Python-3.2 协程：AttributeError：'generator' 对象没有属性 'next'
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Python-3.2 coroutine: AttributeError: 'generator' object has no attribute 'next'
提问by
#!/usr/bin/python3.2
import sys
def match_text(pattern):
line = (yield)
if pattern in line:
print(line)
x = match_text('apple')
x.next()
for line in input('>>>> '):
if x.send(line):
print(line)
x.close()
This is a coroutine but Python3.2 sees it as a generator - why? What is going on here? I'm referring to Python Essential Reference by David Beazeley pg:20.
这是一个协程，但 Python3.2 将其视为生成器 - 为什么？这里发生了什么？我指的是 David Beazeley pg:20 的 Python Essential Reference。
To quote the relevant section:
引用相关部分：
Normally, functions operate on a single set of input arguments. However, a function can
also be written to operate as a task that processes a sequence of inputs sent to
it.This type of function is known as a coroutine and is created by using the yield
statement as an expression (yield) as shown in this example:
def print_matches(matchtext):
print "Looking for", matchtext
while True:
line = (yield) # Get a line of text
if matchtext in line:
print line
To use this function, you first call it, advance it to the first (yield), and then
start sending data to it using send(). For example:
>>> matcher = print_matches("python")
>>> matcher.next() # Advance to the first (yield)
Looking for python
>>> matcher.send("Hello World")
>>> matcher.send("python is cool")
python is cool
>>> matcher.send("yow!")
>>> matcher.close() # Done with the matcher function call
Why doesn't my code work - not that DB's works..
为什么我的代码不起作用 - 不是 DB 的作品..
deathstar> python3.2 xxx
Traceback (most recent call last):
File "xxx", line 9, in <module>
matcher.next() # Advance to the first (yield)
AttributeError: 'generator' object has no attribute 'next'
采纳答案by Karl Knechtel
You're getting thrown off by the error message; type-wise, Python doesn't make a distinction - you can .sendto anything that uses yield, even if it doesn't do anything with the sent value internally.
你被错误信息抛弃了；在类型方面，Python 没有区别 - 您可以.send使用任何使用yield，即使它在内部不对发送的值执行任何操作。
In 3.x, there is no longer a .nextmethod attached to these; instead, use the built-in free function next:
在 3.x 中，不再有.next附加到这些的方法；相反，使用内置的 free 函数next：
next(matcher)
回答by sage
In the case you find yourself patching somebody's code, it seems that the built-in python3 next() function calls the iterator's next() function, so you may be able to find/replace somebody's python2 .next(with the python3-tolerable .__next__(as I just did to make portions of the primefac module work in python3 (among other trivial changes).
如果您发现自己正在修补某人的代码，似乎内置的 python3 next() 函数调用了迭代器的next() 函数，因此您可以像我刚才那样.next(使用 python3-tolerable查找/替换某人的 python2.__next__(使 primefac 模块的一部分在 python3 中工作（以及其他微不足道的更改）。
Here's the reference:
这是参考：
next(iterator[, default])
Retrieve the next item from the iterator by calling its next() method. If default is given, it is returned if the iterator is exhausted, otherwise StopIteration is raised.
下一个（迭代器 [，默认]）
通过调用其next() 方法从迭代器中检索下一项。如果给出默认值，则在迭代器耗尽时返回，否则引发 StopIteration。
回答by Kalaiselvi Rajasekar R
For python version 3.2 the syntax for the next()in-built function should be matcher.__next__()or next(matcher).
对于 python 3.2 版next()，内置函数的语法应该是matcher.__next__()or next(matcher)。
