问题:如何以相同的顺序比较两个具有相同元素的JSON对象相等?

我如何测试python中两个JSON对象是否相等,而忽略列表的顺序?

例如 …

JSON文件a

{
    "errors": [
        {"error": "invalid", "field": "email"},
        {"error": "required", "field": "name"}
    ],
    "success": false
}

JSON文档b

{
    "success": false,
    "errors": [
        {"error": "required", "field": "name"},
        {"error": "invalid", "field": "email"}
    ]
}

a并且b应该比较相等,即使"errors"列表的顺序不同。

How can I test whether two JSON objects are equal in python, disregarding the order of lists?

For example …

JSON document a:

{
    "errors": [
        {"error": "invalid", "field": "email"},
        {"error": "required", "field": "name"}
    ],
    "success": false
}

JSON document b:

{
    "success": false,
    "errors": [
        {"error": "required", "field": "name"},
        {"error": "invalid", "field": "email"}
    ]
}

a and b should compare equal, even though the order of the "errors" lists are different.


回答 0

如果您希望两个具有相同元素但顺序不同的对象相等,那么比较明显的事情就是比较它们的排序后的副本-例如,以JSON字符串a和表示的字典b

import json

a = json.loads("""
{
    "errors": [
        {"error": "invalid", "field": "email"},
        {"error": "required", "field": "name"}
    ],
    "success": false
}
""")

b = json.loads("""
{
    "success": false,
    "errors": [
        {"error": "required", "field": "name"},
        {"error": "invalid", "field": "email"}
    ]
}
""")
>>> sorted(a.items()) == sorted(b.items())
False

…但这是行不通的,因为在每种情况下,"errors"顶层dict的项都是一个列表,其中相同元素的顺序不同,并且sorted()除“一个可迭代的。

为了解决这个问题,我们可以定义一个ordered函数,该函数将对找到的所有列表进行递归排序(并将字典转换(key, value)成对列表,以便它们可排序):

def ordered(obj):
    if isinstance(obj, dict):
        return sorted((k, ordered(v)) for k, v in obj.items())
    if isinstance(obj, list):
        return sorted(ordered(x) for x in obj)
    else:
        return obj

如果我们将此功能应用于ab,结果比较相等:

>>> ordered(a) == ordered(b)
True

If you want two objects with the same elements but in a different order to compare equal, then the obvious thing to do is compare sorted copies of them – for instance, for the dictionaries represented by your JSON strings a and b:

import json

a = json.loads("""
{
    "errors": [
        {"error": "invalid", "field": "email"},
        {"error": "required", "field": "name"}
    ],
    "success": false
}
""")

b = json.loads("""
{
    "success": false,
    "errors": [
        {"error": "required", "field": "name"},
        {"error": "invalid", "field": "email"}
    ]
}
""")
>>> sorted(a.items()) == sorted(b.items())
False

… but that doesn’t work, because in each case, the "errors" item of the top-level dict is a list with the same elements in a different order, and sorted() doesn’t try to sort anything except the “top” level of an iterable.

To fix that, we can define an ordered function which will recursively sort any lists it finds (and convert dictionaries to lists of (key, value) pairs so that they’re orderable):

def ordered(obj):
    if isinstance(obj, dict):
        return sorted((k, ordered(v)) for k, v in obj.items())
    if isinstance(obj, list):
        return sorted(ordered(x) for x in obj)
    else:
        return obj

If we apply this function to a and b, the results compare equal:

>>> ordered(a) == ordered(b)
True

回答 1

另一种方法是使用json.dumps(X, sort_keys=True)选项:

import json
a, b = json.dumps(a, sort_keys=True), json.dumps(b, sort_keys=True)
a == b # a normal string comparison

这适用于嵌套字典和列表。

Another way could be to use json.dumps(X, sort_keys=True) option:

import json
a, b = json.dumps(a, sort_keys=True), json.dumps(b, sort_keys=True)
a == b # a normal string comparison

This works for nested dictionaries and lists.


回答 2

对其进行解码,并将其作为mgilson注释进行比较。

字典的顺序无关紧要,只要键和值匹配即可。(字典在Python中没有顺序)

>>> {'a': 1, 'b': 2} == {'b': 2, 'a': 1}
True

但是顺序在清单中很重要。排序将解决列表的问题。

>>> [1, 2] == [2, 1]
False
>>> [1, 2] == sorted([2, 1])
True

>>> a = '{"errors": [{"error": "invalid", "field": "email"}, {"error": "required", "field": "name"}], "success": false}'
>>> b = '{"errors": [{"error": "required", "field": "name"}, {"error": "invalid", "field": "email"}], "success": false}'
>>> a, b = json.loads(a), json.loads(b)
>>> a['errors'].sort()
>>> b['errors'].sort()
>>> a == b
True

上面的示例适用于问题中的JSON。有关一般解决方案,请参见Zero Piraeus的答案。

Decode them and compare them as mgilson comment.

Order does not matter for dictionary as long as the keys, and values matches. (Dictionary has no order in Python)

>>> {'a': 1, 'b': 2} == {'b': 2, 'a': 1}
True

But order is important in list; sorting will solve the problem for the lists.

>>> [1, 2] == [2, 1]
False
>>> [1, 2] == sorted([2, 1])
True

>>> a = '{"errors": [{"error": "invalid", "field": "email"}, {"error": "required", "field": "name"}], "success": false}'
>>> b = '{"errors": [{"error": "required", "field": "name"}, {"error": "invalid", "field": "email"}], "success": false}'
>>> a, b = json.loads(a), json.loads(b)
>>> a['errors'].sort()
>>> b['errors'].sort()
>>> a == b
True

Above example will work for the JSON in the question. For general solution, see Zero Piraeus’s answer.


回答 3

对于以下两个字典“ dictWithListsInValue”和“ reorderedDictWithReorderedListsInValue”,它们只是彼此的重新排序版本

dictObj = {"foo": "bar", "john": "doe"}
reorderedDictObj = {"john": "doe", "foo": "bar"}
dictObj2 = {"abc": "def"}
dictWithListsInValue = {'A': [{'X': [dictObj2, dictObj]}, {'Y': 2}], 'B': dictObj2}
reorderedDictWithReorderedListsInValue = {'B': dictObj2, 'A': [{'Y': 2}, {'X': [reorderedDictObj, dictObj2]}]}
a = {"L": "M", "N": dictWithListsInValue}
b = {"L": "M", "N": reorderedDictWithReorderedListsInValue}

print(sorted(a.items()) == sorted(b.items()))  # gives false

给我错误的结果即错误。

所以我这样创建了自己的cutstom ObjectComparator:

def my_list_cmp(list1, list2):
    if (list1.__len__() != list2.__len__()):
        return False

    for l in list1:
        found = False
        for m in list2:
            res = my_obj_cmp(l, m)
            if (res):
                found = True
                break

        if (not found):
            return False

    return True


def my_obj_cmp(obj1, obj2):
    if isinstance(obj1, list):
        if (not isinstance(obj2, list)):
            return False
        return my_list_cmp(obj1, obj2)
    elif (isinstance(obj1, dict)):
        if (not isinstance(obj2, dict)):
            return False
        exp = set(obj2.keys()) == set(obj1.keys())
        if (not exp):
            # print(obj1.keys(), obj2.keys())
            return False
        for k in obj1.keys():
            val1 = obj1.get(k)
            val2 = obj2.get(k)
            if isinstance(val1, list):
                if (not my_list_cmp(val1, val2)):
                    return False
            elif isinstance(val1, dict):
                if (not my_obj_cmp(val1, val2)):
                    return False
            else:
                if val2 != val1:
                    return False
    else:
        return obj1 == obj2

    return True


dictObj = {"foo": "bar", "john": "doe"}
reorderedDictObj = {"john": "doe", "foo": "bar"}
dictObj2 = {"abc": "def"}
dictWithListsInValue = {'A': [{'X': [dictObj2, dictObj]}, {'Y': 2}], 'B': dictObj2}
reorderedDictWithReorderedListsInValue = {'B': dictObj2, 'A': [{'Y': 2}, {'X': [reorderedDictObj, dictObj2]}]}
a = {"L": "M", "N": dictWithListsInValue}
b = {"L": "M", "N": reorderedDictWithReorderedListsInValue}

print(my_obj_cmp(a, b))  # gives true

这给了我正确的预期输出!

逻辑很简单:

如果对象的类型为“列表”,则将第一个列表的每个项目与第二个列表的项目进行比较,直到找到为止;如果在通过第二个列表之后未找到该项目,则“找到”为= false。返回“找到的”值

否则,如果要比较的对象的类型为“ dict”,则比较两个对象中所有相应键的存在值。(执行递归比较)

否则,只需调用obj1 == obj2即可。默认情况下,它适用于字符串和数字的对象,并且eq()的定义适当。

(请注意,可以通过删除在object2中找到的项目来进一步改进该算法,以便object1的下一个项目不会将自身与object2中已经找到的项目进行比较。)

For the following two dicts ‘dictWithListsInValue’ and ‘reorderedDictWithReorderedListsInValue’ which are simply reordered versions of each other

dictObj = {"foo": "bar", "john": "doe"}
reorderedDictObj = {"john": "doe", "foo": "bar"}
dictObj2 = {"abc": "def"}
dictWithListsInValue = {'A': [{'X': [dictObj2, dictObj]}, {'Y': 2}], 'B': dictObj2}
reorderedDictWithReorderedListsInValue = {'B': dictObj2, 'A': [{'Y': 2}, {'X': [reorderedDictObj, dictObj2]}]}
a = {"L": "M", "N": dictWithListsInValue}
b = {"L": "M", "N": reorderedDictWithReorderedListsInValue}

print(sorted(a.items()) == sorted(b.items()))  # gives false

gave me wrong result i.e. false .

So I created my own cutstom ObjectComparator like this:

def my_list_cmp(list1, list2):
    if (list1.__len__() != list2.__len__()):
        return False

    for l in list1:
        found = False
        for m in list2:
            res = my_obj_cmp(l, m)
            if (res):
                found = True
                break

        if (not found):
            return False

    return True


def my_obj_cmp(obj1, obj2):
    if isinstance(obj1, list):
        if (not isinstance(obj2, list)):
            return False
        return my_list_cmp(obj1, obj2)
    elif (isinstance(obj1, dict)):
        if (not isinstance(obj2, dict)):
            return False
        exp = set(obj2.keys()) == set(obj1.keys())
        if (not exp):
            # print(obj1.keys(), obj2.keys())
            return False
        for k in obj1.keys():
            val1 = obj1.get(k)
            val2 = obj2.get(k)
            if isinstance(val1, list):
                if (not my_list_cmp(val1, val2)):
                    return False
            elif isinstance(val1, dict):
                if (not my_obj_cmp(val1, val2)):
                    return False
            else:
                if val2 != val1:
                    return False
    else:
        return obj1 == obj2

    return True


dictObj = {"foo": "bar", "john": "doe"}
reorderedDictObj = {"john": "doe", "foo": "bar"}
dictObj2 = {"abc": "def"}
dictWithListsInValue = {'A': [{'X': [dictObj2, dictObj]}, {'Y': 2}], 'B': dictObj2}
reorderedDictWithReorderedListsInValue = {'B': dictObj2, 'A': [{'Y': 2}, {'X': [reorderedDictObj, dictObj2]}]}
a = {"L": "M", "N": dictWithListsInValue}
b = {"L": "M", "N": reorderedDictWithReorderedListsInValue}

print(my_obj_cmp(a, b))  # gives true

which gave me the correct expected output!

Logic is pretty simple:

If the objects are of type ‘list’ then compare each item of the first list with the items of the second list until found , and if the item is not found after going through the second list , then ‘found’ would be = false. ‘found’ value is returned

Else if the objects to be compared are of type ‘dict’ then compare the values present for all the respective keys in both the objects. (Recursive comparison is performed)

Else simply call obj1 == obj2 . It by default works fine for the object of strings and numbers and for those eq() is defined appropriately .

(Note that the algorithm can further be improved by removing the items found in object2, so that the next item of object1 would not compare itself with the items already found in the object2)


回答 4

您可以编写自己的equals函数:

  • 在以下情况下,字典是相等的:1)所有键都相等,2)所有值都相等
  • 如果满足以下条件,则列表相等:所有项目均相同且顺序相同
  • 如果原语相等 a == b

因为您处理JSON,你就会有标准的Python类型:dictlist等等,所以你可以做硬类型检查if type(obj) == 'dict':,等等。

粗略示例(未经测试):

def json_equals(jsonA, jsonB):
    if type(jsonA) != type(jsonB):
        # not equal
        return False
    if type(jsonA) == dict:
        if len(jsonA) != len(jsonB):
            return False
        for keyA in jsonA:
            if keyA not in jsonB or not json_equal(jsonA[keyA], jsonB[keyA]):
                return False
    elif type(jsonA) == list:
        if len(jsonA) != len(jsonB):
            return False
        for itemA, itemB in zip(jsonA, jsonB):
            if not json_equal(itemA, itemB):
                return False
    else:
        return jsonA == jsonB

You can write your own equals function:

  • dicts are equal if: 1) all keys are equal, 2) all values are equal
  • lists are equal if: all items are equal and in the same order
  • primitives are equal if a == b

Because you’re dealing with json, you’ll have standard python types: dict, list, etc., so you can do hard type checking if type(obj) == 'dict':, etc.

Rough example (not tested):

def json_equals(jsonA, jsonB):
    if type(jsonA) != type(jsonB):
        # not equal
        return False
    if type(jsonA) == dict:
        if len(jsonA) != len(jsonB):
            return False
        for keyA in jsonA:
            if keyA not in jsonB or not json_equal(jsonA[keyA], jsonB[keyA]):
                return False
    elif type(jsonA) == list:
        if len(jsonA) != len(jsonB):
            return False
        for itemA, itemB in zip(jsonA, jsonB):
            if not json_equal(itemA, itemB):
                return False
    else:
        return jsonA == jsonB

回答 5

对于其他想要调试两个JSON对象(通常有一个引用和一个target)的人,可以使用以下解决方案。它将列出从目标到引用的不同/不匹配路径的“ 路径 ”。

level 选项用于选择您要研究的深度。

show_variables 可以打开该选项以显示相关变量。

def compareJson(example_json, target_json, level=-1, show_variables=False):
  _different_variables = _parseJSON(example_json, target_json, level=level, show_variables=show_variables)
  return len(_different_variables) == 0, _different_variables

def _parseJSON(reference, target, path=[], level=-1, show_variables=False):  
  if level > 0 and len(path) == level:
    return []
  
  _different_variables = list()
  # the case that the inputs is a dict (i.e. json dict)  
  if isinstance(reference, dict):
    for _key in reference:      
      _path = path+[_key]
      try:
        _different_variables += _parseJSON(reference[_key], target[_key], _path, level, show_variables)
      except KeyError:
        _record = ''.join(['[%s]'%str(p) for p in _path])
        if show_variables:
          _record += ': %s <--> MISSING!!'%str(reference[_key])
        _different_variables.append(_record)
  # the case that the inputs is a list/tuple
  elif isinstance(reference, list) or isinstance(reference, tuple):
    for index, v in enumerate(reference):
      _path = path+[index]
      try:
        _target_v = target[index]
        _different_variables += _parseJSON(v, _target_v, _path, level, show_variables)
      except IndexError:
        _record = ''.join(['[%s]'%str(p) for p in _path])
        if show_variables:
          _record += ': %s <--> MISSING!!'%str(v)
        _different_variables.append(_record)
  # the actual comparison about the value, if they are not the same, record it
  elif reference != target:
    _record = ''.join(['[%s]'%str(p) for p in path])
    if show_variables:
      _record += ': %s <--> %s'%(str(reference), str(target))
    _different_variables.append(_record)

  return _different_variables

For others who’d like to debug the two JSON objects (usually, there is a reference and a target), here is a solution you may use. It will list the “path” of different/mismatched ones from target to the reference.

level option is used for selecting how deep you would like to look into.

show_variables option can be turned on to show the relevant variable.

def compareJson(example_json, target_json, level=-1, show_variables=False):
  _different_variables = _parseJSON(example_json, target_json, level=level, show_variables=show_variables)
  return len(_different_variables) == 0, _different_variables

def _parseJSON(reference, target, path=[], level=-1, show_variables=False):  
  if level > 0 and len(path) == level:
    return []
  
  _different_variables = list()
  # the case that the inputs is a dict (i.e. json dict)  
  if isinstance(reference, dict):
    for _key in reference:      
      _path = path+[_key]
      try:
        _different_variables += _parseJSON(reference[_key], target[_key], _path, level, show_variables)
      except KeyError:
        _record = ''.join(['[%s]'%str(p) for p in _path])
        if show_variables:
          _record += ': %s <--> MISSING!!'%str(reference[_key])
        _different_variables.append(_record)
  # the case that the inputs is a list/tuple
  elif isinstance(reference, list) or isinstance(reference, tuple):
    for index, v in enumerate(reference):
      _path = path+[index]
      try:
        _target_v = target[index]
        _different_variables += _parseJSON(v, _target_v, _path, level, show_variables)
      except IndexError:
        _record = ''.join(['[%s]'%str(p) for p in _path])
        if show_variables:
          _record += ': %s <--> MISSING!!'%str(v)
        _different_variables.append(_record)
  # the actual comparison about the value, if they are not the same, record it
  elif reference != target:
    _record = ''.join(['[%s]'%str(p) for p in path])
    if show_variables:
      _record += ': %s <--> %s'%(str(reference), str(target))
    _different_variables.append(_record)

  return _different_variables

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