python - 将 2 个字典与公共(public)键组合

Question

我有两个字典，需要合并其中相似键的值。这是一个例子:

dict1 = {'key1':[value11,value12,value13] , 'key2':[value21,value22,value23]}
dict2 = {'key1':[value14,value15] , 'key2':[value24,value25]}

我用过:

dict3 = {}
for key in (dict1.viewkeys() | dict2.keys()):
    if key in dict1: dict3.setdefault(key, []).append(dict1[key])
    if key in dict2: dict3.setdefault(key, []).append(dict2[key])

这给了我:

dict3 = {'key1':[[value11,value12,value13],[value14,value15]] , 'key2':[[value21,value22,value23],[value24,value25]]}

我想要的是一个简单的，比如:

期望的输出:

 dict3 = {'key1':[value11,value12,value13,value14,value15] , 'key2':[value21,value22,value23,value24,value25]}

Answer 1

您需要做的就是将 append 修改为 extend，然后它将添加列表的元素，而不是添加列表本身。查看list docs有关 append 和 extend 之间区别的更多详细信息。

dict1 = {'key1':['value11','value12','value13'] , 'key2':['value21','value22','value23']}
dict2 = {'key1':['value14','value15'] , 'key2':['value24','value25']}

dict3 = {}
for key in set().union(dict1, dict2):
    if key in dict1: dict3.setdefault(key, []).extend(dict1[key])
    if key in dict2: dict3.setdefault(key, []).extend(dict2[key])

print(dict3)
# {'key2': ['value21', 'value22', 'value23', 'value24', 'value25'], 'key1': ['value11', 'value12', 'value13', 'value14', 'value15']}

或者，您可以使用 collections.defaultdict默认设置为 list，如下所示。

from collections import defaultdict
dict3 = defaultdict(list)

for key in set().union(dict1, dict2):
  for dic in [dict1, dict2]:
    if key in dic:
      dict3[key] += dic[key]