两个字典（键和值）的递归差异？
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【中文标题】两个字典（键和值）的递归差异？【英文标题】：Recursive diff of two dictionaries (keys and values)? 【发布时间】：2011-08-19 17:03:00 【问题描述】：所以我有一个 python 字典，叫它d1，以及以后那个字典的一个版本，叫它d2。我想找出d1 和d2 之间的所有变化。换句话说，添加、删除或更改的所有内容。棘手的一点是值可以是整数、字符串、列表或字典，因此它需要是递归的。这是我目前所拥有的：
def dd(d1, d2, ctx=""):
print "Changes in " + ctx
for k in d1:
if k not in d2:
print k + " removed from d2"
for k in d2:
if k not in d1:
print k + " added in d2"
continue
if d2[k] != d1[k]:
if type(d2[k]) not in (dict, list):
print k + " changed in d2 to " + str(d2[k])
else:
if type(d1[k]) != type(d2[k]):
print k + " changed to " + str(d2[k])
continue
else:
if type(d2[k]) == dict:
dd(d1[k], d2[k], k)
continue
print "Done with changes in " + ctx
return
除非值是一个列表，否则它工作得很好。如果没有在if(type(d2) == list) 之后重复这个函数的巨大的、稍微改变的版本，我无法想出一个优雅的方式来处理列表。
有什么想法吗？
编辑：这与 this post 不同，因为键可以更改
【问题讨论】：
示例：list1 = [0, 1, 2, 3, 4, 5, 6, 7]、list2 = [0, 2, 3, 4, 5, 6, 7, 8]。你期望什么输出？
如果它们在 2 个不同的字典中使用相同的键，我认为：1 被删除；添加了 8 个（在同一键下）。如果它们在不同的键下，那么它们就是不同的元素。
这很快就会变得棘手。顺序重要吗？如果8 移到前面会怎样：[8, 1, 2, 3, 4, 5, 6, 7]，订单是否计算在内，或者只有在场/缺席（一组）？列表是否可以包含嵌套字典，而字典又包含列表等？
你能举一个输出失败的例子吗？
@samplebias：是的。列表可以包含字典，字典可以包含....它的海龟一路向下。我真的不需要元组，但在这一点上，这并没有多大帮助
【参考方案1】：
如果您想要递归地进行差异化，我已经为 python 编写了一个包： https://github.com/seperman/deepdiff
安装
从 PyPi 安装：
pip install deepdiff
示例用法
导入
>>> from deepdiff import DeepDiff
>>> from pprint import pprint
>>> from __future__ import print_function # In case running on Python 2
相同的对象返回空
>>> t1 = 1:1, 2:2, 3:3
>>> t2 = t1
>>> print(DeepDiff(t1, t2))
项目类型已更改
>>> t1 = 1:1, 2:2, 3:3
>>> t2 = 1:1, 2:"2", 3:3
>>> pprint(DeepDiff(t1, t2), indent=2)
'type_changes': 'root[2]': 'newtype': <class 'str'>,
'newvalue': '2',
'oldtype': <class 'int'>,
'oldvalue': 2
物品的价值发生了变化
>>> t1 = 1:1, 2:2, 3:3
>>> t2 = 1:1, 2:4, 3:3
>>> pprint(DeepDiff(t1, t2), indent=2)
'values_changed': 'root[2]': 'newvalue': 4, 'oldvalue': 2
添加和/或删除项目
>>> t1 = 1:1, 2:2, 3:3, 4:4
>>> t2 = 1:1, 2:4, 3:3, 5:5, 6:6
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff)
'dic_item_added': ['root[5]', 'root[6]'],
'dic_item_removed': ['root[4]'],
'values_changed': 'root[2]': 'newvalue': 4, 'oldvalue': 2
字符串区别
>>> t1 = 1:1, 2:2, 3:3, 4:"a":"hello", "b":"world"
>>> t2 = 1:1, 2:4, 3:3, 4:"a":"hello", "b":"world!"
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
'values_changed': 'root[2]': 'newvalue': 4, 'oldvalue': 2,
"root[4]['b']": 'newvalue': 'world!',
'oldvalue': 'world'
字符串差异2
>>> t1 = 1:1, 2:2, 3:3, 4:"a":"hello", "b":"world!\nGoodbye!\n1\n2\nEnd"
>>> t2 = 1:1, 2:2, 3:3, 4:"a":"hello", "b":"world\n1\n2\nEnd"
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
'values_changed': "root[4]['b']": 'diff': '--- \n'
'+++ \n'
'@@ -1,5 +1,4 @@\n'
'-world!\n'
'-Goodbye!\n'
'+world\n'
' 1\n'
' 2\n'
' End',
'newvalue': 'world\n1\n2\nEnd',
'oldvalue': 'world!\n'
'Goodbye!\n'
'1\n'
'2\n'
'End'
>>>
>>> print (ddiff['values_changed']["root[4]['b']"]["diff"])
---
+++
@@ -1,5 +1,4 @@
-world!
-Goodbye!
+world
1
2
End
类型改变
>>> t1 = 1:1, 2:2, 3:3, 4:"a":"hello", "b":[1, 2, 3]
>>> t2 = 1:1, 2:2, 3:3, 4:"a":"hello", "b":"world\n\n\nEnd"
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
'type_changes': "root[4]['b']": 'newtype': <class 'str'>,
'newvalue': 'world\n\n\nEnd',
'oldtype': <class 'list'>,
'oldvalue': [1, 2, 3]
列表差异
>>> t1 = 1:1, 2:2, 3:3, 4:"a":"hello", "b":[1, 2, 3, 4]
>>> t2 = 1:1, 2:2, 3:3, 4:"a":"hello", "b":[1, 2]
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
'iterable_item_removed': "root[4]['b'][2]": 3, "root[4]['b'][3]": 4
列出差异2：
>>> t1 = 1:1, 2:2, 3:3, 4:"a":"hello", "b":[1, 2, 3]
>>> t2 = 1:1, 2:2, 3:3, 4:"a":"hello", "b":[1, 3, 2, 3]
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
'iterable_item_added': "root[4]['b'][3]": 3,
'values_changed': "root[4]['b'][1]": 'newvalue': 3, 'oldvalue': 2,
"root[4]['b'][2]": 'newvalue': 2, 'oldvalue': 3
忽略顺序或重复列出差异：（使用与上述相同的字典）
>>> t1 = 1:1, 2:2, 3:3, 4:"a":"hello", "b":[1, 2, 3]
>>> t2 = 1:1, 2:2, 3:3, 4:"a":"hello", "b":[1, 3, 2, 3]
>>> ddiff = DeepDiff(t1, t2, ignore_order=True)
>>> print (ddiff)
包含字典的列表：
>>> t1 = 1:1, 2:2, 3:3, 4:"a":"hello", "b":[1, 2, 1:1, 2:2]
>>> t2 = 1:1, 2:2, 3:3, 4:"a":"hello", "b":[1, 2, 1:3]
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
'dic_item_removed': ["root[4]['b'][2][2]"],
'values_changed': "root[4]['b'][2][1]": 'newvalue': 3, 'oldvalue': 1
套装：
>>> t1 = 1, 2, 8
>>> t2 = 1, 2, 3, 5
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (DeepDiff(t1, t2))
'set_item_added': ['root[3]', 'root[5]'], 'set_item_removed': ['root[8]']
命名元组：
>>> from collections import namedtuple
>>> Point = namedtuple('Point', ['x', 'y'])
>>> t1 = Point(x=11, y=22)
>>> t2 = Point(x=11, y=23)
>>> pprint (DeepDiff(t1, t2))
'values_changed': 'root.y': 'newvalue': 23, 'oldvalue': 22
自定义对象：
>>> class ClassA(object):
... a = 1
... def __init__(self, b):
... self.b = b
...
>>> t1 = ClassA(1)
>>> t2 = ClassA(2)
>>>
>>> pprint(DeepDiff(t1, t2))
'values_changed': 'root.b': 'newvalue': 2, 'oldvalue': 1
添加了对象属性：
>>> t2.c = "new attribute"
>>> pprint(DeepDiff(t1, t2))
'attribute_added': ['root.c'],
'values_changed': 'root.b': 'newvalue': 2, 'oldvalue': 1
【讨论】：
感谢@LukasN.P.Egger @MohitC 你能在github上开一张票，写下语法错误在哪里吗？ 它在导入行本身。文件“/usr/lib/python2.6/site-packages/deepdiff/__init__.py”，第 1 行，在str(DeepDiff(t1, t2)) == "" ？我只需要知道它们是否相等......【参考方案2】：
这是一个受Winston Ewert启发的实现
def recursive_compare(d1, d2, level='root'):
if isinstance(d1, dict) and isinstance(d2, dict):
if d1.keys() != d2.keys():
s1 = set(d1.keys())
s2 = set(d2.keys())
print(':<20 + - '.format(level, s1-s2, s2-s1))
common_keys = s1 & s2
else:
common_keys = set(d1.keys())
for k in common_keys:
recursive_compare(d1[k], d2[k], level='.'.format(level, k))
elif isinstance(d1, list) and isinstance(d2, list):
if len(d1) != len(d2):
print(':<20 len1=; len2='.format(level, len(d1), len(d2)))
common_len = min(len(d1), len(d2))
for i in range(common_len):
recursive_compare(d1[i], d2[i], level='[]'.format(level, i))
else:
if d1 != d2:
print(':<20 != '.format(level, d1, d2))
if __name__ == '__main__':
d1='a':[0,2,3,8], 'b':0, 'd':'da':7, 'db':[99,88]
d2='a':[0,2,4], 'c':0, 'd':'da':3, 'db':7
recursive_compare(d1, d2)
将返回：
root + 'b' - 'c'
root.a len1=4; len2=3
root.a[2] 3 != 4
root.d.db [99, 88] != 7
root.d.da 7 != 3
【讨论】：
如果值是具有不同元素顺序的字典列表，这将失败，不是吗？【参考方案3】：一种选择是将您遇到的任何列表转换为以索引为键的字典。例如：
# add this function to the same module
def list_to_dict(l):
return dict(zip(map(str, range(len(l))), l))
# add this code under the 'if type(d2[k]) == dict' block
elif type(d2[k]) == list:
dd(list_to_dict(d1[k]), list_to_dict(d2[k]), k)
这是您在 cmets 中提供的示例字典的输出：
>>> d1 = "name":"Joe", "Pets":["name":"spot", "species":"dog"]
>>> d2 = "name":"Joe", "Pets":["name":"spot", "species":"cat"]
>>> dd(d1, d2, "base")
Changes in base
Changes in Pets
Changes in 0
species changed in d2 to cat
Done with changes in 0
Done with changes in Pets
Done with changes in base
请注意，这将逐个索引进行比较，因此需要进行一些修改才能很好地添加或删除列表项。
【讨论】：
【参考方案4】：只是一个想法：您可以尝试一种面向对象的方法，在该方法中派生自己的字典类，以跟踪对其所做的任何更改（并报告它们）。与尝试比较两个字典相比，这似乎有很多优势……最后会注明一个。
为了展示如何做到这一点，这里有一个相当完整且经过最少测试的示例实现，它应该适用于 Python 2 和 3：
import sys
_NUL = object() # unique object
if sys.version_info[0] > 2:
def iterkeys(d, **kw):
return iter(d.keys(**kw))
else:
def iterkeys(d, **kw):
return d.iterkeys(**kw)
class TrackingDict(dict):
""" Dict subclass which tracks all changes in a _changelist attribute. """
def __init__(self, *args, **kwargs):
super(TrackingDict, self).__init__(*args, **kwargs)
self.clear_changelist()
for key in sorted(iterkeys(self)):
self._changelist.append(AddKey(key, self[key]))
def clear_changelist(self): # additional public method
self._changelist = []
def __setitem__(self, key, value):
modtype = ChangeKey if key in self else AddKey
super(TrackingDict, self).__setitem__(key, value)
self._changelist.append(modtype(key, self[key]))
def __delitem__(self, key):
super(TrackingDict, self).__delitem__(key)
self._changelist.append(RemoveKey(key))
def clear(self):
deletedkeys = self.keys()
super(TrackingDict, self).clear()
for key in sorted(deletedkeys):
self._changelist.append(RemoveKey(key))
def update(self, other=_NUL):
if other is not _NUL:
otherdict = dict(other) # convert to dict if necessary
changedkeys = set(k for k in otherdict if k in self)
super(TrackingDict, self).update(other)
for key in sorted(iterkeys(otherdict)):
if key in changedkeys:
self._changelist.append(ChangeKey(key, otherdict[key]))
else:
self._changelist.append(AddKey(key, otherdict[key]))
def setdefault(self, key, default=None):
if key not in self:
self[key] = default # will append an AddKey to _changelist
return self[key]
def pop(self, key, default=_NUL):
if key in self:
ret = self[key] # save value
self.__delitem__(key)
return ret
elif default is not _NUL: # default specified
return default
else: # not there & no default
self[key] # allow KeyError to be raised
def popitem(self):
key, value = super(TrackingDict, self).popitem()
self._changelist.append(RemoveKey(key))
return key, value
# change-tracking record classes
class DictMutator(object):
def __init__(self, key, value=_NUL):
self.key = key
self.value = value
def __repr__(self):
return '%s(%r%s)' % (self.__class__.__name__, self.key,
'' if self.value is _NUL else ': '+repr(self.value))
class AddKey(DictMutator): pass
class ChangeKey(DictMutator): pass
class RemoveKey(DictMutator): pass
if __name__ == '__main__':
import traceback
import sys
td = TrackingDict('one': 1, 'two': 2)
print('changelist: '.format(td._changelist))
td['three'] = 3
print('changelist: '.format(td._changelist))
td['two'] = -2
print('changelist: '.format(td._changelist))
td.clear()
print('changelist: '.format(td._changelist))
td.clear_changelist()
td['newkey'] = 42
print('changelist: '.format(td._changelist))
td.setdefault('another') # default None value
print('changelist: '.format(td._changelist))
td.setdefault('one more', 43)
print('changelist: '.format(td._changelist))
td.update(zip(('another', 'one', 'two'), (17, 1, 2)))
print('changelist: '.format(td._changelist))
td.pop('newkey')
print('changelist: '.format(td._changelist))
try:
td.pop("won't find")
except KeyError:
print("KeyError as expected:")
traceback.print_exc(file=sys.stdout)
print('...and no change to _changelist:')
print('changelist: '.format(td._changelist))
td.clear_changelist()
while td:
td.popitem()
print('changelist: '.format(td._changelist))
注意，与字典的 before 和 after 状态的简单比较不同，该类将告诉您添加的键然后删除——换句话说，它会保留完整的历史记录，直到其 _changelist 被清除。
输出：
changelist: [AddKey('one': 1), AddKey('two': 2)]
changelist: [AddKey('one': 1), AddKey('two': 2), AddKey('three': 3)]
changelist: [AddKey('one': 1), AddKey('two': 2), AddKey('three': 3), ChangeKey('two': -2)]
changelist: [AddKey('one': 1), AddKey('two': 2), AddKey('three': 3), ChangeKey('two': -2), RemoveKey('one'), RemoveKey('three'), RemoveKey('two')]
changelist: [AddKey('newkey': 42)]
changelist: [AddKey('newkey': 42), AddKey('another': None)]
changelist: [AddKey('newkey': 42), AddKey('another': None), AddKey('one more': 43)]
changelist: [AddKey('newkey': 42), AddKey('another': None), AddKey('one more': 43), ChangeKey('another': 17), AddKey('one': 1), AddKey('two': 2)]
changelist: [AddKey('newkey': 42), AddKey('another': None), AddKey('one more': 43), ChangeKey('another': 17), AddKey('one': 1), AddKey('two': 2), RemoveKey('newkey')]
KeyError as expected:
Traceback (most recent call last):
File "trackingdict.py", line 122, in <module>
td.pop("won't find")
File "trackingdict.py", line 67, in pop
self[key] # allow KeyError to be raised
KeyError: "won't find"
...and no change to _changelist:
changelist: [AddKey('newkey': 42), AddKey('another': None), AddKey('one more': 43), ChangeKey('another': 17), AddKey('one': 1), AddKey('two': 2), RemoveKey('newkey')]
changelist: [RemoveKey('one'), RemoveKey('two'), RemoveKey('another'), RemoveKey('one more')]
【讨论】：
【参考方案5】：您的函数应该首先检查其参数的类型，编写函数以便它可以处理列表、字典、整数和字符串。这样您就不必复制任何内容，只需递归调用即可。
伪代码：
def compare(d1, d2):
if d1 and d2 are dicts
compare the keys, pass values to compare
if d1 and d2 are lists
compare the lists, pass values to compare
if d1 and d2 are strings/ints
compare them
【讨论】：
【参考方案6】：正如 Serge 所建议的那样，我发现这个解决方案有助于快速获得关于两个字典是否匹配“一直向下”的布尔返回：
import json
def match(d1, d2):
return json.dumps(d1, sort_keys=True) == json.dumps(d2, sort_keys=True)
【讨论】：
【参考方案7】：在递归对象时考虑使用hasattr(obj, '__iter__')。如果一个对象实现了__iter__ 方法，您就知道可以对其进行迭代。
【讨论】：
【参考方案8】：自己动手实践和学习很有趣，但我发现对于不平凡的任务，现成和维护的包通常效果更好。
考虑转换为 json 并使用一些像样的“语义”json 比较器，例如 https://www.npmjs.com/package/compare-json 或在线 http://jsondiff.com。需要字符串化数字键。
如果你真的需要，可以尝试将 jsondiff 翻译成 python。
Conversion from javascript to Python code?
【讨论】：
【参考方案9】：你可以试试下面的简单实现
def recursive_compare(obj1, obj2):
""" Compare python objects recursively, support type:
"int, float, long, basestring, set, datetime, date, dict, Sequence"
Example:
>>> recursive_compare([1, 2, 3], [1, 2, 3])
>>> True
>>> recursive_compare([1, 2, 3], [1, 2, 4])
>>> False
>>> recursive_compare('a': 1, 'a': 2)
>>> False
"""
def _diff(obj1, obj2):
# exclude type basestring for backward-compatible python2:
# <str, unicode>
if type(obj1) != type(obj2) and not isinstance(obj1, basestring):
return False
elif isinstance(obj1,
(int, float, long, basestring, set, datetime, date)):
if obj1 != obj2:
return False
elif isinstance(obj1, dict):
keys = obj1.viewkeys() & obj2.viewkeys()
if obj1 and len(keys) == 0 \
or keys.difference(set(obj1.keys())) \
or keys.difference(set(obj2.keys())):
return False
for k in keys:
if _diff(obj1[k], obj2[k]) is False:
return False
elif isinstance(obj1, collections.Sequence):
# require sorted sequence object
if len(obj1) != len(obj2):
return False
for i in range(len(obj1)):
if _diff(obj1[i], obj2[i]) is False:
return False
else:
raise TypeError('do not support type to compare'.format(
type(obj1)))
return False if _diff(obj1, obj2) is False else True
【讨论】：
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