Comparing Python dictionaries and nested dictionaries
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
rkatkam
A firm believer of "Learn by Doing"! love playing with python, mongodb and definitely only on linux ;)
Updated on July 05, 2022Comments
-
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'
, andd2
has keys'a'
and'newa'
(in other wordsd1
is my old dict andd2
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'
ind2
, I will check whether there is'b'
, if yes check value of'cs'
(changed from10
to30
), if this value is changed I want to print it.Another case is, I want to get key
'newa'
fromd2
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"