python学习--模块-functools

 

最近在用flask 装饰器实现代码提取 遇见了函数名重名的问题  

找到这个  functools.wraps  解决了这个问题  

 

functools

（一）partial

In [1]: int('10', 10)
Out[1]: 10

In [2]: int('10', 2)
Out[2]: 2

In [3]: from functools import partial

In [4]: p_int = partial(int, 10)

In [5]: p_int('10')
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-5-1f075b44e9a3> in <module>()
----> 1 p_int('10')

TypeError: an integer is required

In [6]: p_int = partial(int, base=10)
In [7]: p_int('10')
Out[7]: 10

In [8]: type(p_int)
Out[8]: functools.partial

唯一要注意的是可选参数必须写出参数名。base

（二）update_wrapper

把被封装函数的name、module、doc和 dict都复制到封装函数去

# -*- coding: utf-8 -*-
from functools import update_wrapper

def demo1(func):
    def wrapper(*args, **kwargs):
        return func(*args, **kwargs) + 'python'
    return wrapper

@demo1
def aaa():
    '''name'''
    return 'loco_'

print aaa()          # loco_python
print aaa.__doc__    # None
print aaa.__name__   # wrapper


def demo2(func):
    def wrapper(*args, **kwargs):
        return func(*args, **kwargs) + 'python'
    return update_wrapper(wrapper, func)

@demo2
def aaa():
    '''name'''
    return 'loco_'

print aaa()          # loco_python
print aaa.__doc__    # name
print aaa.__name__   # aaa  函数名

（三）wraps 将update_wrapper封装起来了

# -*- coding: utf-8 -*-
from functools import wraps

def demo2(func):
    @wraps(func)
    def wrapper(*args, **kwargs):
        return func(*args, **kwargs) + 'python'
    return wrapper

@demo2
def aaa():
    '''name'''
    return 'loco_'

print aaa()          # loco_python
print aaa.__doc__    # name
print aaa.__name__   # aaa  函数名