调用外部模块时多处理池变慢

Question

我的脚本正在调用librosa模块来计算短音频的 Mel 频率倒谱系数 (MFCC)。加载音频后，我想尽可能快地计算这些（以及其他一些音频功能） - 因此进行多处理。

问题：多处理变体比顺序慢得多。分析说我的代码花费了超过 90% 的时间在<method 'acquire' of '_thread.lock' objects>. 如果它是许多小任务，这并不奇怪，但在一个测试用例中，我将音频分成 4 个块，然后在单独的进程中处理。我认为开销应该最小，但实际上，它几乎和许多小任务一样糟糕。

据我所知，多处理模块应该分叉几乎所有的东西，并且不应该有任何争夺锁的斗争。然而，结果似乎显示出不同的东西。可能是下面的librosa模块保持某种内部锁吗？

我的分析结果为纯文本：https : //drive.google.com/open?id=17DHfmwtVOJOZVnwIueeoWClUaWkvhTPc

作为图片：https : //drive.google.com/open?id=1KuZyo0CurHd9GjXge5CYQhdWn2Q6OG8Z

重现“问题”的代码：

import time
import numpy as np
import librosa
from functools import partial
from multiprocessing import Pool

n_proc = 4

y, sr = librosa.load(librosa.util.example_audio_file(), duration=60) # load audio sample
y = np.repeat(y, 10) # repeat signal so that we can get more reliable measurements
sample_len = int(sr * 0.2) # We will compute MFCC for short pieces of audio

def get_mfcc_in_loop(audio, sr, sample_len):
    # We split long array into small ones of lenth sample_len
    y_windowed = np.array_split(audio, np.arange(sample_len, len(audio), sample_len))
    for sample in y_windowed:
        mfcc = librosa.feature.mfcc(y=sample, sr=sr)

start = time.time()
get_mfcc_in_loop(y, sr, sample_len)
print('Time single process:', time.time() - start)

# Let's test now feeding these small arrays to pool of 4 workers. Since computing
# MFCCs for these small arrays is fast, I'd expect this to be not that fast
start = time.time()
y_windowed = np.array_split(y, np.arange(sample_len, len(y), sample_len))
with Pool(n_proc) as pool:
    func = partial(librosa.feature.mfcc, sr=sr)
    result = pool.map(func, y_windowed)
print('Time multiprocessing (many small tasks):', time.time() - start)

# Here we split the audio into 4 chunks and process them separately. This I'd expect
# to be fast and somehow it isn't. What could be the cause? Anything to do about it?
start = time.time()
y_split = np.array_split(y, n_proc)
with Pool(n_proc) as pool:
    func = partial(get_mfcc_in_loop, sr=sr, sample_len=sample_len)
    result = pool.map(func, y_split)
print('Time multiprocessing (a few large tasks):', time.time() - start)

我机器上的结果：

• 时间单进程：8.48s
• 时间多处理（许多小任务）：44.20s
• 时间多处理（一些大型任务）：41.99s

任何想法是什么导致它？更好的是，如何让它变得更好？

Answer 1

为了调查发生了什么，我运行top -H并注意到生成了 +60 个线程！就是这样。结果证明librosa和依赖项产生了许多额外的线程，这些线程一起破坏了并行性。

解决方案

过度订阅的问题在joblib docs 中有很好的描述。那我们就用它吧。

import time
import numpy as np
import librosa
from joblib import Parallel, delayed

n_proc = 4

y, sr = librosa.load(librosa.util.example_audio_file(), duration=60) # load audio sample
y = np.repeat(y, 10) # repeat signal so that we can get more reliable measurements
sample_len = int(sr * 0.2) # We will compute MFCC for short pieces of audio

def get_mfcc_in_loop(audio, sr, sample_len):
    # We split long array into small ones of lenth sample_len
    y_windowed = np.array_split(audio, np.arange(sample_len, len(audio), sample_len))
    for sample in y_windowed:
        mfcc = librosa.feature.mfcc(y=sample, sr=sr)

start = time.time()
y_windowed = np.array_split(y, np.arange(sample_len, len(y), sample_len))
Parallel(n_jobs=n_proc, backend='multiprocessing')(delayed(get_mfcc_in_loop)(audio=data, sr=sr, sample_len=sample_len) for data in y_windowed)
print('Time multiprocessing with joblib (many small tasks):', time.time() - start)


y_split = np.array_split(y, n_proc)
start = time.time()
Parallel(n_jobs=n_proc, backend='multiprocessing')(delayed(get_mfcc_in_loop)(audio=data, sr=sr, sample_len=sample_len) for data in y_split)
print('Time multiprocessing with joblib (a few large tasks):', time.time() - start)

结果：

• 使用 joblib 进行时间多处理（许多小任务）：2.66
• 使用 joblib 进行时间多处理（一些大型任务）：2.65

比使用多处理模块快 15 倍。