我需要合并多个字典，例如：
dict1 = {1:{"a":{A}}, 2:{"b":{B}}}
dict2 = {2:{"c":{C}}, 3:{"d":{D}}
与A B C并D成为树的叶子，就像{"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，它代表一个目录/文件结构，节点是文档，叶子是文件。
我想合并它们以获得：
dict3 = {1:{"a":{A}}, 2:{"b":{B},"c":{C}}, 3:{"d":{D}}}
我不确定如何使用 Python 轻松做到这一点。
这实际上非常棘手 - 特别是如果您在事情不一致时想要一个有用的错误消息，同时正确接受重复但一致的条目（这里没有其他答案可以做到......）
假设您没有大量条目，递归函数是最简单的：
def merge(a, b, path=None):
"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
else:
raise Exception('Conflict at %s' % '.'.join(path + [str(key)]))
else:
a[key] = b[key]
return a
# works
print(merge({1:{"a":"A"},2:{"b":"B"}}, {2:{"c":"C"},3:{"d":"D"}}))
# has conflict
merge({1:{"a":"A"},2:{"b":"B"}}, {1:{"a":"A"},2:{"b":"C"}})
请注意，这会发生变异a- 的内容b被添加到a（也被返回）。如果你想保留a你可以这样称呼它merge(dict(a), b)。
agf 指出（如下）你可能有两个以上的字典，在这种情况下你可以使用：
reduce(merge, [dict1, dict2, dict3...])
一切都将被添加到dict1.
注意：我编辑了我的初始答案以改变第一个参数；这使得“减少”更容易解释
PS：在python 3中，您还需要from functools import reduce
这是使用生成器的一种简单方法：
def mergedicts(dict1, dict2):
for k in set(dict1.keys()).union(dict2.keys()):
if k in dict1 and k in dict2:
if isinstance(dict1[k], dict) and isinstance(dict2[k], dict):
yield (k, dict(mergedicts(dict1[k], dict2[k])))
else:
# If one of the values is not a dict, you can't continue merging it.
# Value from second dict overrides one in first and we move on.
yield (k, dict2[k])
# Alternatively, replace this with exception raiser to alert you of value conflicts
elif k in dict1:
yield (k, dict1[k])
else:
yield (k, dict2[k])
dict1 = {1:{"a":"A"},2:{"b":"B"}}
dict2 = {2:{"c":"C"},3:{"d":"D"}}
print dict(mergedicts(dict1,dict2))
这打印：
{1: {'a': 'A'}, 2: {'c': 'C', 'b': 'B'}, 3: {'d': 'D'}}
这个问题的一个问题是 dict 的值可以是任意复杂的数据片段。基于这些和其他答案，我想出了这段代码：
class YamlReaderError(Exception):
pass
def data_merge(a, b):
"""merges b into a and return merged result
NOTE: tuples and arbitrary objects are not handled as it is totally ambiguous what should happen"""
key = None
# ## debug output
# sys.stderr.write("DEBUG: %s to %s\n" %(b,a))
try:
if a is None or isinstance(a, str) or isinstance(a, unicode) or isinstance(a, int) or isinstance(a, long) or isinstance(a, float):
# border case for first run or if a is a primitive
a = b
elif isinstance(a, list):
# lists can be only appended
if isinstance(b, list):
# merge lists
a.extend(b)
else:
# append to list
a.append(b)
elif isinstance(a, dict):
# dicts must be merged
if isinstance(b, dict):
for key in b:
if key in a:
a[key] = data_merge(a[key], b[key])
else:
a[key] = b[key]
else:
raise YamlReaderError('Cannot merge non-dict "%s" into dict "%s"' % (b, a))
else:
raise YamlReaderError('NOT IMPLEMENTED "%s" into "%s"' % (b, a))
except TypeError, e:
raise YamlReaderError('TypeError "%s" in key "%s" when merging "%s" into "%s"' % (e, key, b, a))
return a
我的用例是合并 YAML 文件，我只需要处理可能数据类型的子集。因此我可以忽略元组和其他对象。对我来说，一个明智的合并逻辑意味着
其他一切和不可预见的结果都会导致错误。
字典的字典合并
由于这是规范问题（尽管存在某些非普遍性），因此我提供了规范的 Pythonic 方法来解决此问题。
d1 = {'a': {1: {'foo': {}}, 2: {}}}
d2 = {'a': {1: {}, 2: {'bar': {}}}}
d3 = {'b': {3: {'baz': {}}}}
d4 = {'a': {1: {'quux': {}}}}
这是最简单的递归情况，我会推荐两种简单的方法：
def rec_merge1(d1, d2):
'''return new merged dict of dicts'''
for k, v in d1.items(): # in Python 2, use .iteritems()!
if k in d2:
d2[k] = rec_merge1(v, d2[k])
d3 = d1.copy()
d3.update(d2)
return d3
def rec_merge2(d1, d2):
'''update first dict with second recursively'''
for k, v in d1.items(): # in Python 2, use .iteritems()!
if k in d2:
d2[k] = rec_merge2(v, d2[k])
d1.update(d2)
return d1
我相信我更喜欢第二个而不是第一个，但请记住，第一个的原始状态必须从其起源重建。这是用法：
>>> from functools import reduce # only required for Python 3.
>>> reduce(rec_merge1, (d1, d2, d3, d4))
{'a': {1: {'quux': {}, 'foo': {}}, 2: {'bar': {}}}, 'b': {3: {'baz': {}}}}
>>> reduce(rec_merge2, (d1, d2, d3, d4))
{'a': {1: {'quux': {}, 'foo': {}}, 2: {'bar': {}}}, 'b': {3: {'baz': {}}}}
因此，如果它们以 dicts 结尾，那么这是合并结尾空 dicts 的简单案例。如果没有，它不是那么微不足道。如果是字符串，你如何合并它们？集合可以类似地更新，因此我们可以进行这种处理，但是我们丢失了它们合并的顺序。那么顺序重要吗？
所以代替更多信息，如果两个值都不是字典，最简单的方法是给他们标准的更新处理：即第二个字典的值将覆盖第一个，即使第二个字典的值为 None 并且第一个的值为dict有很多信息。
d1 = {'a': {1: 'foo', 2: None}}
d2 = {'a': {1: None, 2: 'bar'}}
d3 = {'b': {3: 'baz'}}
d4 = {'a': {1: 'quux'}}
from collections.abc import MutableMapping
def rec_merge(d1, d2):
'''
Update two dicts of dicts recursively,
if either mapping has leaves that are non-dicts,
the second's leaf overwrites the first's.
'''
for k, v in d1.items():
if k in d2:
# this next check is the only difference!
if all(isinstance(e, MutableMapping) for e in (v, d2[k])):
d2[k] = rec_merge(v, d2[k])
# we could further check types and merge as appropriate here.
d3 = d1.copy()
d3.update(d2)
return d3
现在
from functools import reduce
reduce(rec_merge, (d1, d2, d3, d4))
返回
{'a': {1: 'quux', 2: 'bar'}, 'b': {3: 'baz'}}
我必须删除字母周围的花括号并将它们放在单引号中，以使其成为合法的 Python（否则它们将在 Python 2.7+ 中设置文字）以及附加缺少的大括号：
dict1 = {1:{"a":'A'}, 2:{"b":'B'}}
dict2 = {2:{"c":'C'}, 3:{"d":'D'}}
现在rec_merge(dict1, dict2)返回：
{1: {'a': 'A'}, 2: {'c': 'C', 'b': 'B'}, 3: {'d': 'D'}}
与原始问题的预期结果相匹配（更改后，例如{A}to 'A'。）
基于@andrew 库克。此版本处理嵌套的字典列表，还允许更新值的选项
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
这个简单的递归过程将在覆盖冲突键的同时将一个字典合并到另一个字典中：
#!/usr/bin/env python2.7
def merge_dicts(dict1, dict2):
""" Recursively merges dict2 into dict1 """
if not isinstance(dict1, dict) or not isinstance(dict2, dict):
return dict2
for k in dict2:
if k in dict1:
dict1[k] = merge_dicts(dict1[k], dict2[k])
else:
dict1[k] = dict2[k]
return dict1
print (merge_dicts({1:{"a":"A"}, 2:{"b":"B"}}, {2:{"c":"C"}, 3:{"d":"D"}}))
print (merge_dicts({1:{"a":"A"}, 2:{"b":"B"}}, {1:{"a":"A"}, 2:{"b":"C"}}))
输出：
{1: {'a': 'A'}, 2: {'c': 'C', 'b': 'B'}, 3: {'d': 'D'}}
{1: {'a': 'A'}, 2: {'b': 'C'}}
基于@andrew Cooke 的回答。它以更好的方式处理嵌套列表。
def deep_merge_lists(original, incoming):
"""
Deep merge two lists. Modifies original.
Recursively call deep merge on each correlated element of list.
If item type in both elements are
a. dict: Call deep_merge_dicts on both values.
b. list: Recursively call deep_merge_lists on both values.
c. any other type: Value is overridden.
d. conflicting types: Value is overridden.
If length of incoming list is more that of original then extra values are appended.
"""
common_length = min(len(original), len(incoming))
for idx in range(common_length):
if isinstance(original[idx], dict) and isinstance(incoming[idx], dict):
deep_merge_dicts(original[idx], incoming[idx])
elif isinstance(original[idx], list) and isinstance(incoming[idx], list):
deep_merge_lists(original[idx], incoming[idx])
else:
original[idx] = incoming[idx]
for idx in range(common_length, len(incoming)):
original.append(incoming[idx])
def deep_merge_dicts(original, incoming):
"""
Deep merge two dictionaries. Modifies original.
For key conflicts if both values are:
a. dict: Recursively call deep_merge_dicts on both values.
b. list: Call deep_merge_lists on both values.
c. any other type: Value is overridden.
d. conflicting types: Value is overridden.
"""
for key in incoming:
if key in original:
if isinstance(original[key], dict) and isinstance(incoming[key], dict):
deep_merge_dicts(original[key], incoming[key])
elif isinstance(original[key], list) and isinstance(incoming[key], list):
deep_merge_lists(original[key], incoming[key])
else:
original[key] = incoming[key]
else:
original[key] = incoming[key]
如果您有未知级别的字典，那么我建议使用递归函数：
def combineDicts(dictionary1, dictionary2):
output = {}
for item, value in dictionary1.iteritems():
if dictionary2.has_key(item):
if isinstance(dictionary2[item], dict):
output[item] = combineDicts(value, dictionary2.pop(item))
else:
output[item] = value
for item, value in dictionary2.iteritems():
output[item] = value
return output
概述
以下方法将 dicts 的深度合并问题细分为：
一个参数化的浅合并函数merge(f)(a,b)，它使用一个函数f来合并两个 dictsa和b
一个递归合并函数f与一起使用merge
执行
可以用多种方式编写用于合并两个（非嵌套）dict 的函数。我个人喜欢
def merge(f):
def merge(a,b):
keys = a.keys() | b.keys()
return {key:f(a.get(key), b.get(key)) for key in keys}
return merge
定义适当的递归合并函数的一种好方法f是使用multipledispatch，它允许定义根据参数类型沿不同路径评估的函数。
from multipledispatch import dispatch
#for anything that is not a dict return
@dispatch(object, object)
def f(a, b):
return b if b is not None else a
#for dicts recurse
@dispatch(dict, dict)
def f(a,b):
return merge(f)(a,b)
例子
要合并两个嵌套的字典，只需使用merge(f)例如：
dict1 = {1:{"a":"A"},2:{"b":"B"}}
dict2 = {2:{"c":"C"},3:{"d":"D"}}
merge(f)(dict1, dict2)
#returns {1: {'a': 'A'}, 2: {'b': 'B', 'c': 'C'}, 3: {'d': 'D'}}
笔记：
这种方法的优点是：
该函数是由较小的函数构建的，每个函数都做一件事情，这使得代码更易于推理和测试
该行为不是硬编码的，但可以根据需要进行更改和扩展，从而提高代码重用性（参见下面的示例）。
定制
一些答案还考虑了包含列表的字典，例如其他（可能嵌套的）字典。在这种情况下，可能需要映射列表并根据位置合并它们。这可以通过向合并函数添加另一个定义来完成f：
import itertools
@dispatch(list, list)
def f(a,b):
return [merge(f)(*arg) for arg in itertools.zip_longest(a, b)]
如果有人想要另一种方法来解决这个问题，这是我的解决方案。
美德：简短，声明性和功能性风格（递归，没有突变）。
潜在缺点：这可能不是您正在寻找的合并。请参阅文档字符串以了解语义。
def deep_merge(a, b):
"""
Merge two values, with `b` taking precedence over `a`.
Semantics:
- If either `a` or `b` is not a dictionary, `a` will be returned only if
`b` is `None`. Otherwise `b` will be returned.
- If both values are dictionaries, they are merged as follows:
* Each key that is found only in `a` or only in `b` will be included in
the output collection with its value intact.
* For any key in common between `a` and `b`, the corresponding values
will be merged with the same semantics.
"""
if not isinstance(a, dict) or not isinstance(b, dict):
return a if b is None else b
else:
# If we're here, both a and b must be dictionaries or subtypes thereof.
# Compute set of all keys in both dictionaries.
keys = set(a.keys()) | set(b.keys())
# Build output dictionary, merging recursively values with common keys,
# where `None` is used to mean the absence of a value.
return {
key: deep_merge(a.get(key), b.get(key))
for key in keys
}
安德鲁厨师的回答有一个小问题：在某些情况下，b当您修改返回的字典时，它会修改第二个参数。具体来说是因为这条线：
if key in a:
...
else:
a[key] = b[key]
如果b[key]是 a dict，它将被简单地分配给a，这意味着对它的任何后续修改dict都会影响a和b。
a={}
b={'1':{'2':'b'}}
c={'1':{'3':'c'}}
merge(merge(a,b), c) # {'1': {'3': 'c', '2': 'b'}}
a # {'1': {'3': 'c', '2': 'b'}} (as expected)
b # {'1': {'3': 'c', '2': 'b'}} <----
c # {'1': {'3': 'c'}} (unmodified)
要解决此问题，该行必须替换为：
if isinstance(b[key], dict):
a[key] = clone_dict(b[key])
else:
a[key] = b[key]
在哪里clone_dict：
def clone_dict(obj):
clone = {}
for key, value in obj.iteritems():
if isinstance(value, dict):
clone[key] = clone_dict(value)
else:
clone[key] = value
return
仍然。这显然不考虑list,set和其他东西，但我希望它说明尝试合并时的陷阱dicts。
为了完整起见，这是我的版本，您可以在其中多次传递dicts：
def merge_dicts(*args):
def clone_dict(obj):
clone = {}
for key, value in obj.iteritems():
if isinstance(value, dict):
clone[key] = clone_dict(value)
else:
clone[key] = value
return
def merge(a, b, 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
else:
raise Exception('Conflict at `{path}\''.format(path='.'.join(path + [str(key)])))
else:
if isinstance(b[key], dict):
a[key] = clone_dict(b[key])
else:
a[key] = b[key]
return a
return reduce(merge, args, {})
这个版本的函数会处理 N 个字典，并且只有字典——不能传递不正确的参数，否则会引发 TypeError。合并本身会导致键冲突，而不是在合并链的下游覆盖字典中的数据，而是创建一组值并附加到该集合；没有数据丢失。
它可能不是页面上最有效的，但它是最彻底的，当您将 2 到 N dicts 合并时，您不会丢失任何信息。
def merge_dicts(*dicts):
if not reduce(lambda x, y: isinstance(y, dict) and x, dicts, True):
raise TypeError, "Object in *dicts not of type dict"
if len(dicts) < 2:
raise ValueError, "Requires 2 or more dict objects"
def merge(a, b):
for d in set(a.keys()).union(b.keys()):
if d in a and d in b:
if type(a[d]) == type(b[d]):
if not isinstance(a[d], dict):
ret = list({a[d], b[d]})
if len(ret) == 1: ret = ret[0]
yield (d, sorted(ret))
else:
yield (d, dict(merge(a[d], b[d])))
else:
raise TypeError, "Conflicting key:value type assignment"
elif d in a:
yield (d, a[d])
elif d in b:
yield (d, b[d])
else:
raise KeyError
return reduce(lambda x, y: dict(merge(x, y)), dicts[1:], dicts[0])
print merge_dicts({1:1,2:{1:2}},{1:2,2:{3:1}},{4:4})
输出：{1: [1, 2], 2: {1: 2, 3: 1}, 4: 4}
由于 dictviews 支持集合操作，我能够大大简化 jterrace 的答案。
def merge(dict1, dict2):
for k in dict1.keys() - dict2.keys():
yield (k, dict1[k])
for k in dict2.keys() - dict1.keys():
yield (k, dict2[k])
for k in dict1.keys() & dict2.keys():
yield (k, dict(merge(dict1[k], dict2[k])))
任何将 dict 与非 dict 组合的尝试（从技术上讲，一个具有“keys”方法的对象和一个没有“keys”方法的对象）都会引发 AttributeError。这包括对函数的初始调用和递归调用。这正是我想要的，所以我离开了它。您可以轻松捕获递归调用引发的 AttributeErrors，然后产生您喜欢的任何值。
简短的甜蜜：
from collections.abc import MutableMapping as Map
def nested_update(d, v):
"""
Nested update of dict-like 'd' with dict-like 'v'.
"""
for key in v:
if key in d and isinstance(d[key], Map) and isinstance(v[key], Map):
nested_update(d[key], v[key])
else:
d[key] = v[key]
dict.update这与 Python 的方法类似（并且是基于）的。它会在原地更新 dict 时返回（如果您愿意，None您可以随时添加）。输入的键将覆盖任何现有的键（它不会尝试解释字典的内容）。return ddvd
它也适用于其他（“dict-like”）映射。
看看toolz包装
import toolz
dict1={1:{"a":"A"},2:{"b":"B"}}
dict2={2:{"c":"C"},3:{"d":"D"}}
toolz.merge_with(toolz.merge,dict1,dict2)
给
{1: {'a': 'A'}, 2: {'b': 'B', 'c': 'C'}, 3: {'d': 'D'}}
我有一个迭代解决方案 - 使用大字典和其中很多（例如 jsons 等）效果更好：
import collections
def merge_dict_with_subdicts(dict1: dict, dict2: dict) -> dict:
"""
similar behaviour to builtin dict.update - but knows how to handle nested dicts
"""
q = collections.deque([(dict1, dict2)])
while len(q) > 0:
d1, d2 = q.pop()
for k, v in d2.items():
if k in d1 and isinstance(d1[k], dict) and isinstance(v, dict):
q.append((d1[k], v))
else:
d1[k] = v
return dict1
请注意，这将使用 d2 中的值来覆盖 d1，以防它们不是两个字典。（与 python 相同dict.update()）
一些测试：
def test_deep_update():
d = dict()
merge_dict_with_subdicts(d, {"a": 4})
assert d == {"a": 4}
new_dict = {
"b": {
"c": {
"d": 6
}
}
}
merge_dict_with_subdicts(d, new_dict)
assert d == {
"a": 4,
"b": {
"c": {
"d": 6
}
}
}
new_dict = {
"a": 3,
"b": {
"f": 7
}
}
merge_dict_with_subdicts(d, new_dict)
assert d == {
"a": 3,
"b": {
"c": {
"d": 6
},
"f": 7
}
}
# test a case where one of the dicts has dict as value and the other has something else
new_dict = {
'a': {
'b': 4
}
}
merge_dict_with_subdicts(d, new_dict)
assert d['a']['b'] == 4
我已经用大约 1200 个字典进行了测试——这种方法需要 0.4 秒，而递归解决方案需要 2.5 秒。
当然，代码将取决于您解决合并冲突的规则。这是一个版本，它可以采用任意数量的参数并将它们递归地合并到任意深度，而不使用任何对象突变。它使用以下规则来解决合并冲突：
{"foo": {...}}优先于{"foo": "bar"}）{"a": 1},{"a", 2}和{"a": 3}按顺序，结果将是{"a": 3}）try:
from collections import Mapping
except ImportError:
Mapping = dict
def merge_dicts(*dicts):
"""
Return a new dictionary that is the result of merging the arguments together.
In case of conflicts, later arguments take precedence over earlier arguments.
"""
updated = {}
# grab all keys
keys = set()
for d in dicts:
keys = keys.union(set(d))
for key in keys:
values = [d[key] for d in dicts if key in d]
# which ones are mapping types? (aka dict)
maps = [value for value in values if isinstance(value, Mapping)]
if maps:
# if we have any mapping types, call recursively to merge them
updated[key] = merge_dicts(*maps)
else:
# otherwise, just grab the last value we have, since later arguments
# take precedence over earlier arguments
updated[key] = values[-1]
return updated
我有两个字典 (a和b)，每个字典都可以包含任意数量的嵌套字典。我想递归合并它们，b优先于a.
将嵌套字典视为树，我想要的是：
a，以便每个叶子的每条路径都b表示为aa是否在相应路径中找到叶子的子树b
b叶子节点保持叶子的不变性。现有的答案对我来说有点复杂，并且在架子上留下了一些细节。我一起破解了以下内容，它通过了我的数据集的单元测试。
def merge_map(a, b):
if not isinstance(a, dict) or not isinstance(b, dict):
return b
for key in b.keys():
a[key] = merge_map(a[key], b[key]) if key in a else b[key]
return a
示例（为清晰起见格式化）：
a = {
1 : {'a': 'red',
'b': {'blue': 'fish', 'yellow': 'bear' },
'c': { 'orange': 'dog'},
},
2 : {'d': 'green'},
3: 'e'
}
b = {
1 : {'b': 'white'},
2 : {'d': 'black'},
3: 'e'
}
>>> merge_map(a, b)
{1: {'a': 'red',
'b': 'white',
'c': {'orange': 'dog'},},
2: {'d': 'black'},
3: 'e'}
b需要维护的路径是：
1 -> 'b' -> 'white'2 -> 'd' -> 'black'3 -> 'e'.a具有以下独特且不冲突的路径：
1 -> 'a' -> 'red' 1 -> 'c' -> 'orange' -> 'dog'所以它们仍然在合并的地图中表示。
另一个答案怎么样？！？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)
这应该有助于将所有项目合并dict2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]
请对其进行测试并告诉我们这是否是您想要的。
编辑：
上述解决方案仅合并了一个级别，但正确解决了OP给出的示例。要合并多个级别，应使用递归。
我一直在测试您的解决方案，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扩展 jterrace 解决方案以像所有其他递归解决方案一样工作的关键。
我能想到的最简单的方法是：
#!/usr/bin/python
from copy import deepcopy
def dict_merge(a, b):
if not isinstance(b, dict):
return b
result = deepcopy(a)
for k, v in b.iteritems():
if k in result and isinstance(result[k], dict):
result[k] = dict_merge(result[k], v)
else:
result[k] = deepcopy(v)
return result
a = {1:{"a":'A'}, 2:{"b":'B'}}
b = {2:{"c":'C'}, 3:{"d":'D'}}
print dict_merge(a,b)
输出：
{1: {'a': 'A'}, 2: {'c': 'C', 'b': 'B'}, 3: {'d': 'D'}}
我在这里有另一个稍微不同的解决方案：
def deepMerge(d1, d2, inconflict = lambda v1,v2 : v2) :
''' merge d2 into d1. using inconflict function to resolve the leaf conflicts '''
for k in d2:
if k in d1 :
if isinstance(d1[k], dict) and isinstance(d2[k], dict) :
deepMerge(d1[k], d2[k], inconflict)
elif d1[k] != d2[k] :
d1[k] = inconflict(d1[k], d2[k])
else :
d1[k] = d2[k]
return d1
默认情况下，它会解决冲突以支持第二个字典中的值，但是您可以轻松地覆盖它，使用一些巫术，您甚至可以从中抛出异常。:)。
class Utils(object):
"""
>>> a = { 'first' : { 'all_rows' : { 'pass' : 'dog', 'number' : '1' } } }
>>> b = { 'first' : { 'all_rows' : { 'fail' : 'cat', 'number' : '5' } } }
>>> Utils.merge_dict(b, a) == { 'first' : { 'all_rows' : { 'pass' : 'dog', 'fail' : 'cat', 'number' : '5' } } }
True
>>> main = {'a': {'b': {'test': 'bug'}, 'c': 'C'}}
>>> suply = {'a': {'b': 2, 'd': 'D', 'c': {'test': 'bug2'}}}
>>> Utils.merge_dict(main, suply) == {'a': {'b': {'test': 'bug'}, 'c': 'C', 'd': 'D'}}
True
"""
@staticmethod
def merge_dict(main, suply):
"""
获取融合的字典，以main为主,suply补充,冲突时以main为准
:return:
"""
for key, value in suply.items():
if key in main:
if isinstance(main[key], dict):
if isinstance(value, dict):
Utils.merge_dict(main[key], value)
else:
pass
else:
pass
else:
main[key] = value
return main
if __name__ == '__main__':
import doctest
doctest.testmod()
嘿，我也有同样的问题，但我想了一个解决方案，我会在这里发布，以防它对其他人也有用，基本上合并嵌套字典并添加值，对我来说我需要计算一些概率所以这个一个效果很好：
#used to copy a nested dict to a nested dict
def deepupdate(target, src):
for k, v in src.items():
if k in target:
for k2, v2 in src[k].items():
if k2 in target[k]:
target[k][k2]+=v2
else:
target[k][k2] = v2
else:
target[k] = copy.deepcopy(v)
通过使用上述方法，我们可以合并：
目标 = {'6,6': {'6,63': 1}, '63,4': {'4,4': 1}, '4,4': {'4,3': 1} , '6,63': {'63,4': 1}}
src = {'5,4': {'4,4': 1}, '5,5': {'5,4': 1}, '4,4': {'4,3': 1} }
这将变成： {'5,5': {'5,4': 1}, '5,4': {'4,4': 1}, '6,6': {'6,63' : 1}, '63,4': {'4,4': 1}, '4,4': {'4,3': 2}, '6,63': {'63,4': 1 }}
还要注意这里的变化：
目标 = {'6,6': {'6,63': 1}, '6,63': {'63,4': 1}, '4,4': {'4,3': 1} , '63,4': {'4,4': 1}}
src = {'5,4': {'4,4': 1}, '4,3': {'3,4': 1}, '4,4': {'4,9': 1} , '3,4': {'4,4': 1}, '5,5': {'5,4': 1}}
合并 = {'5,4': {'4,4': 1}, '4,3': {'3,4': 1}, '6,63': {'63,4': 1} , '5,5': {'5,4': 1}, '6,6': {'6,63': 1}, '3,4': {'4,4': 1}, ' 63,4': {'4,4': 1}, '4,4': {'4,3': 1, '4,9': 1} }
不要忘记添加复制的导入：
import copy
from collections import defaultdict
from itertools import chain
class DictHelper:
@staticmethod
def merge_dictionaries(*dictionaries, override=True):
merged_dict = defaultdict(set)
all_unique_keys = set(chain(*[list(dictionary.keys()) for dictionary in dictionaries])) # Build a set using all dict keys
for key in all_unique_keys:
keys_value_type = list(set(filter(lambda obj_type: obj_type != type(None), [type(dictionary.get(key, None)) for dictionary in dictionaries])))
# Establish the object type for each key, return None if key is not present in dict and remove None from final result
if len(keys_value_type) != 1:
raise Exception("Different objects type for same key: {keys_value_type}".format(keys_value_type=keys_value_type))
if keys_value_type[0] == list:
values = list(chain(*[dictionary.get(key, []) for dictionary in dictionaries])) # Extract the value for each key
merged_dict[key].update(values)
elif keys_value_type[0] == dict:
# Extract all dictionaries by key and enter in recursion
dicts_to_merge = list(filter(lambda obj: obj != None, [dictionary.get(key, None) for dictionary in dictionaries]))
merged_dict[key] = DictHelper.merge_dictionaries(*dicts_to_merge)
else:
# if override => get value from last dictionary else make a list of all values
values = list(filter(lambda obj: obj != None, [dictionary.get(key, None) for dictionary in dictionaries]))
merged_dict[key] = values[-1] if override else values
return dict(merged_dict)
if __name__ == '__main__':
d1 = {'aaaaaaaaa': ['to short', 'to long'], 'bbbbb': ['to short', 'to long'], "cccccc": ["the is a test"]}
d2 = {'aaaaaaaaa': ['field is not a bool'], 'bbbbb': ['field is not a bool']}
d3 = {'aaaaaaaaa': ['filed is not a string', "to short"], 'bbbbb': ['field is not an integer']}
print(DictHelper.merge_dictionaries(d1, d2, d3))
d4 = {"a": {"x": 1, "y": 2, "z": 3, "d": {"x1": 10}}}
d5 = {"a": {"x": 10, "y": 20, "d": {"x2": 20}}}
print(DictHelper.merge_dictionaries(d4, d5))
输出：
{'bbbbb': {'to long', 'field is not an integer', 'to short', 'field is not a bool'},
'aaaaaaaaa': {'to long', 'to short', 'filed is not a string', 'field is not a bool'},
'cccccc': {'the is a test'}}
{'a': {'y': 20, 'd': {'x1': 10, 'x2': 20}, 'z': 3, 'x': 10}}
以下函数将 b 合并为 a。
def mergedicts(a, b):
for key in b:
if isinstance(a.get(key), dict) or isinstance(b.get(key), dict):
mergedicts(a[key], b[key])
else:
a[key] = b[key]
return a
只是另一个轻微的变化：
这是一个纯粹的基于python3集的深度更新函数。它通过一次循环遍历一个级别来更新嵌套字典，并调用自身来更新每个下一级字典值：
def deep_update(dict_original, dict_update):
if isinstance(dict_original, dict) and isinstance(dict_update, dict):
output=dict(dict_original)
keys_original=set(dict_original.keys())
keys_update=set(dict_update.keys())
similar_keys=keys_original.intersection(keys_update)
similar_dict={key:deep_update(dict_original[key], dict_update[key]) for key in similar_keys}
new_keys=keys_update.difference(keys_original)
new_dict={key:dict_update[key] for key in new_keys}
output.update(similar_dict)
output.update(new_dict)
return output
else:
return dict_update
一个简单的例子：
x={'a':{'b':{'c':1, 'd':1}}}
y={'a':{'b':{'d':2, 'e':2}}, 'f':2}
print(deep_update(x, y))
>>> {'a': {'b': {'c': 1, 'd': 2, 'e': 2}}, 'f': 2}
我没有对此进行广泛的测试，因此，我们鼓励您提供反馈。
from collections import defaultdict
dict1 = defaultdict(list)
dict2= defaultdict(list)
dict3= defaultdict(list)
dict1= dict(zip(Keys[ ],values[ ]))
dict2 = dict(zip(Keys[ ],values[ ]))
def mergeDict(dict1, dict2):
dict3 = {**dict1, **dict2}
for key, value in dict3.items():
if key in dict1 and key in dict2:
dict3[key] = [value , dict1[key]]
return dict3
dict3 = mergeDict(dict1, dict2)
#sort keys alphabetically.
dict3.keys()