由于似乎没有人提到deepdiff，为了完整起见，我将在这里添加它。我发现它非常方便获得(嵌套)对象的差异:

安装

pip install deepdiff

示例代码

import deepdiff
import json

dict_1 = {
    "a": 1,
    "nested": {
        "b": 1,
    }
}

dict_2 = {
    "a": 2,
    "nested": {
        "b": 2,
    }
}

diff = deepdiff.DeepDiff(dict_1, dict_2)
print(json.dumps(diff, indent=4))

输出

{
    "values_changed": {
        "root['a']": {
            "new_value": 2,
            "old_value": 1
        },
        "root['nested']['b']": {
            "new_value": 2,
            "old_value": 1
        }
    }
}

注意关于检查结果的漂亮打印:如果两个字典具有相同的属性键(可能与示例中的属性值不同)，上面的代码就可以工作。但是，如果存在一个“extra”属性是字典之一，json.dumps()将失败

TypeError: Object of type PrettyOrderedSet is not JSON serializable

解决方案:使用diff.to_json()和json.loads() / json.dumps()来漂亮地打印:

import deepdiff
import json

dict_1 = {
    "a": 1,
    "nested": {
        "b": 1,
    },
    "extra": 3
}

dict_2 = {
    "a": 2,
    "nested": {
        "b": 2,
    }
}

diff = deepdiff.DeepDiff(dict_1, dict_2)
print(json.dumps(json.loads(diff.to_json()), indent=4))  

输出:

{
    "dictionary_item_removed": [
        "root['extra']"
    ],
    "values_changed": {
        "root['a']": {
            "new_value": 2,
            "old_value": 1
        },
        "root['nested']['b']": {
            "new_value": 2,
            "old_value": 1
        }
    }
}

替代方案:使用pprint，结果是不同的格式:

import pprint

# same code as above

pprint.pprint(diff, indent=4)

输出:

{   'dictionary_item_removed': [root['extra']],
    'values_changed': {   "root['a']": {   'new_value': 2,
                                           'old_value': 1},
                          "root['nested']['b']": {   'new_value': 2,
                                                     'old_value': 1}}}

下面的代码将帮助您比较python中的dict列表

def compate_generic_types(object1, object2):
    if isinstance(object1, str) and isinstance(object2, str):
        return object1 == object2
    elif isinstance(object1, unicode) and isinstance(object2, unicode):
        return object1 == object2
    elif isinstance(object1, bool) and isinstance(object2, bool):
        return object1 == object2
    elif isinstance(object1, int) and isinstance(object2, int):
        return object1 == object2
    elif isinstance(object1, float) and isinstance(object2, float):
        return object1 == object2
    elif isinstance(object1, float) and isinstance(object2, int):
        return object1 == float(object2)
    elif isinstance(object1, int) and isinstance(object2, float):
        return float(object1) == object2

    return True

def deep_list_compare(object1, object2):
    retval = True
    count = len(object1)
    object1 = sorted(object1)
    object2 = sorted(object2)
    for x in range(count):
        if isinstance(object1[x], dict) and isinstance(object2[x], dict):
            retval = deep_dict_compare(object1[x], object2[x])
            if retval is False:
                print "Unable to match [{0}] element in list".format(x)
                return False
        elif isinstance(object1[x], list) and isinstance(object2[x], list):
            retval = deep_list_compare(object1[x], object2[x])
            if retval is False:
                print "Unable to match [{0}] element in list".format(x)
                return False
        else:
            retval = compate_generic_types(object1[x], object2[x])
            if retval is False:
                print "Unable to match [{0}] element in list".format(x)
                return False

    return retval

def deep_dict_compare(object1, object2):
    retval = True

    if len(object1) != len(object2):
        return False

    for k in object1.iterkeys():
        obj1 = object1[k]
        obj2 = object2[k]
        if isinstance(obj1, list) and isinstance(obj2, list):
            retval = deep_list_compare(obj1, obj2)
            if retval is False:
                print "Unable to match [{0}]".format(k)
                return False

        elif isinstance(obj1, dict) and isinstance(obj2, dict):
            retval = deep_dict_compare(obj1, obj2)
            if retval is False:
                print "Unable to match [{0}]".format(k)
                return False
        else:
            retval = compate_generic_types(obj1, obj2)
            if retval is False:
                print "Unable to match [{0}]".format(k)
                return False

    return retval

如果你想知道在两个字典中有多少值是匹配的，你应该这样说:)

也许是这样的:

shared_items = {k: x[k] for k in x if k in y and x[k] == y[k]}
print(len(shared_items))

为什么不只是遍历一个字典，并在过程中检查另一个字典(假设两个字典都有相同的键)?

x = dict(a=1, b=2)
y = dict(a=2, b=2)

for key, val in x.items():
    if val == y[key]:
        print ('Ok', val, y[key])
    else:
        print ('Not', val, y[key])

输出:

Not 1 2
Ok 2 2

功能很好IMO，清晰直观。但为了给你(另一个)答案，我是这么说的:

def compare_dict(dict1, dict2):
    for x1 in dict1.keys():
        z = dict1.get(x1) == dict2.get(x1)
        if not z:
            print('key', x1)
            print('value A', dict1.get(x1), '\nvalue B', dict2.get(x1))
            print('-----\n')

可能对你或任何人都有用。

编辑:

我已经创建了一个递归版本的上面..在其他答案中没有看到吗

def compare_dict(a, b):
    # Compared two dictionaries..
    # Posts things that are not equal..
    res_compare = []
    for k in set(list(a.keys()) + list(b.keys())):
        if isinstance(a[k], dict):
            z0 = compare_dict(a[k], b[k])
        else:
            z0 = a[k] == b[k]

        z0_bool = np.all(z0)
        res_compare.append(z0_bool)
        if not z0_bool:
            print(k, a[k], b[k])
    return np.all(res_compare)

你要做的就是x==y

你这样做不是一个好主意，因为字典里的条目不应该有任何顺序。你可能会比较[('a'，1)，('b'，1)]和[('b'，1)， ('a'，1)](相同的字典，不同的顺序)。

例如，看这个:

>>> x = dict(a=2, b=2,c=3, d=4)
>>> x
{'a': 2, 'c': 3, 'b': 2, 'd': 4}
>>> y = dict(b=2,c=3, d=4)
>>> y
{'c': 3, 'b': 2, 'd': 4}
>>> zip(x.iteritems(), y.iteritems())
[(('a', 2), ('c', 3)), (('c', 3), ('b', 2)), (('b', 2), ('d', 4))]

虽然只有一项不同，但你的算法会发现所有项都是不同的