Question

我有两个词典，需要将相似键的值组合在一起。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])

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：

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，然后添加列表的元素而不是添加列表本身。有关append和extend之间差异的详细信息，请参阅list docs。

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，您可以使用默认设置为list的{{3}}，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]

Answer 2

您可以更简单地做到这一点，但如果您想使用代码，只需将append更改为extend

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

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

print dict3

输出：

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

你可以阅读这个post关于ov追加延伸的差异

Answer 3

这是一个通用方法，您可以根据需要传递任意数量的dict。

>>> def mix_dict(*args):
       res = {}
       for d in args:
           if not isinstance(d, dict):
               continue
           for k, v in d.iteritems():
               res.setdefault(k, [])
               if isinstance(v, list):
                   res[k].extend(v)
               else:
                   res[k].append(v)
       return res
>>> dict1 = {'key1':['value11','value12','value13'] , 'key2':['value21','value22','value23']}
>>> dict2 = {'key1':['value14','value15'] , 'key2':['value24','value25']}
>>> dict3 = mix_dict(dict1, dict2)
>>> print dict3
... {'key1': ['value11', 'value12', 'value13', 'value14', 'value15'],
     'key2': ['value21', 'value22', 'value23', 'value24', 'value25']}

Answer 4

这是另一种方法。

您可以支持将列表的N个词组合并到具有此功能的列表的单个词典中：

def mergeDoLs(*dicts):
    def flatten(LoL):
        return [e for l in LoL for e in l]

    rtr={k:[] for k in set(flatten(d.keys() for d in dicts))}
    for k, v in flatten(d.items() for d in dicts):
        rtr[k].extend(v)
    return rtr

使用：

>>> dict1 = {'key1':['value11','value12','value13'] , 'key2':['value21','value22','value23'], 'key3':[1]}
>>> dict2 = {'key1':['value14','value15'] , 'key2':['value24','value25']}
>>> dict3 = {'key3':[2]}
>>> mergeDoLs(dict1, dict2, dict3) 
{'key3': [1, 2], 'key2': ['value21', 'value22', 'value23', 'value24', 'value25'], 'key1': ['value11', 'value12', 'value13', 'value14', 'value15']}

Answer 5

使用dict.update（）合并两个字典键：property

将2个词典与共同键组合在一起

5 个答案: