在使用Python协程时，如何避免额外的迭代？

Question

下面是Python3.9中的一个协同例程

def coroutine(func):
    def start(*args, **kwargs):
        cr = func(*args, **kwargs)
        next(cr)
        return cr
    return start


@coroutine
def grep(pattern):
    while True:
        line = yield "I have to yield something here?"
        if pattern in line:
            # do something fancy to the line
            yield line
        else:
            raise ValueError("err")

由于line = yield正在接收和发送数据，因此我需要执行额外的next调用才能使其工作：

gg = grep("thing")

item = gg.send("That thing!")
print(item)
next(gg)

item = gg.send("That thing also!")
print(item)
next(gg)

item = gg.send("And what about here this thing.")
print(item)
next(gg)

item = gg.send("Not this.")
print(item)
next(gg)

打印的内容：

That thing!
That thing also!
And what about here this thing.
Traceback (most recent call last):
  File "/home/erasmose/Workspace/clustr/cmapper/temp.py", line 33, in <module>
    item = gg.send("Not this.")
  File "/home/erasmose/Workspace/clustr/cmapper/temp.py", line 16, in grep
    raise ValueError("err")
ValueError: err

如果我删除"next“调用：

gg = grep("thing")

item = gg.send("That thing!")
print(item)

item = gg.send("That thing also!")
print(item)

item = gg.send("And what about here this thing.")
print(item)

item = gg.send("Not this.")
print(item)

输出为：

That thing!
I have to yield something here?
And what about here this thing.
I have to yield something here?

有没有办法避免这些额外的“下一步”调用？

Answer 1

您需要组合这两个yield语句，以便协程在同一个yield中发送和接收

不过，你仍然需要额外的收益才能开始这个过程。

如下所示：

@coroutine
def grep(pattern):
    line = yield "I have to yield something here?"  # receive the first one
    while True:
        if pattern in line:
            line = yield line  # send and receive in one yield statement
        else:
            raise ValueError("err")