Question

Update:

Hello again. My question is, how can I compare values of an dictionary for equality. More Informationen about my Dictionary:

• keys are session numbers

• values of each key are nested lists -> f.e.

  [[1,0],[2,0],[3,1]]

• the length of values for each key arent the same, so it could be that session number 1 have more values then session number 2

• here an example dictionary:

order_session =
{1:[[100,0],[22,1],[23,2]],10:[100,0],[232,0],[10,2],[11,2]],22:[[5,2],[23,2],….],
… }

My Goal:

Step 1: to compare the values of session number 1 with the values of the whole other session numbers in the dictionary for equality

Step 2: take the next session number and compare the values with the other values of the other session numbers, and so on
– finally we have each session number value compared

Step 3: save the result into a list f.e.
output = [[100,0],[23,2], … ] or output = [(100,0),(23,2), … ]

• if you can see a value-pair [100,0] of session 1 and 10 are the same. also the value-pair [23,2] of session 1 and 22 are the same.

Thanks for helping me out.

Update 2

Thank you for all your help and tips to change the nested list of lists into list of tuples, which are quite better to handle it.

I prefer Boaz Yaniv solution 😉
I also like the use of collections.Counter() … unlucky that I use 2.6.4 (Counter works at 2.7) maybe I change to 2.7 sometimes.

Answer 1

If your dictionary is long, you’d want to use sets, for better performance (looking up already-encountered values in lists is going to be quite slow):

def get_repeated_values(sessions):
    known = set()
    already_repeated = set()
    for lst in sessions.itervalues():
        session_set = set(tuple(x) for x in lst)
        repeated = (known & session_set) - already_repeated
        already_repeated |= repeated
        known |= session_set
        for val in repeated:
            yield val

sessions = {1:[[100,0],[22,1],[23,2]],10:[[100,0],[232,0],[10,2],[11,2]],22:[[5,2],[23,2]]}
for x in get_repeated_values(sessions):
    print x

I also suggest (again, for performance reasons) to nest tuples inside your lists instead of lists, if you’re not going to change them on-the-fly. The code I posted here will work either way, but it would be faster if the values are already tuples.