python字典中部分匹配键的最大值

Question

I have the following dictionary where the keys are 'month,country:ID' and values are just totals:

ID_dict = {'11,United Kingdom:14416': 129.22, '11,United Kingdom:17001': 357.6, 
'12,United States:14035': 90000.0, '12,United Kingdom:17850': 241.16,'12,United 
States:14099': 90000.0, '12,France:12583': 252.0, '12,United Kingdom:13047': 
215.13, '01,Germany:12662': 78.0, '01,Germany:12600': 14000}

The actual dictionary will be much larger than this one.

I am trying to return the key for each 'month, country' that contains the highest total. If there is a tie the ID's would be separated by a comma. Example Output based on dictionary above:

'11,United Kingdom:17001'
'12,United Kingdom:17850'
'12,United States:14035, 14099'
'12,France:12583'
'01,Germany:12600'

I can get the strings of the highest values using the following code:

highest = max(ID_dict.values())
print([k for k, v in ID_dict.items() if v == highest])

But really struggling to get past this point. I was experimenting using re.match and re.search but was not getting very far with those.

Answer 1

You can find the maximum for each month, country pair, store this relation in a dictionary. Then create a dictionary that have as keys the pairs (month, country) and as values a list of IDs that have value equal to the maximum for that (month, country) pair:

import re

ID_dict = {'11,United Kingdom:14416': 129.22, '11,United Kingdom:17001': 357.6,
           '12,United States:14035': 90000.0, '12,United Kingdom:17850': 241.16, '12,United States:14099': 90000.0,
           '12,France:12583': 252.0, '12,United Kingdom:13047': 215.13, '01,Germany:12662': 78.0,
           '01,Germany:12600': 14000}

table = {tuple(re.split(',|:', key)[:2]): value for key, value in sorted(ID_dict.items(), key=lambda e: e[1])}

result = {}
for key, value in ID_dict.items():
    splits = re.split(',|:', key)
    if value == table[tuple(splits[:2])]:
        result.setdefault(tuple(splits[:2]), []).append(splits[2])

for key, value in result.items():
    print('{}:{}'.format(','.join(key), ', '.join(value)))

Output

01,Germany:12600
12,United States:14099, 14035
12,United Kingdom:17850
11,United Kingdom:17001
12,France:12583

The above approach is O(nlogn) because it uses sorted , to make it O(n) you can change the dictionary comprehension by this loop:

table = {}
for s, v in ID_dict.items():
    key = tuple(re.split(',|:', s)[:2])
    table[key] = max(table.get(key, v), v)

Answer 2

The following code creates a new dictionary with 'month,country' keys and lists of (value, IDnum) as the values. It then sorts each list, and collects all the IDnums that correspond to the highest value.

ID_dict = {
    '11,United Kingdom:14416': 129.22, '11,United Kingdom:17001': 357.6, 
    '12,United States:14035': 90000.0, '12,United Kingdom:17850': 241.16,
    '12,United States:14099': 90000.0, '12,France:12583': 252.0, 
    '12,United Kingdom:13047': 215.13, '01,Germany:12662': 78.0, 
    '01,Germany:12600': 14000
}

# Create a new dict with 'month,country' keys 
# and lists of (value, IDnum) as the values
new_data = {}
for key, val in ID_dict.items():
    newkey, idnum = key.split(':')
    new_data.setdefault(newkey, []).append((val, idnum))

# Sort the values for each 'month,country' key,
# and get the IDnums corresponding to the highest values
for key, val in new_data.items():
    val = sorted(val, reverse=True)
    highest = val[0][0]
    # Collect all IDnums that have the highest value
    ids = []
    for v, idnum in val:
        if v != highest:
            break
        ids.append(idnum)
    print(key + ':' + ', '.join(ids))

output

11,United Kingdom:17001
12,United States:14099, 14035
12,United Kingdom:17850
12,France:12583
01,Germany:12600