Comparing Python dictionaries and nested dictionaries

39,011

Solution 1

comparing 2 dictionaries using recursion:

Edited for python 3 (works for python 2 as well):

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 old answer:

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])

Output:

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

Solution 2

Modified user3's code to make it even better

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

Output:

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)

Solution 3

why not use deepdiff library.

see it at: 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]'}}

of course it is more powerful, check the doc for more.

Solution 4

This should provide what you need with helpful functions:

For 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

For Python 3 (or higher):

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)

Solution 5

For python 3 or higher, Code for comparing any data.

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
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rkatkam
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rkatkam

A firm believer of "Learn by Doing"! love playing with python, mongodb and definitely only on linux ;)

Updated on July 05, 2022

Comments

  • rkatkam
    rkatkam almost 2 years

    I know there are several similar questions out there, but my question is quite different and difficult for me. I have two dictionaries:

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

    i.e. d1 has key 'a', and d2 has keys 'a' and 'newa' (in other words d1 is my old dict and d2 is my new dict).

    I want to iterate over these dictionaries such that, if the key is same check for its value (nested dict), e.g. when I find key 'a' in d2, I will check whether there is 'b', if yes check value of 'cs' (changed from 10 to 30), if this value is changed I want to print it.

    Another case is, I want to get key 'newa' from d2 as the newly added key.

    Hence, after iterating through these 2 dicts, this is the expected output:

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

    I have the following code with me, I am trying with many loops which are not working though, but is not a good option too, hence I am looking to find whether I can get expected output with a recursive piece of code.

    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"