Question

我有一个source可观察对象，我订阅logger观察者用于记录目的。

我还订阅了source，因此我可以执行计算。当我的计算完成后，我已完成source并且我想处理logger：

             +-------------------+
             |                   |
   +---------+ source observable +--------+
   |         |                   |        |
   |         +-------------------+        |
   |                                      |
   |                                      |
+--v---------------+         +------------v--------+
|                  |         |                     |
|     logger       |         |    computations     |
|    (observer)    |         |    (observable)     |
+-------^----------+         +-----------+---------+
        |                                |
        |                                |
        |        dispose logger          |
        +--------------------------------+
            when computations completed

然而，logger并未在正确的时间处理 - 通常会发生一两个额外的滴答声：

MWE

from rx import Observable

# Some source
source = Observable.interval(1)

# Create logger for source
logged = []
logger = source.subscribe(logged.append)

# Now do stuff/computations with source
calculated = source.map(lambda x: x**2).take_while(lambda x: x < 20)

# Output computed values and stop logging when we're done with our computation
calculated.subscribe(print, print, logger.dispose)

# I expect only values that passed through our computation to have been logged
# The last value should be 5 because 5**2 = 25 which is larger than 20
# which in turn causes our computation to terminate
assert logged == [0, 1, 2, 3, 4, 5], logged

但我明白了：

Traceback (most recent call last):
  File "C:\Program Files (x86)\Python27\lib\site-packages\IPython\core\interactiveshell.py", line 3035, in run_code
    exec(code_obj, self.user_global_ns, self.user_ns)
  File "<ipython-input-54-e8cb1fb583bf>", line 1, in <module>
    assert logged == [0, 1, 2, 3, 4, 5], logged
AssertionError: [0, 1, 2, 3, 4, 5, 6, 7]

7如何记录？我们的计算应在source发出5之后终止，此时logger将被处理掉。

我做错了什么？

Answer 1

这是线程同步问题。 interval()运算符启动新线程以在指定的时间间隔内调用on_next()。处理完订阅后，需要一些时间，直到其他线程检测到此信号并停止工作。并且一毫秒接近它所需的时间。

为了记录通过反应链的消息，将记录功能直接插入该链是更可靠的：

logged = []
def logger(x):
    logged.append(x)
    return x

calculated = source \
    .map(logger) \
    .map(lambda x: x**2) \
    .take_while(lambda x: x < 20) \
    .subscribe(print, print)

如何在完成另一个可观察者时处置一个观察者？

MWE

1 个答案: