Python Difference between 2 lists

You can do something like this but it's not that pretty:

def sub(d1, d2):
    d2_ = d2.copy()
    res = []
    for elem in d1:
        if elem in d2_:
            d2_.remove(elem)
        else:
            res.append(elem)
    return res

Then you can use sub(d1, d2) + sub(d2, d1) to get diff(d1, d2).

Assuming your values, not just the keys, in each dictionary are hashable, you can make frozensets of each dicts .items(), collect them into a collections.Counter, and perform an xor-like operation on the Counters to get the disparate counts. Then you just flatten out the Counter again, and convert the frozensets back to dicts:

from collections import Counter

def diff(l1, l2):
    c1 = Counter(frozenset(d.items()) for d in l1)
    c2 = Counter(frozenset(d.items()) for d in l2)
    c1xc2 = (c1 - c2) + (c2 - c1)  # Equivalent to c1 ^ c2 if they were sets, but we need counts, and Counter doesn't support ^
    # elements() flattens the Counter to iterable of repeated values, map(dict, ...)
    # converts frozensets back to dicts
    return list(map(dict, c1xc2.elements()))

The simpler operation of computing d1 - d2 or d2 - d1 is left as an exercise; the key in all cases is converting your list of dicts to Counters of frozensets of the dict's .items(), performing the operation you care about with -, then converting back to a list of dicts.

This should be fairly efficient (roughly O(n) on combined input sizes); improving on it would require a lot more custom code that's frankly not worth the bother.

Not elegant, but this seems to work:

def diff(d1, d2):
  found = []
  to_return = []
  for item in d1:
    if item in found:
      continue
    found += [item]
    to_return += [item] * (len(list(filter(lambda x : x == item, d1))) - len(list(filter(lambda x : x == item, d2))))
  return to_return