[英]Merge multiple dictionaries when their keys have a list of values Python
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. 我碰到过一篇帖子,其中有完整而正确的方法来合并两个字典,而每个字典的每个键都有一个值列表。
The input of the program is something like this: 该程序的输入如下所示:
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']}
code of merge function is: 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
With 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']}
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. 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,每次尝试从“字典”列表中获取2个项目,然后尝试将其合并。
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
So is there a more pythonic way to achieve that? 那么,有没有更Python的方法来实现这一目标?
您有一个错字:您用dict
表示dicts
。
There is no need to iterate over pairs of dictionaries, as set().union() operation can take multiple arguments: 无需迭代对字典,因为set()。union()操作可以采用多个参数:
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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