Question

我想用Python制作一个节拍器。

我有2个声音文件。每秒应该播放第一个声音文件，每隔四秒钟播放第二个声音文件。

我找到了扭曲的模块，但无法弄清楚如何传递参数并迭代它。所以我有我想要迭代的变量计数器，但它不能使用以下代码：

import simpleaudio as sa
from twisted.internet import task
from twisted.internet import reactor


def beat(bpm, counter):
    timeout = 60/bpm
    l = task.LoopingCall(play_beat, counter=counter)
    l.start(timeout)
    reactor.run()


def play_beat(counter):
    counter += 1
    print(counter) #This prints always 2, I am expecting an iteration like 2, 3, 4, 5, 6 ...
    if counter % 4 == 0:
        wave_obj = sa.WaveObject.from_wave_file("wav/beat_end.wav")
    else:
        wave_obj = sa.WaveObject.from_wave_file("wav/beat_start.wav")
    play_obj = wave_obj.play()
    play_obj.wait_done()
    pass

beat(60, 1)

Answer 1

这是另一种解决方案。

this.something = 'fruit'

值得注意的是，我已删除import simpleaudio as sa from twisted.internet import task from twisted.internet import reactor def beat(bpm, wave_obj): timeout = 60.0 / bpm l = task.LoopingCall(play_beat, wave_obj) l.start(timeout) def play_beat(wave_obj): wave_obj.play() beat_start = sa.WaveObject.from_wave_file(...) beat_end = sa.WaveObject.from_wave_file(...) bpm = 60.0 bps = bpm / 60 reactor.callLater(0 / bps, beat, bpm / 4, beat_start) reactor.callLater(1 / bps, beat, bpm / 4, beat_start) reactor.callLater(2 / bps, beat, bpm / 4, beat_start) reactor.callLater(3 / bps, beat, bpm / 4, beat_end) reactor.run()电话。如果你这样阻塞反应堆线程，你可能会对基于时间的事件安排不太可靠。

扭曲的Python节拍器：如何传递参数进行迭代？

1 个答案: