如何使用生成器形成多个管道？

Question

我正在使用 python，并且正在尝试找到一种将多个生成器优雅地链接在一起的方法。问题的一个例子是有一个根生成器，它提供某种数据，并且每个值都像级联一样传递给它的“子级”，而级联又可以修改它们接收到的对象。我可以走这条路：

for x in gen1:
    gen2(x)
    gen3(x)

但它丑陋且不优雅。我正在考虑一种更实用的做事方式。

Answer 1

您可以将生成器转换为协程，以便它们可以send() 并从彼此接收值（使用(yield) 表达式）。这将使每个人都有机会更改他们收到的值，和/或将它们传递给下一个生成器/协程（或完全忽略它们）。

请注意，在下面的示例代码中，我使用了一个名为 coroutine 的装饰器来“启动”生成器/协程函数。这会导致它们在第一个 yield 表达式/语句之前执行。这是 youtube 视频中显示的一个稍微修改的版本，该视频是 Dave Beazley 在 PyCon 2009 上发表的题为 A Curious Course on Coroutines and Concurrency 的非常有启发性的演讲。

正如您应该能够从生成的输出中看到的那样，每个管道正在处理数据值，这些管道通过单个 send() 配置到头协程，然后有效地将其“多路复用”到每个管道。由于每个子协程也这样做，因此可以建立一个精心制作的进程“树”。

import sys

def coroutine(func):
    """ Decorator to "prime" generators used as coroutines. """
    def start(*args,**kwargs):
        cr = func(*args,**kwargs)  # Create coroutine generator function.
        next(cr)                   # Advance to just before its first yield.
        return cr
    return start

def pipe(name, value, divisor, coroutines):
    """ Utility function to send values to list of coroutines. """
    print('  {}: {} is divisible by {}'.format(name, value, divisor))
    for cr in coroutines:
        cr.send(value)

def this_func_name():
    """ Helper function that returns name of function calling it. """
    frame = sys._getframe(1)
    return frame.f_code.co_name


@coroutine
def gen1(*coroutines):
    while True:
        value = (yield)     # Receive values sent here via "send()".
        if value % 2 == 0:  # Only pipe even values.
            pipe(this_func_name(), value, 2, coroutines)

@coroutine
def gen2(*coroutines):
    while True:
        value = (yield)     # Receive values sent here via "send()".
        if value % 4 == 0:  # Only pipe values divisible by 4.
            pipe(this_func_name(), value, 4, coroutines)

@coroutine
def gen3(*coroutines):
    while True:
        value = (yield)     # Receive values sent here via "send()".
        if value % 6 == 0:  # Only pipe values divisible by 6.
            pipe(this_func_name(), value, 6, coroutines)

# Create and link together some coroutine pipelines.
g3 = gen3()
g2 = gen2()
g1 = gen1(g2, g3)

# Send values through both pipelines (g1 -> g2, and g1 -> g3) of coroutines.
for value in range(17):
    print('piping {}'.format(value))
    g1.send(value)

输出：

piping 0
  gen1: 0 is divisible by 2
  gen2: 0 is divisible by 4
  gen3: 0 is divisible by 6
piping 1
piping 2
  gen1: 2 is divisible by 2
piping 3
piping 4
  gen1: 4 is divisible by 2
  gen2: 4 is divisible by 4
piping 5
piping 6
  gen1: 6 is divisible by 2
  gen3: 6 is divisible by 6
piping 7
piping 8
  gen1: 8 is divisible by 2
  gen2: 8 is divisible by 4
piping 9
piping 10
  gen1: 10 is divisible by 2
piping 11
piping 12
  gen1: 12 is divisible by 2
  gen2: 12 is divisible by 4
  gen3: 12 is divisible by 6
piping 13
piping 14
  gen1: 14 is divisible by 2
piping 15
piping 16
  gen1: 16 is divisible by 2
  gen2: 16 is divisible by 4

Answer 2

管道可能看起来更像这样：

for x in gen3(gen2(gen1())):
    print x

例如：

for i, x in enumerate(range(10)):
    print i, x

没有办法在 Python 中分叉（或“tee”）管道。如果您需要多个管道，则必须复制它们：gen2(gen1()) 和 gen3(gen1())。

Answer 3

Dave Beazley gave this example in a talk he did in 2008。目标是总结在 Apache Web 服务器日志中传输了多少字节的数据。假设日志格式如：

81.107.39.38 -  ... "GET /ply/ HTTP/1.1" 200 7587
81.107.39.38 -  ... "GET /favicon.ico HTTP/1.1" 404 133
81.107.39.38 -  ... "GET /admin HTTP/1.1" 403 -

传统（非发电机）解决方案可能如下所示：

with open("access-log") as wwwlog:
    total = 0
    for line in wwwlog:
        bytes_as_str = line.rsplit(None,1)[1]
        if bytes_as_str != '-':
            total += int(bytes_as_str)
print("Total: {}".format(total))

为此使用生成器表达式的生成器管道可以可视化为：

access-log => wwwlog => bytecolumn => bytes => sum() => total

可能看起来像：

with open("access-log") as wwwlog:
    bytecolumn = (line.rsplit(None,1)[1] for line in wwwlog)
    bytes = (int(x) for x in bytecolumn if x != '-')
print("Total: {}".format(sum(bytes)))

Dave Beazley 的幻灯片和更多示例可在 on his website 获取。 His later presentations elucidate this further。

如果不确切知道您要做什么，很难说更多，因此我们可以评估您正在做的每件事是否甚至需要自定义生成器（生成器表达式/理解可以在没有需要声明生成器函数）。

Answer 4

下面是一个简洁的例子：

def negate_nums(g):
    for x in g:
        yield -x

def square_nums(g):
    for x in g:
        yield x ** 2

def half_num(g):
    for x in g:
        yield x / 2.0

def compose_gens(first_gen,*rest_gens):
    newg = first_gen(compose_gens(*rest_gens)) if rest_gens else first_gen
    return newg

for x in compose_gens(negate_nums,square_nums,half_num,range(10)):
    print(x)

在这里，您正在编写生成器，以便在最终的 compose_gens 调用中从右到左调用它们。您可以通过反转 args 将其更改为管道。