有没有办法枚举共享相同索引的列表或词典？

Question

I want to enumerate a list or dictionary, if the items share the same value or are equal I want them to use 1 index.

This is my code:

def print_data(league_results):
    for i, (team, points) in 
    enumerate(sorted(league_results.items(), key=lambda x: x[1], 
    reverse=True), 1):
        if points == 1:
          i = i
        print('{}. {}, {}'.format(i, team.strip(''), points))

Output

1. Man City, 6
2. Liverpool, 5
3. Everton, 1
4. Stoke, 1
5. Fulham, 0

This is my desired output: If a team share a point they should have the same index. Everton & Stoke must share the same index.

1. Man City, 6
2. Liverpool, 5
3. Everton, 1 
3. Stoke, 1
4. Fulham, 0

Answer 1

You can use itertools.groupby :

def print_data(league_results):
    grouped_results = itertools.groupby(
        sorted(league_results.items(), key=lambda x: x[1], reverse=True), 
        key=lambda x: x[1]
    )
    for i, (_, teams) in enumerate(grouped_results, 1):
        for team, points in teams:
            print('{}. {}, {}'.format(i, team.strip(''), points))

Results:

1. Man City, 6
2. Liverpool, 5
3. Everton, 1
3. Stoke, 1
4. Fulham, 0

Here you have the live example

Explanation:

groupby will make an iterable with matching between the amount of points and the teams (it will group them if they are sorted) giving us an iterable of (points, [(team, points), ...]), ... . Hence when enumerating it you will enumerate this groups, that will be our index i . Since we want the enumeration to start from 1 instead of 0 we simply ask the function to start from it enumerate(iterable, start_enumerate_value)