill': 33.0, u'knowledge': 51.66, u'interest': 5.0}
Unexpected: Future {u'skill': 50.0, u'knowledge': 77.99, u'interest': 8.0}
Desired: Presentation {u'skill': 22.0, u'knowledge': 54.0, u'interest': 32.0}
Unexpected: Presentation {u'skill': 30.0, u'knowledge': 74.0, u'interest': 44.0}
Desired: Homerun {u'skill': 15.0, u'knowledge': 37.0, u'interest': 22.0}
Unexpected: Homerun {u'skill': 29.0, u'knowledge': 71.0, u'interest': 42.0}
Desired: Visual_perception {u'skill': 22.46, u'knowledge': 31.35, u'interest': 10.0}
Unexpected: Visual_perception {u'skill': 24.46, u'knowledge': 34.35, u'interest': 16.0}

以第二顺序添加字典似乎会使首先放置的字典的值加倍(将它们添加两次？)。我不明白为什么会这样。如何独立于添加顺序可靠地发生所需的添加行为？

还有一件事我不明白:当我创建一个新的 StatementDict() 并用相同的值填充它时，为什么 small_test1a 的值会改变？

运行以下行会导致 small_test1a 在循环的最后一次迭代中发生变化:

small_test1b = StatementDict()
for origin in ("tudelft", "website", "linkedin", "shareworks"):
    small_test1b.update(few_statements[origin])
print "\nDoes .update() function?", small_test1a == small_test1b
print small_test1a

附言使用我的实际数据，根本没有添加。相反，将保留第一个放置的值。这与更新普通字典不同，后者的值被覆盖。不幸的是，我无法使用少量测试数据重现此行为。

Answer 1

当你这样做时:

self[ann_id] = lvl_dict

您将 self[ann_id] 设为该特定词典的另一个名称(例如，“tudelft”的名称)。然后，当您执行后续操作时:

self[ann_id]['skill'] += lvl_dict['skill']

您根据当前版本修改先前的 lvl_dict(即，在本例中，根据“网站”更改“tudelft”的版本)。

对此的最小修复是 copy第一本词典。但是，我可能会尝试使用 collections.defaultdict，这样您就可以完全消除 if ann_id in self: 测试。当 defaultdict 创建新字典时，它将是一个新实例，因此不会修改任何现有字典。

---

在下面的评论中使用 defaultdict 和 lambda 函数的示例:

from collections import defaultdict

class StatementDict(defaultdict):
    def __init__(self):
        defaultdict.__init__(self,
            lambda: {'skill': 0.0, 'knowledge': 0.0, 'interest': 0.0})

    def add(self, statement):
        ... as before ...