Sorting a List of Dictionaries by Multiple Keys

from operator import itemgetter

users = [
    {'name': 'Alice', 'age': 30, 'city': 'New York'},
    {'name': 'Bob', 'age': 25, 'city': 'Los Angeles'},
    {'name': 'Charlie', 'age': 30, 'city': 'Chicago'},
    {'name': 'David', 'age': 25, 'city': 'New York'}
]

print("Original users:")
for user in users:
    print(user)

# Sort by 'age' (ascending)
sorted_by_age = sorted(users, key=itemgetter('age'))
print("
Sorted by age:")
for user in sorted_by_age:
    print(user)

# Sort by 'age' (descending)
sorted_by_age_desc = sorted(users, key=itemgetter('age'), reverse=True)
print("
Sorted by age (descending):")
for user in sorted_by_age_desc:
    print(user)

# Sort by 'age' (ascending) then by 'name' (ascending)
sorted_by_age_and_name = sorted(users, key=itemgetter('age', 'name'))
print("
Sorted by age then name:")
for user in sorted_by_age_and_name:
    print(user)

# Sort by 'city' (ascending) then 'age' (descending) - requires lambda for mixed order
# For mixed order, sorting twice is often clearer and sometimes more performant for complex keys
# First sort by secondary key in desired order
temp_sorted = sorted(users, key=itemgetter('age'), reverse=True)
# Then sort by primary key, preserving secondary order for ties
final_sorted = sorted(temp_sorted, key=itemgetter('city'))
print("
Sorted by city (asc) then age (desc):")
for user in final_sorted:
    print(user)

# Alternatively with a lambda (less efficient for simple key access, but more flexible for complex keys)
sorted_with_lambda = sorted(users, key=lambda x: (x['city'], -x['age'])) # Negative age for descending
print("
Sorted by city (asc) then age (desc) with lambda:")
for user in sorted_with_lambda:
    print(user)