如何将两个字典列表合并为字典：以指定字典值为键分组元素

Let's break down how to solve this problem—we need to evenly distribute the students from lst1 into the teams defined in lst2, then structure the result into a dictionary where each team name is a key pointing to its assigned students.

Step-by-Step Solution (Even Split Case) #

First, let's cover the core scenario where the number of students is perfectly divisible by the number of teams (like your example with 4 students and 2 teams). Here's a clean, readable implementation:

lst1 = [{'st_name': 'ram', 'st_email_id': 'ram@abc.com'}, 
        {'st_name': 'Raj', 'st_email_id': 'raj@abc.com'}, 
        {'st_name': 'jatin', 'st_email_id': 'jatin@abc.com'},
        {'st_name': 'tom', 'st_email_id': 'tom@abc.com'}]
lst2 = [{'team_name': 'Team1'}, {'team_name': 'Team2'}]

# Calculate how many students each team gets
total_students = len(lst1)
total_teams = len(lst2)
students_per_team = total_students // total_teams

# Split the student list into equal-sized groups
student_groups = [lst1[i * students_per_team : (i + 1) * students_per_team] for i in range(total_teams)]

# Map each team to its student group (convert team name to lowercase as in your desired output)
final_result = {team['team_name'].lower(): group for team, group in zip(lst2, student_groups)}

print(final_result)

Output #

Running this code will give you exactly the format you need:

{
    "team1": [
        {'st_name': 'ram', 'st_email_id': 'ram@abc.com'},
        {'st_name': 'Raj', 'st_email_id': 'raj@abc.com'}
    ],
    "team2": [
        {'st_name': 'jatin', 'st_email_id': 'jatin@abc.com'},
        {'st_name': 'tom', 'st_email_id': 'tom@abc.com'}
    ]
}

Handling Uneven Student Counts #

If you ever have a scenario where the number of students isn't perfectly divisible by the number of teams (e.g., 5 students and 2 teams), we can adjust the code to distribute the extra students evenly across the first few teams. Here's a more robust version using itertools.islice:

from itertools import islice

lst1 = [{'st_name': 'ram', 'st_email_id': 'ram@abc.com'}, 
        {'st_name': 'Raj', 'st_email_id': 'raj@abc.com'}, 
        {'st_name': 'jatin', 'st_email_id': 'jatin@abc.com'},
        {'st_name': 'tom', 'st_email_id': 'tom@abc.com'},
        {'st_name': 'sarah', 'st_email_id': 'sarah@abc.com'}]
lst2 = [{'team_name': 'Team1'}, {'team_name': 'Team2'}]

def split_uneven_list(lst, num_groups):
    it = iter(lst)
    base_size = len(lst) // num_groups
    extra = len(lst) % num_groups
    for i in range(num_groups):
        # Assign one extra student to the first 'extra' groups
        group_size = base_size + (1 if i < extra else 0)
        yield list(islice(it, group_size))

total_teams = len(lst2)
student_groups = list(split_uneven_list(lst1, total_teams))
final_result = {team['team_name'].lower(): group for team, group in zip(lst2, student_groups)}

print(final_result)

Quick Notes #

• We convert team names to lowercase with .lower() to match your sample output—if you want to keep the original capitalization (e.g., "Team1" instead of "team1"), just remove that method call.
• The first solution works for clean even splits, while the second handles edge cases where student counts don't divide evenly by team numbers.

内容的提问来源于stack exchange，提问作者user14794753