What's the cleanest way to compare 2 dictionary?

[John Machin]

John Henry wrote:
> Hi list,
>
> I am sure there are many ways of doing comparision but I like to see
> what you would do if you have 2 dictionary sets (containing lots of
> data - like 20000 keys and each key contains a dozen or so of records)
> and you want to build a list of differences about these two sets.
>
> I like to end up with 3 lists: what's in A and not in B, what's in B
> and not in A, and of course, what's in both A and B.
>
> What do you think is the cleanest way to do it?  (I am sure you will
> come up with ways that astonishes me  :=) )
>

Paddy has already pointed out a necessary addition to your requirement
definition: common keys with different values.

Here's another possible addition: you say that "each key contains a
dozen or so of records". I presume that you mean like this:

a = {1: ['rec1a', 'rec1b'], 42: ['rec42a', 'rec42b']} # "dozen" -> 2 to
save typing :-)

Now that happens if the other dictionary contains:

b = {1: ['rec1a', 'rec1b'], 42: ['rec42b', 'rec42a']}

Key 42 would be marked as different by Paddy's classification, but the
values are the same, just not in the same order. How do you want to
treat that? avalue == bvalue? sorted(avalue) == sorted(bvalue)? Oh, and
are you sure the buckets don't contain duplicates? Maybe you need
set(avalue) == set(bvalue). What about 'rec1a' vs 'Rec1a' vs 'REC1A'?

All comparisons are equal, but some comparisons are more equal than
others :-)

Cheers,
John