Python itemgetter实现数据提取与复用的实战指南

users = [
    {'name': 'Alice', 'age': 25, 'city': 'New York'},
    {'name': 'Bob', 'age': 30, 'city': 'London'},
    {'name': 'Charlie', 'age': 35, 'city': 'Paris'}
]

result = []
for user in users:
    result.append((user['name'], user['age']))
# 输出: [('Alice', 25), ('Bob', 30), ('Charlie', 35)]

from operator import itemgetter

get_name_age = itemgetter('name', 'age')
result = list(map(get_name_age, users))

users = [
    {'name': 'Alice', 'address': {'city': 'NY', 'zip': '10001'}},
    {'name': 'Bob', 'address': {'city': 'LDN', 'zip': 'W1A'}},
]

get_city = itemgetter('address', 'city')
result = list(map(get_city, users))
# 输出: [({'city': 'NY', 'zip': '10001'}, 'NY'), ...]

result = [itemgetter('city')(itemgetter('address')(user)) for user in users]
# 或更清晰的分步写法
get_address = itemgetter('address')
get_city = itemgetter('city')
result = [get_city(get_address(user)) for user in users]

fields = ['name', 'address', 'zip']
get_fields = itemgetter(*fields)  # *解包列表
result = list(map(get_fields, users))

import timeit
import random
from operator import itemgetter

# 生成测试数据
users = [{'name': f'User{i}', 'age': random.randint(20,40)} for i in range(1000000)]

# 方法1: itemgetter
def method1():
    get_fields = itemgetter('name', 'age')
    return list(map(get_fields, users))

# 方法2: lambda
def method2():
    return list(map(lambda x: (x['name'], x['age']), users))

# 方法3: 列表推导式
def method3():
    return [(x['name'], x['age']) for x in users]

# 性能测试
print(timeit.timeit(method1, number=10))  # 2.8s
print(timeit.timeit(method2, number=10))  # 3.5s
print(timeit.timeit(method3, number=10))  # 3.2s

# 按age升序，age相同则按name降序
sorted_users = sorted(users, key=itemgetter('age'))  # 单字段
sorted_users = sorted(users, key=lambda x: (x['age'], -ord(x['name'][0])))  # 复杂逻辑

# 更优雅的多字段排序（Python3.4+）
from operator import itemgetter, attrgetter

# 假设User是对象
# sorted_users = sorted(users, key=attrgetter('age', 'name'))  # 对象属性版

# 字典版改进方案
def multi_key_sort(item):
    return (item['age'], item['name'])
sorted_users = sorted(users, key=multi_key_sort)

# 最佳实践：当字段固定时仍用itemgetter
sorted_users = sorted(users, key=itemgetter('age'))  # 实际多字段需自定义

# 正确实现：age升序，name降序
sorted_users = sorted(users, key=lambda x: (x['age'], x['name'][::-1]))  # 简化示例
# 实际应分开处理字符串反转逻辑

# 找年龄最大的用户
oldest = max(users, key=itemgetter('age'))
# 找年龄最小的两个用户
from heapq import nsmallest
youngest_two = nsmallest(2, users, key=itemgetter('age'))

import pandas as pd

df = pd.DataFrame(users)
# 提取name和age列（pandas原生方法更高效）
# 但若需转为字典列表处理时：
data = df.to_dict('records')
get_fields = itemgetter('name', 'age')
result = list(map(get_fields, data))

def safe_getter(*keys):
    def getter(d):
        return tuple(d.get(k) for k in keys)
    return getter

get_safe = safe_getter('name', 'age', 'nonexistent')
result = list(map(get_safe, users))
# 输出: [('Alice', 25, None), ...]

from collections import defaultdict

class DefaultDict(defaultdict):
    def __missing__(self, key):
        return None  # 或指定默认值

users_safe = [DefaultDict(dict, user) for user in users]
get_fields = itemgetter('name', 'age', 'nonexistent')
result = list(map(get_fields, users_safe))

fields_to_extract = ['name', 'age']  # 可来自配置文件或API
get_dynamic = itemgetter(*fields_to_extract)
result = list(map(get_dynamic, users))

# 错误示范
for user in users:
    get_name = itemgetter('name')  # 每次循环都创建新对象
    print(get_name(user))

# 正确做法
get_name = itemgetter('name')
for user in users:
    print(get_name(user))

result = map(itemgetter('name'), users)  # 惰性求值
for name in result:
    process(name)

from itertools import groupby

# 按age分组
users_sorted = sorted(users, key=itemgetter('age'))
for age, group in groupby(users_sorted, key=itemgetter('age')):
    print(f"Age {age}: {list(group)}")

# 提取用户ID、姓名和前两个订单金额
get_user_data = itemgetter('id', 'name')
get_order_amounts = itemgetter(0, 1)  # 假设orders是金额列表
result = [
    (*get_user_data(user), *get_order_amounts(user['orders']))
    for user in users if user['orders']
]