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。
请您参考如下方法:
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 ...
