我有以下两个函数,它们使用两个字典并递归地添加它们的值。
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()
)
如果我提供以下输入,那么输出就可以了,我期待:
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
如果第二个字典中没有找到密钥,则返回第一个字典中的密钥值。所以它在一个字典项上运行,第二个字典中的其余键怎么办?如何更新上面的脚本,以便输出:
output: {'abc': {'missing': 1, 'modified': 0, 'additional': 2}, 'cde': {'missing': 1, 'modified': 1, 'additional': 2}}
答案 0 :(得分:3)
如果我理解你想做什么,所有这些都可以通过以下代码实现:
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)
会给出所需的结果。
请注意,如果使用不兼容的类型(例如
)调用它会引发AttributeError
dict_sum({'a': 1}, 2)
编辑专门处理列表(使用唯一元素创建列表)。
答案 1 :(得分:2)
制作一个怪物发电机,你似乎喜欢它:)
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()))