在Python中递归地合并字典

Question

我有两个函数，它们可以接收两个字典，并递归地将它们的值相加。

def recursive_dict_sum_helper(v1, d2, k):
    try: v2 = d2[k]
    except KeyError: return v1 #problem is here if key not found it just return value but what about rest of the keys which is in d2??

    if not v1: return v2
    # "add" two values: if they can be added with '+', then do so,
    # otherwise expect dictionaries and treat them appropriately.
    try:
        if type(v1) == list and type(v2) == list:
            v1.extend(v2)
            return list(set(v1))
        else:
            return v1 + v2
    except: return recursive_dict_sum(v1, v2)

def recursive_dict_sum(d1, d2):
    if len(d1) < len(d2):
        temp = d1
        d1 = d2
        d2 = temp
    # Recursively produce the new key-value pair for each
    # original key-value pair, and make a dict with the results.
    return dict(
        (k, recursive_dict_sum_helper(v, d2, k))
        for (k, v) in d1.items()
    )

如果我输入以下内容，输出结果是我预期的：

a = {'abc': {'missing': 1, 'modified': 0, 'additional': 2}}
b = {'abc': {'missing': 1, 'modified': 1, 'additional': 2}}

mn = recursive_dict_sum(a, b)

output: mn = {'abc': {'missing': 2, 'modified': 1, 'additional': 4}}

但是如果输入是：

a = {'abc': {'missing': 1, 'modified': 0, 'additional': 2}}
b = {'cde': {'missing': 1, 'modified': 1, 'additional': 2}}

output: {'abc': {'missing': 1, 'modified': 0, 'additional': 2}} #which is wrong

如果在第二个字典中找不到某个键，它就会返回第一个字典中的键值。那么这就只处理了第一个字典的内容，第二个字典里的其他键怎么办呢？我该如何更新上面的脚本，以便输出结果是：

output: {'abc': {'missing': 1, 'modified': 0, 'additional': 2}, 'cde': {'missing': 1, 'modified': 1, 'additional': 2}}

Answer 1

做成了一个超级生成器，听起来你喜欢这个样子 :)

def recursive_dict_sum(d1, d2):
    return dict((k, ((d1[k] if k in d1 else d2[k])
                       if k not in d1 or k not in d2
                      else (d1[k] + d2[k] if not isinstance(d1[k], dict)
                                        else recursive_dict_sum(d1[k], d2[k]))))
                for k in set(d1.keys() + d2.keys()))

Answer 2

如果我理解你想做的事情没错，下面的代码就能实现你的需求：

def dict_sum(d1, d2):
    if d1 is None: return d2
    if d2 is None: return d1
    if isinstance(d1, list) and isinstance(d2, list):
        return list(set(d1 + d2))
    try:
        return d1 + d2
    except TypeError:
        # assume d1 and d2 are dictionaries
        keys = set(d1.iterkeys()) | set(d2.iterkeys())
        return dict((key, dict_sum(d1.get(key), d2.get(key))) for key in keys)

dict_sum(a, b) 会给出你想要的结果。

不过要注意，如果用不兼容的类型来调用它，比如：

dict_sum({'a': 1}, 2)

补充说明：要特别处理列表（创建一个只包含唯一元素的列表）。