Is there a better way to compare dictionary values

Question

I am currently using the following function to compare dictionary values. Is there a faster or better way to do it?

match = True
for keys in dict1:
    if dict1[keys] != dict2[keys]:
        match = False
        print keys
        print dict1[keys],
        print  '->' ,
        print dict2[keys]

Edit: Both the dicts contain the same keys.

Answer 1

If you're just comparing for equality, you can just do this:

if not dict1 == dict2:
    match = False

Otherwise, the only major problem I see is that you're going to get a KeyError if there is a key in dict1 that is not in dict2, so you may want to do something like this:

for key in dict1:
    if not key in dict2 or dict1[key] != dict2[key]:
        match = False

You could compress this into a comprehension to just get the list of keys that don't match too:

mismatch_keys = [key for key in x if not key in y or x[key] != y[key]]
match = not bool(mismatch_keys) #If the list is not empty, they don't match 
for key in mismatch_keys:
    print key
    print '%s -> %s' % (dict1[key],dict2[key])

The only other optimization I can think of might be to use "len(dict)" to figure out which dict has fewer entries and loop through that one first to have the shortest loop possible.

Answer 2

>>> a = {'x': 1, 'y': 2}
>>> b = {'y': 2, 'x': 1}
>>> print a == b
True
>>> c = {'z': 1}
>>> print a == c
False
>>>

Answer 3

Uhm, you are describing dict1 == dict2 ( check if boths dicts are equal )

But what your code does is all( dict1[k]==dict2[k] for k in dict1 ) ( check if all entries in dict1 are equal to those in dict2 )

Answer 4

If the dicts have identical sets of keys and you need all those prints for any value difference, there isn't much you can do; maybe something like:

diffkeys = [k for k in dict1 if dict1[k] != dict2[k]]
for k in diffkeys:
  print k, ':', dict1[k], '->', dict2[k]

pretty much equivalent to what you have, but you might get nicer presentation for example by sorting diffkeys before you loop on it.

Answer 5

You can use sets for this too

>>> a = {'x': 1, 'y': 2}
>>> b = {'y': 2, 'x': 1}
>>> set(a.iteritems())-set(b.iteritems())
set([])
>>> a['y']=3
>>> set(a.iteritems())-set(b.iteritems())
set([('y', 3)])
>>> set(b.iteritems())-set(a.iteritems())
set([('y', 2)])
>>> set(b.iteritems())^set(a.iteritems())
set([('y', 3), ('y', 2)])