我需要合并多个字典，这是我有例如:
dict1 = {1:{"a":{A}}, 2:{"b":{B}}}
dict2 = {2:{"c":{C}}, 3:{"d":{D}}}
A、B、C和D是树的叶子，比如{"info1":"value"， "info2":"value2"}
字典的级别(深度)未知，可能是{2:{"c":{"z":{"y":{c}}}}}
在我的例子中，它表示一个目录/文件结构，节点是文档，叶子是文件。
我想将它们合并得到:
dict3 = {1:{"a":{A}}, 2:{"b":{B},"c":{C}}, 3:{"d":{D}}}
我不确定如何用Python轻松做到这一点。
当前回答
换个答案怎么样?!?这也避免了突变/副作用:
def merge(dict1, dict2):
output = {}
# adds keys from `dict1` if they do not exist in `dict2` and vice-versa
intersection = {**dict2, **dict1}
for k_intersect, v_intersect in intersection.items():
if k_intersect not in dict1:
v_dict2 = dict2[k_intersect]
output[k_intersect] = v_dict2
elif k_intersect not in dict2:
output[k_intersect] = v_intersect
elif isinstance(v_intersect, dict):
v_dict2 = dict2[k_intersect]
output[k_intersect] = merge(v_intersect, v_dict2)
else:
output[k_intersect] = v_intersect
return output
dict1 = {1:{"a":{"A"}}, 2:{"b":{"B"}}}
dict2 = {2:{"c":{"C"}}, 3:{"d":{"D"}}}
dict3 = {1:{"a":{"A"}}, 2:{"b":{"B"},"c":{"C"}}, 3:{"d":{"D"}}}
assert dict3 == merge(dict1, dict2)
其他回答
我一直在测试你的解决方案，并决定在我的项目中使用这个:
def mergedicts(dict1, dict2, conflict, no_conflict):
for k in set(dict1.keys()).union(dict2.keys()):
if k in dict1 and k in dict2:
yield (k, conflict(dict1[k], dict2[k]))
elif k in dict1:
yield (k, no_conflict(dict1[k]))
else:
yield (k, no_conflict(dict2[k]))
dict1 = {1:{"a":"A"}, 2:{"b":"B"}}
dict2 = {2:{"c":"C"}, 3:{"d":"D"}}
#this helper function allows for recursion and the use of reduce
def f2(x, y):
return dict(mergedicts(x, y, f2, lambda x: x))
print dict(mergedicts(dict1, dict2, f2, lambda x: x))
print dict(reduce(f2, [dict1, dict2]))
将函数作为参数传递是将jterrace解决方案扩展为所有其他递归解决方案的关键。
这应该有助于将所有项从dict2合并到dict1:
for item in dict2:
if item in dict1:
for leaf in dict2[item]:
dict1[item][leaf] = dict2[item][leaf]
else:
dict1[item] = dict2[item]
请测试一下，告诉我们这是否是你想要的。
编辑:
上述解决方案只合并了一个级别，但正确地解决了op给出的例子。如果合并多个级别，应该使用递归。
换个答案怎么样?!?这也避免了突变/副作用:
def merge(dict1, dict2):
output = {}
# adds keys from `dict1` if they do not exist in `dict2` and vice-versa
intersection = {**dict2, **dict1}
for k_intersect, v_intersect in intersection.items():
if k_intersect not in dict1:
v_dict2 = dict2[k_intersect]
output[k_intersect] = v_dict2
elif k_intersect not in dict2:
output[k_intersect] = v_intersect
elif isinstance(v_intersect, dict):
v_dict2 = dict2[k_intersect]
output[k_intersect] = merge(v_intersect, v_dict2)
else:
output[k_intersect] = v_intersect
return output
dict1 = {1:{"a":{"A"}}, 2:{"b":{"B"}}}
dict2 = {2:{"c":{"C"}}, 3:{"d":{"D"}}}
dict3 = {1:{"a":{"A"}}, 2:{"b":{"B"},"c":{"C"}}, 3:{"d":{"D"}}}
assert dict3 == merge(dict1, dict2)
你可以使用toolz包中的merge函数，例如:
>>> import toolz
>>> dict1 = {1: {'a': 'A'}, 2: {'b': 'B'}}
>>> dict2 = {2: {'c': 'C'}, 3: {'d': 'D'}}
>>> toolz.merge_with(toolz.merge, dict1, dict2)
{1: {'a': 'A'}, 2: {'c': 'C'}, 3: {'d': 'D'}}
基于@andrew cooke。这个版本处理字典的嵌套列表，还允许选项更新值
def merge(a, b, path=None, update=True):
"http://stackoverflow.com/questions/7204805/python-dictionaries-of-dictionaries-merge"
"merges b into a"
if path is None: path = []
for key in b:
if key in a:
if isinstance(a[key], dict) and isinstance(b[key], dict):
merge(a[key], b[key], path + [str(key)])
elif a[key] == b[key]:
pass # same leaf value
elif isinstance(a[key], list) and isinstance(b[key], list):
for idx, val in enumerate(b[key]):
a[key][idx] = merge(a[key][idx], b[key][idx], path + [str(key), str(idx)], update=update)
elif update:
a[key] = b[key]
else:
raise Exception('Conflict at %s' % '.'.join(path + [str(key)]))
else:
a[key] = b[key]
return a
