Python中late-binding-closures

2018-01-03 @望京

示例1

>>> b = []

>>> for i in range(5):

...     b.append(lambda :i)

...

>>> for j in b:

...     print j()

...

4

4

4

4

4

>>> for m in b:

...     print type(m),m

...

<type 'function'> <function <lambda> at 0x7fe4aae708c0>

<type 'function'> <function <lambda> at 0x7fe4aae70938>

<type 'function'> <function <lambda> at 0x7fe4aae709b0>

<type 'function'> <function <lambda> at 0x7fe4aae70b18>

<type 'function'> <function <lambda> at 0x7fe4aae70b90>

>>>

>>>

为什么不是输出 0~4？

Closures in Python are late-binding,

meaning that each lambda function in the list will only evaluate the variable i when invoked,

and not when defined. 

That's why all functions return the same value, i.e. the last value of ì (which is 4).

late-binding-closures in Python http://docs.python-guide.org/en/latest/writing/gotchas/#late-binding-closures

怎么修改使之输出 0~4？ b.append(lambda i=i:i)

>>> b = []

>>> for i in range(5):

...     b.append(lambda i=i:i)

...

>>> for j in b:

...     print j()

...

0

1

2

3

4

>>>

或者使用 functools.partial ：

>>> b = []

>>> for i in range(5):

...     from functools import partial

...     b.append(partial(lambda x:x, i))

...

>>> for j in b:

...     print j()

...

0

1

2

3

4

>>>

参考：https://stackoverflow.com/questions/38369470/lambdas-from-a-list-comprehension-are-returning-a-lambda-when-called

lambda补充

>>> foo = [2, 18, 9, 22, 17, 24, 8, 12, 27]

>>>

>>> print filter(lambda x: x % 3 == 0, foo)   # 过滤

[18, 9, 24, 12, 27]

>>>

>>> print map(lambda x: x * 2 + 10, foo)      # map

[14, 46, 28, 54, 44, 58, 26, 34, 64]

>>>

>>> print reduce(lambda x, y: x + y, foo)     # 求和

139

>>>

示例2

>>> def func(arg=[]):

...     arg.append(1)

...     print arg

...

>>> func()

[1]

>>> func()

[1, 1]

>>>

修改1：在函数调用的时候传入参数

>>> def func(arg=[]):

...     arg.append(1)

...     print arg

...

>>> func([])

[1]

>>> func([])

[1]

>>>

修改2：默认参数改为 None （更安全的做法）

>>>

>>> def func(arg=None):

...     if not arg:

...         arg = []

...     arg.append(1)

...     print arg

...

>>>

>>> func()

[1]

>>> func()

[1]

>>>

示例3

>>>

>>> def func(x, l=[]):

...     for i in range(x):

...             l.append(i*i)

...     print l

...

>>> func(2)

[0, 1]

>>> func(3)

[0, 1, 0, 1, 4]

>>>

2018-03-12 https://zhuanlan.zhihu.com/p/33376761

今天在地铁上看到公众号推荐这个文章，也是late-binding问题，再补充下：

def foo():

    temp = [lambda x : i*x for i in range(4)]

    return temp

for bar in foo():

    print(bar(2))

# 输出

6

6

6

6

其实可以改写成下面这样，这样就比较容易看出是闭包了：lambda本身是一个函数，调用了外面的变量 i

temp = []

for i in range(4):

    temp.append(lambda x : i*x)

for bar in temp:

    print(bar(2))

解决方法1： [lambda x,i=i : i*x for i in range(4)]

def foo():

    temp = [lambda x,i=i : i*x for i in range(4)]

    return temp

for bar in foo():

    print(bar(2))

解决方法2：使用 functools.partial

from functools import partial

from operator import mul

def foo():

    temp = [partial(mul,i) for i in range(4)]

    return temp

for bar in foo():

    print(bar(2))

解决方法3：把temp改成生成器

def foo():

    temp = (lambda x : i*x for i in range(4))

    return temp

for bar in foo():

    print(bar(2))

# 或者用 next() 方式调用

# bar = foo()

# print(next(bar)(2))

# print(next(bar)(2))

# print(next(bar)(2))

# print(next(bar)(2))

# 或者用 next() 方式调用

# bar = foo()

# print(bar.__next__()(2))

# print(bar.__next__()(2))

# print(bar.__next__()(2))

# print(bar.__next__()(2))

解决方法4：使用yield

def foo():

    for i in range(4):

        yield lambda x : i*x

for bar in foo():

    print(bar(2))

补充：

构造生成器的两种方式：

    使用类似列表生成式的方式生成 (2*n + 1 for n in range(3, 11))

    使用包含yield的函数来生成

如果计算过程比较简单，可以直接把列表生成式改成generator；

但是，如果计算过程比较复杂，就只能通过包含yield的函数来构造generator。

学无止境，戒骄戒躁。

巴特西

Python中late-binding-closures

2018-01-03 @望京

示例1

lambda补充

示例2

示例3

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