Question

我有一本这种格式的字典：

{
    'a': ([1,2,3],[-1,-2,-3]),
    'b': ([1,2,3],[-1,-2,-3]),
    'z': ([],[-1])

}

还有一个：

{
    'a': ([4,5],[]),
    'c': ([1,2,3],[-4]),
    'z': ([],[-3])
}

我想将它们都合并为这种格式：

{
    'a': ([1,2,3,4,5],[-1,-2,-3]),
    'b': ([1,2,3],[-1,-2,-3]),
    'c': ([1,2,3],[-4]),
    'z': ([],[-1,-3])
}

我将如何去做？我对Python也很陌生，我想知道这是否是表示数据的最佳方法，是否应该使用某种数据结构代替字典。

Answer 1

尝试一下：

d1 = {'a': ([1,2,3],[-1,-2,-3]),'b': ([1,2,3],[-1,-2,-3]),'z': ([],[-1])}
d2 = {'a': ([4,5],[]),'c': ([1,2,3],[-4]),'z': ([],[-3])}
d3 = {}
d4 = {**d1, **d2}  # This will work for Python 3.5+
for k in d4:
    if k in d2 and k in d1:
        tm = (d1[k][0]+ d2[k][0], d1[k][1]+d2[k][1])
        d3[k] = tm
    elif k in d2 and k not in d1:
        d3[k] = d2[k]
    else:
        d3[k] = d1[k]

输出：

d3 = {'a': ([1, 2, 3, 4, 5], [-1, -2, -3]), 'b': ([1, 2, 3], [-1, -2, -3]), 'z': ([], [-1, -3]), 'c': ([1, 2, 3], [-4])}

Answer 2

您可以使用处理合并的帮助器功能将此问题分为两个步骤。

dict1 = {
    'a': ([1,2,3],[-1,-2,-3]),
    'b': ([1,2,3],[-1,-2,-3]),
    'z': ([],[-1])

}

dict2 = {
    'a': ([4,5],[]),
    'c': ([1,2,3],[-4]),
    'z': ([],[-3])
}

def ordered_list_merge(lst1, lst2):
     ''' 
     Merges two lists, and does not add duplicates from lst2 into lst1
     '''
    resulting_list = lst1.copy() #to ensure list 1 is not mutated by changes to resulting_list
    resulting_list.extend(x for x in lst2 if x not in resulting_list)
    return resulting_list

import copy
result_dict = copy.deepcopy(dict1) #to ensure result_dict is an entirely new object, and mutations on result_dict do not affect dict1

for k, v in dict2.items():
    if k not in result_dict:
        result_dict[k] = v
    else:
        result_dict[k] = tuple(ordered_list_merge(lst1, lst2)
                            for lst1, lst2 in 
                            zip(result_dict[k], v))
print(result_dict) 
#Output:
{'a': ([1, 2, 3, 4, 5], [-1, -2, -3]),
 'b': ([1, 2, 3], [-1, -2, -3]),
 'z': ([], [-1, -3]),
 'c': ([1, 2, 3], [-4])}

请注意，字典本质上是无序的（或者记住python 3.7+中的插入顺序），不应依赖于顺序。使用排序来获取元组列表，如果顺序也很重要，则使用OrderedDict。

如何合并具有两个列表作为值的字典？

2 个答案: