python - 在 Python 中使用装饰器将函数变成生成器

Question

装饰器是否有办法将下面的函数转换为生成器？

@decorator_that_makes_func_into_generator
def countdown(n):
    while n > 0:
        print n,
        n = n - 1

如有必要，可以修改该功能。请注意，该函数没有 yield 语句，否则它已经是一个生成器。

Answer 1

如果您无法更改countdown 的源，则必须捕获输出:

import sys
from io import StringIO

def decorator_that_makes_func_into_generator(func):
    def wrapper(*a, **ka):
        # Temporarily redirect all output to StringIO instance (intr)
        ts, intr = sys.stdout, StringIO()
        sys.stdout = intr
        func(*a, **ka)
        # Restore normal stdout from backup (ts)
        sys.stdout = ts
        # Get output from intr, split it by whitespace and use it as generator
        yield from intr.getvalue().split()

    return wrapper

@decorator_that_makes_func_into_generator
def countdown(n):
    while n > 0:
        print(n)
        n = n - 1

print(countdown(5), list(countdown(5)))
# <generator object wrapper at 0x01E09058> ['5', '4', '3', '2', '1']

如果您可以更改函数，并且如果您想从countdown(list 或其他序列类型)返回一些内容，并且然后从返回的对象创建一个生成器，装饰器看起来像

def decorator_that_makes_func_into_generator(func):
    def wrapper(*a, **ka):
        yield from func(*a, **ka)
    return wrapper

注意:很棒的 yield from 是在 Python 3.3 中引入的，在旧版本上使用普通循环代替:

for x in func(*a, **ka): 
    yield x

例子:

@decorator_that_makes_func_into_generator
def countdown(n):
    res = []
    while n > 0:
        res.append(n)
        n = n - 1
    return res

print(type(countdown(5)), list(countdown(5))) 
# Output: <class 'generator'> [5, 4, 3, 2, 1]

没有什么能阻止您将 decorator_that_makes_func_into_generator 应用于生成器:

@decorator_that_makes_func_into_generator
def countdown(n):
    while n > 0:
        yield n
        n = n - 1

print(type(countdown(5)), list(countdown(5))) 
# Outputs <class 'generator'> [5, 4, 3, 2, 1]