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。
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。
优点:简洁、声明性和函数式风格(递归,没有突变)。
潜在缺点:这可能不是你想要的合并。查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
}
Short-n-sweet:
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]
这类似于(并且构建在)Python的字典上。更新方法。它返回None(如果你喜欢,你总是可以添加返回d),因为它在原地更新dict d。v中的键将覆盖d中任何现有的键(它不会尝试解释字典的内容)。
它也适用于其他(“类字典”)映射。
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。
优点:简洁、声明性和函数式风格(递归,没有突变)。
潜在缺点:这可能不是你想要的合并。查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
}
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)