两个字典(键和值)的递归比较？
两个字典(键和值)的递归比较？
提问于 2011-05-06 04:32:23
所以我有一个python字典，称为d1，稍后还有一个版本，称为d1。我想找出d1和d2之间的所有变化。换句话说，所有添加、删除或更改的内容。棘手的一点是，值可以是it、string、list或dict，因此它需要是递归的。这就是我到目前为止所知道的：
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它工作得很好，除非值是一个列表。我想不出一个优雅的方法来处理列表，除非在if(type(d2) == list)之后重复这个函数的一个巨大的、略微变化的版本。
有什么想法吗？
编辑:这不同于this post，因为键可以改变
回答 8
Stack Overflow用户
回答已采纳
发布于 2011-05-06 05:24:38
一种选择是将您运行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)以下是您在注释中给出的示例字典的输出：
>>> 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处理添加或删除的列表项。
Stack Overflow用户
发布于 2014-10-03 08:43:25
如果您希望以递归方式实现差异，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
正在导入
>>> from deepdiff import DeepDiff
>>> from pprint import pprint
>>> from __future__ import print_function # In case running on Python 2同一对象返回空
>>> t1 = {1:1, 2:2, 3:3}
>>> t2 = t1
>>> print(DeepDiff(t1, t2))
{}项目的类型已更改
>>> 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}}}项的值已更改
>>> 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}}}字符串差异
>>> 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类型更改
>>> 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}}列表差异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}}}忽略顺序或重复项的列表差异：(使用与上面相同的字典)
>>> 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}}}集合：
>>> 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}}}已添加对象属性：
>>> t2.c = "new attribute"
>>> pprint(DeepDiff(t1, t2))
{'attribute_added': ['root.c'],
'values_changed': {'root.b': {'newvalue': 2, 'oldvalue': 1}}}Stack Overflow用户
发布于 2018-12-17 23:45:11
下面是一个受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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/5903720
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