获取给定元组键的字典中的最大值，其中元组的元素可能位于不同的位置

Question

I have the following dictionary:

{('2019-07-243541760601284691', '2019-07-243541760603812661'): 1086,
('2019-07-243541760601314711', '2019-07-243541760603996721'): 662,
('2019-07-243541760603794841', '2019-07-243541760600899921'): 483,
('2019-07-243541760603794841', '2019-07-243541760601224211'): 70,
('2019-07-243541760603794841', '2019-07-243541760607368321'): 54,
('2019-07-243541760600899921', '2019-07-243541760601224211'): 93,
('2019-07-243541760600899921', '2019-07-243541760607368321'): 74,
('2019-07-243541760601224211', '2019-07-243541760607368321'): 490,
('2019-07-243541760613553761', '2019-07-243541760602348611'): 484,
('2019-07-243541760602450401', '2019-07-243541760602927941'): 1118,
('2019-07-243541760603292161', '2019-07-243541760606108621'): 732}

The keys are tuples of unique ids. As you can see certain ids are repeated within both columns (eg. lines 3 to 5 have the same first id, and element 2 from line 3 is also present in lines 6 and 7). I want to find a way to return the maximum value for each given unique id.

If it helps the output I am looking for would look like this:

{('2019-07-243541760601284691', '2019-07-243541760603812661'): 1086,
('2019-07-243541760601314711', '2019-07-243541760603996721'): 662,
('2019-07-243541760603794841', '2019-07-243541760600899921'): 483,
('2019-07-243541760601224211', '2019-07-243541760607368321'): 490,
('2019-07-243541760613553761', '2019-07-243541760602348611'): 484,
('2019-07-243541760602450401', '2019-07-243541760602927941'): 1118,
('2019-07-243541760603292161', '2019-07-243541760606108621'): 732}

In other words each unique id should appear at most in one key in the dictionary.

Answer 1

You can change the behavior of the built in types in Python. For your case it's really easy to create a dict subclass that will store duplicated values in lists under the same key automatically. You can then get the max value of the list.

class Dictlist(dict):
    def __setitem__(self, key, value):
        try:
            self[key]
        except KeyError:
            super(Dictlist, self).__setitem__(key, [])
        self[key].append(value)

Output example:

>>> d = dictlist.Dictlist()
>>> d['test'] = 1
>>> d['test'] = 2
>>> d['test'] = 3
>>> d
{'test': [1, 2, 3]}
>>> d['other'] = 100
>>> d
{'test': [1, 2, 3], 'other': [100]}

So when you want to return the max value of a key, just pass the key as the parameters and this will give you the max value in that list.

max(dict[key_value])

Answer 2

First I sort by values in a list L . Then I iterate over L , and I store each new tuple's id in a set ids_set , and if they were not in this set yet, then I put them into my dict final .

d = {('2019-07-243541760601284691', '2019-07-243541760603812661'): 1086,
('2019-07-243541760601314711', '2019-07-243541760603996721'): 662,
('2019-07-243541760603794841', '2019-07-243541760600899921'): 483,
('2019-07-243541760603794841', '2019-07-243541760601224211'): 70,
('2019-07-243541760603794841', '2019-07-243541760607368321'): 54,
('2019-07-243541760600899921', '2019-07-243541760601224211'): 93,
('2019-07-243541760600899921', '2019-07-243541760607368321'): 74,
('2019-07-243541760601224211', '2019-07-243541760607368321'): 490,
('2019-07-243541760613553761', '2019-07-243541760602348611'): 484,
('2019-07-243541760602450401', '2019-07-243541760602927941'): 1118,
('2019-07-243541760603292161', '2019-07-243541760606108621'): 732}

L = sorted(d.items(), key=lambda item: item[1], reverse=True)
# L = [(('2019-07-243541760602450401', '2019-07-243541760602927941'), 1118),...

final = {}
ids_set = set()
for i in L:
    t = i[0] # tuple
    val = i[1]
    if (t[0] or t[1]) not in ids_set:
        ids_set.update(t)
        final[t] = val

for i, v in final.items():
    print(i, v)

Output :

('2019-07-243541760602450401', '2019-07-243541760602927941') 1118
('2019-07-243541760601284691', '2019-07-243541760603812661') 1086
('2019-07-243541760603292161', '2019-07-243541760606108621') 732
('2019-07-243541760601314711', '2019-07-243541760603996721') 662
('2019-07-243541760601224211', '2019-07-243541760607368321') 490
('2019-07-243541760613553761', '2019-07-243541760602348611') 484
('2019-07-243541760603794841', '2019-07-243541760600899921') 483

Answer 3

d = {('2019-07-243541760601284691', '2019-07-243541760603812661'): 1086,
('2019-07-243541760601314711', '2019-07-243541760603996721'): 662,
('2019-07-243541760603794841', '2019-07-243541760600899921'): 483,
('2019-07-243541760603794841', '2019-07-243541760601224211'): 70,
('2019-07-243541760603794841', '2019-07-243541760607368321'): 54,
('2019-07-243541760600899921', '2019-07-243541760601224211'): 93,
('2019-07-243541760600899921', '2019-07-243541760607368321'): 74,
('2019-07-243541760601224211', '2019-07-243541760607368321'): 490,
('2019-07-243541760613553761', '2019-07-243541760602348611'): 484,
('2019-07-243541760602450401', '2019-07-243541760602927941'): 1118,
('2019-07-243541760603292161', '2019-07-243541760606108621'): 732}

# flatten the dict so that each id in the tuple has value
# but first make sure we sort on values so that if the key appears twice
# the latest element should be the one with max value (dict only keeps one item per key)
sorted_dict = dict(item for item in sorted(d.items(), key=lambda x: x[-1]))
flat_dict = {**{k[0]:v for k,v in d.items()}, **{k[1]:v for k,v in sorted_dict .items()}}

# get the unique keys you want
keys, already_used = [], []
for k in d:
    if k[0] in already_used or k[1] in already_used:
        continue
    keys.append(k)
    already_used.extend(k)

# now create the new_dict
new_dict = {k:max(flat_dict[k[0]], flat_dict[k[1]]) for k in keys}

Output

>>> new_dict
{('2019-07-243541760601284691', '2019-07-243541760603812661'): 1086,
 ('2019-07-243541760601314711', '2019-07-243541760603996721'): 662,
 ('2019-07-243541760603794841', '2019-07-243541760600899921'): 483,
 ('2019-07-243541760601224211', '2019-07-243541760607368321'): 490,
 ('2019-07-243541760613553761', '2019-07-243541760602348611'): 484,
 ('2019-07-243541760602450401', '2019-07-243541760602927941'): 1118,
 ('2019-07-243541760603292161', '2019-07-243541760606108621'): 732}