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。
还有一个轻微的变化:
下面是一个纯粹的基于python3集的深度更新函数。它通过一次循环遍历一层来更新嵌套字典,并调用自己来更新下一层的字典值:
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
举个简单的例子:
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。
还有一个轻微的变化:
下面是一个纯粹的基于python3集的深度更新函数。它通过一次循环遍历一层来更新嵌套字典,并调用自己来更新下一层的字典值:
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
举个简单的例子:
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}
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()
以下是来自@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]
这个版本的函数将处理N个字典,并且只处理字典——不能传递不恰当的参数,否则将引发TypeError。合并本身解释了键冲突,它不是覆盖来自合并链下的字典的数据,而是创建一组值并追加到该值;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,当你合并2到N字典时,你不会丢失任何信息。
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}