问题
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\"
回答1:
comparing 2 dictionaries using recursion:
Edited for python 3:
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 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])
print ("comparing d1 to d2:")
print (findDiff(d1,d2))
print ("comparing d2 to d1:")
print (findDiff(d2,d1))
python 2 code:
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.keys():
if not d2.has_key(k):
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]
print "comparing d1 to d2:"
print findDiff(d1,d2)
print "comparing d2 to d1:"
print findDiff(d2,d1)
Output:
comparing d1 to d2:
a->b :
- cs : 10
+ cs : 30
None
comparing d2 to d1:
a->b :
- cs : 30
+ cs : 10
a :
newa as key not in d2
None
回答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)
回答3:
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)
回答4:
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
来源：https://stackoverflow.com/questions/27265939/comparing-python-dictionaries-and-nested-dictionaries