operator笔记

# operator.itemgetter(*items) # 获取item

>>> from operator import itemgetter
# list使用下标进行返回
>>> a
[('a', 1), ('b', 2), ('c', 3), ('d', 4)]
>>> itemgetter(1)(a) # 获取a中下标为1的元组
('b', 2)
>>> itemgetter(1,3)(a) # 同时获取多个元组
(('b', 2), ('d', 4))
# dict使用键来返回
>>> a = {'one': 1, 'two':2, 'three':3}
>>> itemgetter('one')(a)
1
>>> itemgetter('one', 'two')(a)
(1, 2)

# 使用attrgetter()获取其属性
# 《python cookbook》 书籍的例子

from operator import attrgetter
class User(object):
def __init__(self, uid):
self.uid = uid
def __repr__(self):
return 'User({})'.format(self.uid)
users = [User(25), User(14), User(100)]
print(sorted(users, key=attrgetter('uid')))
"""
D:笔记python电子书Python3>python index.py
[User(14), User(25), User(100)]
"""

# 使用attrgetter可以使用lambda来进行代替：

class User(object):
def __init__(self, uid):
self.uid = uid
def __repr__(self):
return 'User({})'.format(self.uid)
users = [User(25), User(14), User(100)]
print(sorted(users, key=lambda u: u.uid))
"""
D:笔记python电子书Python3>python index.py
[User(14), User(25), User(100)]
"""

《python cookbook》书籍中提到attrgetter()的速度相对来说比lambda快一些，具体可使用pytest进行测试,这里不做详细测试