比較 Python 字典和嵌套字典

Question

我知道那里有幾個類似的問題，但我的問題對我來說完全不同且困難。 我有兩本詞典：

d1 = {'a': {'b': {'cs': 10}, 'd': {'cs': 20}}}
d2 = {'a': {'b': {'cs': 30}, 'd': {'cs': 20}}, 'newa': {'q': {'cs': 50}}}

即d1有鍵'a' ，而d2有鍵'a'和'newa' （換句話說， d1是我的舊字典， d2是我的新字典）。

我想迭代這些字典，如果鍵是相同的，檢查它的值（嵌套字典），例如當我在d2找到鍵'a'時，我將檢查是否有'b' ，如果是，檢查值'cs' （從10更改為30 ），如果此值更改，我想打印它。

另一種情況是，我想從d2獲取密鑰'newa'作為新添加的密鑰。

因此，在遍歷這 2 個 dicts 之后，這是預期的輸出：

"d2" has new key "newa"
Value of "cs" is changed from 10 to 30 of key "b" which is of key "a"

我有以下代碼，我正在嘗試許多無法正常工作的循環，但也不是一個好的選擇，因此我正在尋找是否可以通過遞歸代碼獲得預期的輸出。

for k, v in d1.iteritems():
    for k1, v1 in d2.iteritems():
        if k is k1:
            print k
            for k2 in v:
                for k3 in v1:
                    if k2 is k3:
                        print k2, "sub key matched"

        else:
            print "sorry no match found"

Answer 1

使用遞歸比較 2 個字典：

為 python 3 編輯（也適用於 python 2）：

d1= {'a':{'b':{'cs':10},'d':{'cs':20}}}
d2= {'a':{'b':{'cs':30} ,'d':{'cs':20}},'newa':{'q':{'cs':50}}}

def findDiff(d1, d2, path=""):
    for k in d1:
        if k in d2:
            if type(d1[k]) is dict:
                findDiff(d1[k],d2[k], "%s -> %s" % (path, k) if path else k)
            if d1[k] != d2[k]:
                result = [ "%s: " % path, " - %s : %s" % (k, d1[k]) , " + %s : %s" % (k, d2[k])]
                print("\n".join(result))
        else:
            print ("%s%s as key not in d2\n" % ("%s: " % path if path else "", k))

print("comparing d1 to d2:")
findDiff(d1,d2)
print("comparing d2 to d1:")
findDiff(d2,d1)

Python 2 舊答案：

def findDiff(d1, d2, path=""):
    for k in d1:
        if (k not in d2):
            print (path, ":")
            print (k + " as key not in d2", "\n")
        else:
            if type(d1[k]) is dict:
                if path == "":
                    path = k
                else:
                    path = path + "->" + k
                findDiff(d1[k],d2[k], path)
            else:
                if d1[k] != d2[k]:
                    print (path, ":")
                    print (" - ", k," : ", d1[k])
                    print (" + ", k," : ", d2[k])

輸出：

comparing d1 to d2:
a -> b: 
 - cs : 10
 + cs : 30
comparing d2 to d1:
a -> b: 
 - cs : 30
 + cs : 10

Answer 2

修改 user3 的代碼以使其更好

d1= {'as': 1, 'a':
        {'b':
            {'cs':10,
             'qqq': {'qwe':1}
            },
            'd': {'csd':30}
        }
    }
d2= {'as': 3, 'a':
        {'b':
            {'cs':30,
             'qqq': 123
            },
            'd':{'csd':20}
        },
        'newa':
        {'q':
            {'cs':50}
        }
    }

def compare_dictionaries(dict_1, dict_2, dict_1_name, dict_2_name, path=""):
    """Compare two dictionaries recursively to find non mathcing elements

    Args:
        dict_1: dictionary 1
        dict_2: dictionary 2

    Returns:

    """
    err = ''
    key_err = ''
    value_err = ''
    old_path = path
    for k in dict_1.keys():
        path = old_path + "[%s]" % k
        if not dict_2.has_key(k):
            key_err += "Key %s%s not in %s\n" % (dict_2_name, path, dict_2_name)
        else:
            if isinstance(dict_1[k], dict) and isinstance(dict_2[k], dict):
                err += compare_dictionaries(dict_1[k],dict_2[k],'d1','d2', path)
            else:
                if dict_1[k] != dict_2[k]:
                    value_err += "Value of %s%s (%s) not same as %s%s (%s)\n"\
                        % (dict_1_name, path, dict_1[k], dict_2_name, path, dict_2[k])

    for k in dict_2.keys():
        path = old_path + "[%s]" % k
        if not dict_1.has_key(k):
            key_err += "Key %s%s not in %s\n" % (dict_2_name, path, dict_1_name)

    return key_err + value_err + err


a = compare_dictionaries(d1,d2,'d1','d2')
print a

輸出：

Key d2[newa] not in d1
Value of d1[as] (1) not same as d2[as] (3)
Value of d1[a][b][cs] (10) not same as d2[a][b][cs] (30)
Value of d1[a][b][qqq] ({'qwe': 1}) not same as d2[a][b][qqq] (123)
Value of d1[a][d][csd] (30) not same as d2[a][d][csd] (20)

Answer 3

這應該提供您需要的有用功能：

對於 Python 2.7

def isDict(obj):
    return obj.__class__.__name__ == 'dict'

def containsKeyRec(vKey, vDict):
    for curKey in vDict:
        if curKey == vKey or (isDict(vDict[curKey]) and containsKeyRec(vKey, vDict[curKey])):
            return True
    return False

def getValueRec(vKey, vDict):
    for curKey in vDict:
        if curKey == vKey:
            return vDict[curKey]
        elif isDict(vDict[curKey]) and getValueRec(vKey, vDict[curKey]):
            return containsKeyRec(vKey, vDict[curKey])
    return None

d1= {'a':{'b':{'cs':10},'d':{'cs':20}}}
d2= {'a':{'b':{'cs':30} ,'d':{'cs':20}},'newa':{'q':{'cs':50}}}

for key in d1:
    if containsKeyRec(key, d2):
        print "dict d2 contains key: " + key
        d2Value = getValueRec(key, d2)
        if d1[key] == d2Value:
            print "values are equal, d1: " + str(d1[key]) + ", d2: " + str(d2Value)
        else:
            print "values are not equal, d1: " + str(d1[key]) + ", d2: " + str(d2Value)

    else:
        print "dict d2 does not contain key: " + key

對於 Python 3（或更高版本）：

def id_dict(obj):
    return obj.__class__.__name__ == 'dict'


def contains_key_rec(v_key, v_dict):
    for curKey in v_dict:
        if curKey == v_key or (id_dict(v_dict[curKey]) and contains_key_rec(v_key, v_dict[curKey])):
            return True
    return False


def get_value_rec(v_key, v_dict):
    for curKey in v_dict:
        if curKey == v_key:
            return v_dict[curKey]
        elif id_dict(v_dict[curKey]) and get_value_rec(v_key, v_dict[curKey]):
            return contains_key_rec(v_key, v_dict[curKey])
    return None


d1 = {'a': {'b': {'cs': 10}, 'd': {'cs': 20}}}
d2 = {'a': {'b': {'cs': 30}, 'd': {'cs': 20}}, 'newa': {'q': {'cs': 50}}}

for key in d1:
if contains_key_rec(key, d2):
    d2_value = get_value_rec(key, d2)
    if d1[key] == d2_value:
        print("values are equal, d1: " + str(d1[key]) + ", d2: " + str(d2_value))
        pass
    else:
        print("values are not equal:\n"
              "list1: " + str(d1[key]) + "\n" +
              "list2: " + str(d2_value))

else:
    print("dict d2 does not contain key: " + key)

Answer 4

為什么不使用 deepdiff 庫。

見： https : //github.com/seperman/deepdiff

>>> from deepdiff import DeepDiff
>>> t1 = {1:1, 3:3, 4:4}
>>> t2 = {1:1, 3:3, 5:5, 6:6}
>>> ddiff = DeepDiff(t1, t2)
>>> print(ddiff)
{'dictionary_item_added': {'root[5]', 'root[6]'}, 'dictionary_item_removed': {'root[4]'}}

當然它更強大，查看文檔了解更多。

Answer 5

對於 python 3 或更高版本，用於比較任何數據的代碼。

def do_compare(data1, data2, data1_name, data2_name, path=""):
    if operator.eq(data1, data2) and not path:
        log.info("Both data have same content")
    else:
        if isinstance(data1, dict) and isinstance(data2, dict):
            compare_dict(data1, data2, data1_name, data2_name, path)
        elif isinstance(data1, list) and isinstance(data2, list):
            compare_list(data1, data2, data1_name, data2_name, path)
        else:
            if data1 != data2:
                value_err = "Value of %s%s (%s) not same as %s%s (%s)\n"\
                            % (data1_name, path, data1, data2_name, path, data2)
                print (value_err)
        # findDiff(data1, data2)

def compare_dict(data1, data2, data1_name, data2_name, path):
    old_path = path
    for k in data1.keys():
        path = old_path + "[%s]" % k
        if k not in data2:
            key_err = "Key %s%s not in %s\n" % (data1_name, path, data2_name)
            print (key_err)
        else:
            do_compare(data1[k], data2[k], data1_name, data2_name, path)
    for k in data2.keys():
        path = old_path + "[%s]" % k
        if k not in data1:
            key_err = "Key %s%s not in %s\n" % (data2_name, path, data1_name)
            print (key_err)

def compare_list(data1, data2, data1_name, data2_name, path):
    data1_length = len(data1)
    data2_length = len(data2)
    old_path = path
    if data1_length != data2_length:
        value_err = "No: of items in %s%s (%s) not same as %s%s (%s)\n"\
                            % (data1_name, path, data1_length, data2_name, path, data2_length)
        print (value_err)
    for index, item in enumerate(data1):
        path = old_path + "[%s]" % index
        try:
            do_compare(data1[index], data2[index], data1_name, data2_name, path)
        except IndexError:
            pass

Answer 6

添加一個添加更多功能的版本：

• 可以比較任意嵌套的類似 JSON 的字典和列表
• 允許您指定要忽略的鍵（例如在片狀單元測試中）
• 允許您指定帶有數值的鍵，只要它們在彼此的特定百分比范圍內，這些鍵就會被視為相等

如果您如下定義deep_diff函數並在@rkatkam 的示例中調用它，您將得到：

>>> deep_diff(d1, d2)

{'newa': (None, {'q': {'cs': 50}}), 'a': {'b': {'cs': (10, 30)}}}

下面是函數定義：

def deep_diff(x, y, parent_key=None, exclude_keys=[], epsilon_keys=[]):
    """
    Take the deep diff of JSON-like dictionaries

    No warranties when keys, or values are None

    """
    # pylint: disable=unidiomatic-typecheck

    EPSILON = 0.5
    rho = 1 - EPSILON

    if x == y:
        return None

    if parent_key in epsilon_keys:
        xfl, yfl = float_or_None(x), float_or_None(y)
        if xfl and yfl and xfl * yfl >= 0 and rho * xfl <= yfl and rho * yfl <= xfl:
            return None

    if not (isinstance(x, (list, dict)) and (isinstance(x, type(y)) or isinstance(y, type(x)))):
        return x, y

    if isinstance(x, dict):
        d = type(x)()  # handles OrderedDict's as well
        for k in x.keys() ^ y.keys():
            if k in exclude_keys:
                continue
            if k in x:
                d[k] = (deepcopy(x[k]), None)
            else:
                d[k] = (None, deepcopy(y[k]))

        for k in x.keys() & y.keys():
            if k in exclude_keys:
                continue

            next_d = deep_diff(
                x[k], y[k], parent_key=k, exclude_keys=exclude_keys, epsilon_keys=epsilon_keys
            )
            if next_d is None:
                continue

            d[k] = next_d

        return d if d else None

    # assume a list:
    d = [None] * max(len(x), len(y))
    flipped = False
    if len(x) > len(y):
        flipped = True
        x, y = y, x

    for i, x_val in enumerate(x):
        d[i] = (
            deep_diff(
                y[i], x_val, parent_key=i, exclude_keys=exclude_keys, epsilon_keys=epsilon_keys
            )
            if flipped
            else deep_diff(
                x_val, y[i], parent_key=i, exclude_keys=exclude_keys, epsilon_keys=epsilon_keys
            )
        )

    for i in range(len(x), len(y)):
        d[i] = (y[i], None) if flipped else (None, y[i])

    return None if all(map(lambda x: x is None, d)) else d

Answer 7

添加非遞歸解決方案。

  # Non Recursively traverses through a large nested dictionary
  # Uses a queue of dicts_to_process to keep track of what needs to be traversed rather than using recursion.
  # Slightly more complex than the recursive version, but arguably better as there is no risk of stack overflow from
  # too many levels of recursion
  def get_dict_diff_non_recursive(dict1, dict2):
      dicts_to_process=[(dict1,dict2,"")]
      while dicts_to_process:
          d1,d2,current_path = dicts_to_process.pop()
          for key in d1.keys():
              current_path = os.path.join(current_path, f"{key}")
              #print(f"searching path {current_path}")
              if key not in d2 or d1[key] != d2[key]:
                  print(f"difference at {current_path}")
              if type(d1[key]) == dict:
                  dicts_to_process.append((d1[key],d2[key],current_path))
              elif type(d1[key]) == list and d1[key] and type(d1[key][0]) == dict:
                  for i in range(len(d1[key])):
                      dicts_to_process.append((d1[key][i], d2[key][i],current_path))

Answer 8

我不喜歡我在許多線程中找到的許多答案......他們中的很多人建議使用非常強大的deepdiff不要誤會我，但它只是沒有給我我想要的輸出，而不僅僅是一個字符串的差異，或者一個新構建的看起來很奇怪的字典，其中包含從原始鍵的嵌套鍵中收集的新鍵......但實際上返回一個帶有原始鍵和增量值的真實字典。

我的用例是發送較小的有效負載，如果 MQTT 網絡沒有區別，則不發送。

我發現的解決方案部分從此鏈接中被盜，但對其進行了修改以僅提供增量。 然后我遞歸地解析它，如果它嵌套來構建最終的 diff 字典，則再次調用diff_dict() 。 結果證明它比那里的許多示例簡單得多。 僅供參考，它不關心排序。

我的解決方案：

def diff_dict(d1, d2):
    d1_keys = set(d1.keys())
    d2_keys = set(d2.keys())
    shared_keys = d1_keys.intersection(d2_keys)
    shared_deltas = {o: (d1[o], d2[o]) for o in shared_keys if d1[o] != d2[o]}
    added_keys = d2_keys - d1_keys
    added_deltas = {o: (None, d2[o]) for o in added_keys}
    deltas = {**shared_deltas, **added_deltas}
    return parse_deltas(deltas)


def parse_deltas(deltas: dict):
    res = {}
    for k, v in deltas.items():
        if isinstance(v[0], dict):
            tmp = diff_dict(v[0], v[1])
            if tmp:
                res[k] = tmp
        else:
            res[k] = v[1]
    return res

例子：

original = {
    'int': 1,
    'float': 0.1000,
    'string': 'some string',
    'bool': True,
    'nested1': {
        'int': 2,
        'float': 0.2000,
        'string': 'some string2',
        'bool': True,
        'nested2': {
            'string': 'some string3'
        }
    }
}
new = {
    'int': 2,
    'string': 'some string',
    'nested1': {
        'int': 2,
        'float': 0.5000,
        'string': 'new string',
        'bool': False,
        'nested2': {
            'string': 'new string nested 2 time'
        }
    },
    'test_added': 'added_val'
}

print(diff_dict(original, new))

輸出：

{'int': 2, 'nested1': {'string': 'new string', 'nested2': {'string': 'new string nested 2 time'}, 'bool': False, 'float': 0.5}, 'test_added': 'added_val'}