Comparing Python dictionaries and nested dictionaries

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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

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