延迟绑定的闭包
def create_multipliers():
return [lambda x : i * x for i in range(5)]
for multiplier in create_multipliers():
print(multiplier(2))
""" output
8
8
8
8
8
"""
错误的解读：与 lambda 有关
def create_multipliers():
multipliers = []
for i in range(5):
def multiplier(x):
return i * x
multipliers.append(multiplier)
return multipliers
for multiplier in create_multipliers():
print(multiplier(2))
""" output
8
8
8
8
8
"""
解决方案：
1、预先计算函数默认值
def create_multipliers():
return [lambda x, i=i : i * x for i in range(5)]
for multiplier in create_multipliers():
print(multiplier(2))
""" output
0
2
4
6
8
"""
2、functools.partial 函数
from functools import partial
from operator import mul
def create_multipliers():
return [partial(mul, i) for i in range(5)]
for multiplier in create_multipliers():
print(multiplier(2))
""" output
0
2
4
6
8
"""
可变的默认参数
def append_to(element, []):
to.append(element)
return to
first_list = append_to(10)
print(first_list)
""" output
[10]
"""
second_list = append_to(20)
print(second_list)
""" output
[10, 20]
"""
新列表仅会在函数定义时被创建一次，后续每次函数调用都是用同一个列表
解决方案：
def append_to(element, to=None):
if to is None:
to = []
to.append(element)
return to
创建一个包含 N 个相同对象的列表
create_list = [[] for __ in range(4)]
许多 Python 风格指南建议使用单下划线 _ 抛弃值，其问题在于它通常被用作 gettext.gettext() 函数的别名，同时在交互模式下用来保存上一次操作的值。
