Question

我正在使用Python 2.7并且已经查看了几个解决方案，如果您知道要合并的字典数量有多少，但我可以在2到5之间找到任何字典。

我有一个循环，它生成一个具有相同键但不同值的dict。我想将新值添加到上一个。

如：

for num in numbers:
    dict = (function which outputs a dictionary)
    [merge with dictionary from previous run of the loop]

所以如果：

dict (from loop one) = {'key1': 1,
 'key2': 2,
 'key3': 3}

和

dict (from loop two) = {'key1': 4,
 'key2': 5,
 'key3': 6}

结果是dict：

dict = {'key1': [1,4]
 'key2': [2,5],
 'key3': [3,6]}

Answer 1

使用defaultdict：

In [18]: def gen_dictionaries():
    ...:     yield {'key1': 1, 'key2': 2, 'key3': 3}
    ...:     yield {'key1': 4, 'key2': 5, 'key3': 6}
    ...:

In [19]: from collections import defaultdict

In [20]: final = defaultdict(list)

In [21]: for d in gen_dictionaries():
    ...:     for k, v in d.iteritems():
    ...:         final[k].append(v)
    ...:

In [22]: final
Out[22]: defaultdict(list, {'key1': [1, 4], 'key2': [2, 5], 'key3': [3, 6]})

Answer 2

以通用方式实现此目的的一种方法是使用set找到dict的两个键的并集，然后使用字典理解来获取期望的词典：

>>> dict_list = [d1, d2]  # list of all the dictionaries which are to be joined

 # set of the keys present in all the dicts; 
>>> common_keys = set(dict_list[0]).union(*dict_list[1:])

>>> {k: [d[k] for d in dict_list if k in d] for k in common_keys}
{'key3': [3, 6], 'key2': [2, 5], 'key1': [1, 4]}

其中d1和d2是：

d1 = {'key1': 1,
      'key2': 2,
      'key3': 3}

d2 = {'key1': 4,
      'key2': 5,
      'key3': 6}

解释：此处，dict_list是您要合并的所有dict个对象的列表。然后我在所有dict对象中创建common_keys 键集。最后，我正在使用字典理解创建一个新字典（使用嵌套列表理解和过滤器）。

根据OP的评论，由于所有的词组都拥有相同的键，我们可以跳过set的用法。因此代码可以写成：

>>> dict_list = [d1, d2]  

>>> {k: [d[k] for d in dict_list] for k in dict_list[0]}
{'key3': [3, 6], 'key2': [2, 5], 'key1': [1, 4]}

Answer 3

//if foo is a function returning void but is declared in global space and not part of another class
TakesFun(bind(foo));   
//if foo is a function returning void but is declared in class called ClassX and the function is required to be called for object "obj".
TakesFun(bind(ClassX::foo, obj));   
//if foo is a function taking an integer as argument and returning void but is declared in class called ClassX and the function is required to be called for object "obj".
TakesFun(bind(ClassX::foo, obj, 5)); //5 is the argument supplied to function foo

输出如下：

dict1 = {'m': 2, 'n':4, 'o':7, 'p':90} dict2 = {'m': 1, 'n': 3} dict3 = {} for key,value in dict1.iteritems(): if key not in dict3: dict3[key] = list() dict3[key].append(value) for key,value in dict2.iteritems(): if key not in dict3: dict3[key] = list() dict3[key].append(value) print(dict3)

加入多个dicts以创建具有值的新列表作为原始字典值的列表

3 个答案: