dictionary - Python: find difference in dicts -

dicta = {'a':1, 'b':2, 'c':3} dictb = {'a':2, 'b':2, 'c':4}  if dicta == dictb:     print "dicts same" else:     # print diffs     key in dicta.iterkeys():         try:             if dicta[key] != dictb[key]:                 print "different value key: %s" % key         except keyerror:             print "dictb has no key: %s" % key 

this gets inefficient if number of items in dicta , dictb huge

any faster way this?

i thinking somehow using sets not sure.

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this might duplicate seems people iterating in answers in other similar questions

you can use dict view objetcs:

keys views set-like since entries unique , hashable. if values hashable, (key, value) pairs unique , hashable, items view set-like. (values views not treated set-like since entries not unique.) these set operations available (“other” refers either view or set):

dictview & other return intersection of dictview , other object new set.  dictview | other return union of dictview , other object new set.  dictview - other return difference between dictview , other object (all elements in dictview aren’t in other) new set.  dictview ^ other return symmetric difference (all elements either in dictview or other, not in both) of dictview , other object new set.  diff = dicta.viewkeys() - dictb.viewkeys()  print(diff)    set([])  print(dicta.viewitems() - dictb.viewitems()) set([('a', 1), ('c', 3)]) 

or sets:

print(set(dicta.iteritems()).difference(dictb.iteritems())) 

the restriction have memory