Question

我有两本词典

a = {'home': {'name': 'Team1', 'score': 0}, 'away': {'name': 'Team2', 'score': 0}}
b = {'home': {'name': 'Team1', 'score': 2}, 'away': {'name': 'Team2', 'score': 0}}

密钥永远不会改变，但我想让['home'] ['得分']改变

这么简单吗？

Answer 1

作为一个下意识的初步回应：

a = {'home': {'name': 'Team1', 'score': 0}, 'away': {'name': 'Team2', 'score': 0}}
b = {'home': {'name': 'Team1', 'score': 2}, 'away': {'name': 'Team2', 'score': 0}}

def valchange(d1, d2, parent=''):
    changes=[]
    for k in d1.keys():
        if type(d1[k])==type({}):
            changes.extend(valchange(d1[k], d2[k], k))
        else:
            if d1[k]!=d2[k]:
                if parent=='':
                    changes.append(k + ' has changed ')
                else:
                    changes.append(parent + '.' + k + ' has changed')
    return changes

print valchange(a,b)

>>>
['home.score has changed']

Answer 2

您可以使用我的python包： https://github.com/seperman/deepdiff

它不仅仅涉及递归字典差异：

安装

从PyPi安装：

pip install deepdiff

如果您是Python3，还需要安装：

pip install future six

使用示例

>>> from deepdiff import DeepDiff
>>> from pprint import pprint
>>> from __future__ import print_function

相同的对象返回空

>>> t1 = {1:1, 2:2, 3:3}
>>> t2 = t1
>>> ddiff = DeepDiff(t1, t2)
>>> print (ddiff.changes)
    {}

项目类型已更改

>>> t1 = {1:1, 2:2, 3:3}
>>> t2 = {1:1, 2:"2", 3:3}
>>> ddiff = DeepDiff(t1, t2)
>>> print (ddiff.changes)
    {'type_changes': ["root[2]: 2=<type 'int'> vs. 2=<type 'str'>"]}

项目的价值已更改

>>> t1 = {1:1, 2:2, 3:3}
>>> t2 = {1:1, 2:4, 3:3}
>>> ddiff = DeepDiff(t1, t2)
>>> print (ddiff.changes)
    {'values_changed': ['root[2]: 2 ====>> 4']}

添加和/或删除了项目

>>> 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.changes)
    {'dic_item_added': ['root[5, 6]'],
     'dic_item_removed': ['root[4]'],
     'values_changed': ['root[2]: 2 ====>> 4']}

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.changes, indent = 2)
    { 'values_changed': [ 'root[2]: 2 ====>> 4',
                          "root[4]['b']:\n--- \n+++ \n@@ -1 +1 @@\n-world\n+world!"]}
>>>
>>> print (ddiff.changes['values_changed'][1])
    root[4]['b']:
    --- 
    +++ 
    @@ -1 +1 @@
    -world
    +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.changes, indent = 2)
    { 'values_changed': [ "root[4]['b']:\n--- \n+++ \n@@ -1,5 +1,4 @@\n-world!\n-Goodbye!\n+world\n 1\n 2\n End"]}
>>>
>>> print (ddiff.changes['values_changed'][0])
    root[4]['b']:
    --- 
    +++ 
    @@ -1,5 +1,4 @@
    -world!
    -Goodbye!
    +world
     1
     2
     End

输入类型

>>> 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.changes, indent = 2)
    { 'type_changes': [ "root[4]['b']: [1, 2, 3]=<type 'list'> vs. world\n\n\nEnd=<type 'str'>"]}

列表差异

>>> 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, 2]}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff.changes, indent = 2)
    { 'list_removed': ["root[4]['b']: [3]"]}

列表差异2：请注意，它没有考虑到订单

>>> # Note that it DOES NOT take order into account
... 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]}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff.changes, indent = 2)
    { }

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.changes, indent = 2)
    { 'dic_item_removed': ["root[4]['b'][2][2]"],
      'values_changed': ["root[4]['b'][2][1]: 1 ====>> 3"]}

Answer 3

这是一个非常简单的解决方案。它返回一个列表列表，其中包含不同元素的所有第一级和第二级字典键。希望这就是你想要的：）

a = {'home':{'name': 'Team1', 'score': 0}, 'away':{'name': 'Team2', 'score': 0}}
b = {'home':{'name': 'Team1', 'score': 2}, 'away':{'name': 'Team2', 'score': 0}}

diffs = []
for i in a:
    for j in a[i]:
        if a[i][j] != b[i][j]:
            diffs += [i, j]

print diffs

干杯！

Answer 4

我不确定这是否是你需要的，但是我发现了一个在python数据结构之间产生差异的库。它被称为datadiff，它可以让你了解两个字典之间的所有不同之处。我不确定是否会给你“商店”的差异，也许它会为整个“家”条目显示不同的价值。你必须尝试一下。

python中的Diff多维字典

4 个答案:

安装

使用示例