将此Python结构的值合并到单个字典中的更快方法是什么?

all_classes

我有一本字典,像这样:

groups_and_classes = {'group_1': {'class_A': [1, 2, 3],
                                  'class_B': [1, 3, 5, 7], 
                                  'class_c': [1, 2], # ...many more items like this
                                 },
                      'group_2': {'class_A': [11, 12, 13],
                                  'class_C': [5, 6, 7, 8, 9]
                                 }, # ...and many more items like this
                     }

函数从中创建新对象 groups_and_classes 像这样(创建这个的函数经常被调用):

all_classes = {'class_A': [1, 2, 3, 11, 12, 13],
               'class_B': [1, 3, 5, 7, 9],
               'class_C': [1, 2, 5, 6, 7, 8, 9]
              }

现在,有一个循环是这样做的:

all_classes = {}
for group in groups_and_classes.values():
    for c, vals in group.iteritems():
        for v in vals:
            if all_classes.has_key(c):
                if v not in all_classes[c]:
                    all_classes[c].append(v)
            else:
                all_classes[c] = [v]

到目前为止,我把代码改成使用 set list 因为名单的顺序无关紧要 值必须是唯一的:

all_classes = {}
for group in groups_and_classes.values():
    for c, vals in group.iteritems():
        try:
            all_classes[c].update(set(vals))
        except KeyError:
            all_classes[c] = set(vals)

所有课程

问题 :有没有更有效的方法 (除了同时建造外 是构建的,并且在调用此函数的任何地方都会发生更改)?