当它们的键具有值列表时合并多个字典 [英] Merge multiple dictionaries when their keys have a list of values Python

问题描述

当每本字典的每个键都有一个值列表时，我遇到了一个帖子，该帖子具有完整且正确的方式合并两个字典的方式.

I came across a post that had a complete and correct way of merging two dictionaries, when each dictionary has for each key a list of values.

程序的输入如下:

d1:  {'apple': ['5', '65'], 'blue': ['9', '10', '15', '43'], 'candle': ['15'], 'is': ['5', '6', '13', '45', '96'], 'yes': ['1', '2', '3', '11'], 'zone': ['5', '6', '9', '10', '12', '14', '18', '19', '23', '24', '25', '29', '45']}
d2:  {'apple': ['43', '47'], 'appricote': ['1', '2', '3', '4', '5', '6'], 'candle': ['1', '2', '4', '5', '6', '9'], 'delta': ['14', '43', '47'], 'dragon': ['23', '24', '25', '26'], 'eclipse': ['11', '13', '15', '19'], 'island': ['1', '34', '35']}

合并功能的代码为:

def merge_dictionaries(dict1, dict2):

result = {}
new_result = {}
for key in set().union(*(dict1, dict2)):
    result[key] = sorted(dict1.get(key, []) + dict2.get(key, []), key=int)
    # Delete any double value for each key
    new_result[key] = [ii for n, ii in enumerate(result[key]) if ii not in result[key][:n]]

return new_result

结果:

Merged:  {'is': ['5', '6', '13', '45', '96'], 'dragon': ['23', '24', '25', '26'], 'apple': ['5', '43', '47', '65'], 'appricote': ['1', '2', '3', '4', '5', '6'], 'delta': ['14', '43', '47'], 'eclipse': ['11', '13', '15', '19'], 'yes': ['1', '2', '3', '11'], 'zone': ['5', '6', '9', '10', '12', '14', '18', '19', '23', '24', '25', '29', '45'], 'blue': ['9', '10', '15', '43'], 'island': ['1', '34', '35'], 'candle': ['1', '2', '4', '5', '6', '9', '15']}

我现在想要实现的是扩展这种方式来合并动态数量的此类字典.这是我的尝试，我每次都会从"dicts"列表中获得2个项目，然后尝试将它们合并.

What I want to achieve now is to extend this way for merging a dynamic number of dictionaries of this kind.This is my try in which I get from the list of "dicts" 2 items each time and try to merge them.

def merge_multiple_dictionaries(dicts):
''' This is just to print the input I get
l = len(dicts)
print("Len of dicts: ", l)
for j in range(0, l):
    print("j is: ", dicts[j])
'''
result = {}
new_result = {}
for k in range(0, l):
     if k is (len(dicts)-1):
        break
    print("k: ", k)
    for key in set().union(*(dicts[k], dicts[k+1])):
        result[key] = sorted(dicts[k].get(key, []) + dicts[k+1].get(key, []), key=int)
        # Delete any double value for each key
        new_result[key] = [ii for n, ii in enumerate(result[key]) if ii not in result[key][:n]]
# result = OrderedDict(sorted([(k, v) for k, v in result.items()]))

return new_result

那么还有一种更Python化的方法来实现这一目标吗?

So is there a more pythonic way to achieve that?

推荐答案

由于set().union()操作可以使用多个参数，因此无需遍历字典对.

There is no need to iterate over pairs of dictionaries, as set().union() operation can take multiple arguments:

def merge_dictionaries(*args):
    result = {}
    for key in set().union(*args):
        # let's convert the value union to set before sorting to get rid of the duplicates
        result[key] = sorted(set.union(*[set(d.get(key, [])) for d in args]), key=int)
    return result

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