Comparing 2 lists consisting of dictionaries with unique keys in python

Question

I have 2 lists, both of which contain same number of dictionaries. Each dictionary has a unique key. There is a match for each dictionary of the first list in the second list, that is a dictionary with a unique key exists in the other list. But the other elements of such 2 dictionaries may vary. For example:

list_1 = [             {                 'unique_id': '001',                 'key1': 'AAA',                 'key2': 'BBB',                 'key3': 'EEE'              },              {                 'unique_id': '002',                 'key1': 'AAA',                 'key2': 'CCC',                 'key3': 'FFF'              }          ]   list_2 = [              {                 'unique_id': '001',                 'key1': 'AAA',                 'key2': 'DDD',                 'key3': 'EEE'              },              {                 'unique_id': '002',                 'key1': 'AAA',                 'key2': 'CCC',                 'key3': 'FFF'              }          ] 

I want to compare all elements of 2 matching dictionaries. If any of the elements are not equal, I want to print the none-equal elements.

Would you please help?

Answer 1

Assuming that the dicts line up like in your example input, you can use the zip() function to get a list of associated pairs of dicts, then you can use any() to check if there is a difference:

>>> list_1 = [{'unique_id':'001', 'key1':'AAA', 'key2':'BBB', 'key3':'EEE'},                {'unique_id':'002', 'key1':'AAA', 'key2':'CCC', 'key3':'FFF'}] >>> list_2 = [{'unique_id':'001', 'key1':'AAA', 'key2':'DDD', 'key3':'EEE'},               {'unique_id':'002', 'key1':'AAA', 'key2':'CCC', 'key3':'FFF'}] >>> pairs = zip(list_1, list_2) >>> any(x != y for x, y in pairs) True 

Or to get the differing pairs:

>>> [(x, y) for x, y in pairs if x != y] [({'key3': 'EEE', 'key2': 'BBB', 'key1': 'AAA', 'unique_id': '001'}, {'key3': 'EEE', 'key2': 'DDD', 'key1': 'AAA', 'unique_id': '001'})] 

You can even get the keys which don't match for each pair:

>>> [[k for k in x if x[k] != y[k]] for x, y in pairs if x != y] [['key2']] 

Possibly together with the associated values:

>>> [[(k, x[k], y[k]) for k in x if x[k] != y[k]] for x, y in pairs if x != y] [[('key2', 'BBB', 'DDD')]] 

NOTE: In case you're input lists are not sorted yet, you can do that easily as well:

>>> from operator import itemgetter >>> list_1, list_2 = [sorted(l, key=itemgetter('unique_id'))                        for l in (list_1, list_2)] 

Answer 2

The fastest and most comprehensive way would be, to use two sets of tuples:

set_list1 = set(tuple(sorted(d.items())) for d in sorted(list1)) set_list2 = set(tuple(sorted(d.items())) for d in sorted(list2))      

(if your list is already sorted, simply remove the list sort to save performance)

Find overlapping using intersection:

set_overlapping = set_list1.intersection(set_list2) 

Find difference using symmetric_difference

set_difference = set_list1.symmetric_difference(set_list2) 

Convert tuple back to dict

 for tuple_element in set_difference:      list_dicts_difference.append(dict((x, y) for x, y in tuple_element))