在 Python 中，如何抑制选择代码的控制台输出？ [复制]

Question

我有一小部分代码：

stream = pyaudio.PyAudio().open(
    format   = pyaudio.PyAudio().get_format_from_width(1),
    channels = 1, 
    rate     = bitrate, 
    output   = True
)

当我运行它时，它会产生几行控制台输出：

ALSA lib pcm_dsnoop.c:614:(snd_pcm_dsnoop_open) unable to open slave
ALSA lib pcm_dmix.c:1024:(snd_pcm_dmix_open) unable to open slave
ALSA lib pcm.c:2267:(snd_pcm_open_noupdate) Unknown PCM cards.pcm.rear
ALSA lib pcm.c:2267:(snd_pcm_open_noupdate) Unknown PCM cards.pcm.center_lfe
ALSA lib pcm.c:2267:(snd_pcm_open_noupdate) Unknown PCM cards.pcm.side
ALSA lib pcm_dmix.c:1024:(snd_pcm_dmix_open) unable to open slave
Cannot connect to server socket err = No such file or directory
Cannot connect to server request channel
jack server is not running or cannot be started

如何更改围绕此特定函数调用的代码，以使上述表单的控制台输出不出现，同时不使其他代码的控制台输出消失？

我正在设想以上下文管理器的形式解决这个问题，可能有点像下面这样：

@contextlib.contextmanager
def silence():
    # <insert magic to direct stdout and stderr to null>
    yield
    # <insert magic to direct stdout and stderr to console>

with silence():
    stream = pyaudio.PyAudio().open(
        format   = pyaudio.PyAudio().get_format_from_width(1),
        channels = 1, 
        rate     = bitrate, 
        output   = True
    )

Answer 1

装饰器可以工作。也许是这样的：

def silence(func):
    def silenced():
        // silence output
        func()
        // change output to normal
    return silenced


@silence
def openStream(args):
    stream = pyaudio.PyAudio().open(
        format   = pyaudio.PyAudio().get_format_from_width(1),
        channels = 1, 
        rate     = bitrate, 
        output   = True
    )
    return stream