两个字典（键和值）的递归差异？

Question

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

Answer 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}}}

Answer 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

Answer 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。

Answer 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')]

Answer 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

Answer 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)

Answer 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。

Answer 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?

Answer 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