Question

我遇到了一个帖子，它有一个完整而正确的方法来合并两个字典，当每个字典为每个键都有一个值列表时。

程序的输入是这样的：

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']}

合并功能代码是：

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这类词典。这是我从“＃dic; dicts＆＃34;每次2项，并尝试合并它们。

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

那么有更多的pythonic方式来实现吗？

Answer 1

您有一个拼写错误：您使用dict表示dicts。

Answer 2

没有必要迭代成对的字典，如set（）。union（）操作可以带多个参数：

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

当其键具有值列表Python

2 个答案: