具有Keras预测的Python多处理功能

时间:2018-08-02 15:12:57

标签: python keras multiprocessing python-multiprocessing

上下文

一个Keras模型(link here,为了进行MWE)需要并行预测很多测试数据。

我将 cube 定义为numpy.ndarray的3D uint。它的每个垂直切片都是一个,它是 npixels = 128的高度, nbins = 128的深度。

每个预测都会对去噪列(大小相同)中的一列进行变换。

我提供了三种方法:单线程,多处理和pathos包多处理。两种多线程方法都不起作用,我也不知道原因。

代码

import keras
import numpy as np
import threading
import pathos.multiprocessing
import multiprocessing


def __res_sum_squares(y_true, y_pred):
    squared_diff = (y_true - y_pred) ** 2
    return keras.backend.sum(squared_diff)


__npixels, __nbins = 128, 128
__shape_col = (__npixels, __nbins)
__shape_nn = (1, __npixels, __nbins, 1)
__model = keras.models.load_model('./model.h5', compile=True, custom_objects={'res_sum_squares': __res_sum_squares})

__max_parallel_predictions = 4
__sema = threading.BoundedSemaphore(value=__max_parallel_predictions)


def __mt_pathos_manager(col_ratio):
    return __denoise(col_ratio[0], col_ratio[1])


def __denoise_frame_mt_pathos(frame_ratios):
    results = pathos.multiprocessing.ProcessingPool().map(__mt_pathos_manager, frame_ratios)
    return results


def __denoise_frame_mt_multiprocessing(frame_ratios):
    pool = multiprocessing.Pool()
    results = pool.map(__denoise, map(lambda col_ratio: col_ratio, frame_ratios))
    pool.close()
    pool.join()
    return results


def __denoise(col, ratio=None):
    """
        :param col: the source column
        :param ratio: logging purposes
        :return: the denoised column
    """
    really_predict = True
    if type(col) is tuple:
        col, ratio = col[0], col[1]
    col_denoise = np.reshape(col, __shape_nn)

    print("{} acquiring".format(ratio))
    __sema.acquire()
    print("{} acquired".format(ratio))
    #  ~    ~  ~  ~  ~  ~  ~  ~  ~  ~ CRITICAL SECTION START ~  ~  ~  ~  ~  ~  ~  ~  ~  ~
    col_denoise = __model.predict(col_denoise) if really_predict else col_denoise
    #  ~    ~  ~  ~  ~  ~  ~  ~  ~  ~ CRITICAL SECTION END   ~  ~  ~  ~  ~  ~  ~  ~  ~  ~
    print("{} releasing".format(ratio))
    __sema.release()
    print("{} released".format(ratio))

    return np.reshape(col_denoise, __shape_col)


def denoise_cube(cube, mp=False, mp_pathos=False):
    """
        :param cube: a numpy 3D array of ncols * npixels * nbins
        :param mp: use multiprocessing
        :param mp_pathos: use pathos multiprocessing
        :return: the denoised cube
    """
    ncol = cube.shape[0]
    ratios = [(ic * 100.0) / ncol for ic in range(0, ncol)]
    frame_ratios = zip(cube, ratios)

    if mp:
        if mp_pathos:
            l_cols_denoised = __denoise_frame_mt_pathos(frame_ratios)
        else:
            l_cols_denoised = __denoise_frame_mt_multiprocessing(frame_ratios)
    else:
        l_cols_denoised = [__denoise(col, ratio) for col, ratio in frame_ratios]
    return l_cols_denoised


if __name__ == "__main__":

    test_cube = np.random.rand(1000, __npixels, __nbins)

    # Single threaded impl: works fine
    denoise_cube(test_cube, mp=False)
    # Multiprocessing Pool: blocks at the eighth "acquired" print
    denoise_cube(test_cube, mp=True, mp_pathos=False)
    # Pathos multiprocessing Pool: blocks at the eighth "acquired" print
    denoise_cube(test_cube, mp=True, mp_pathos=True)

分析

我猜到的第一件事是,经过8次调用(=测试机上cpu内核的数量),对__model.predict()的抢购以某种方式被阻止了。 因此,我放置了一个threading.BoundedSemaphore,访问次数少于8。什么都没有。

单线程按预期工作:

0.0 acquiring
0.0 acquired
0.0 releasing
0.0 released
< ............ >
99.9 acquiring
99.9 acquired
99.9 releasing
99.9 released

多重处理(两个版本)都没有。

0.0 acquiring
0.0 acquired
3.2 acquiring
3.2 acquired
6.4 acquiring
6.4 acquired
9.6 acquiring
9.6 acquired
12.8 acquiring
12.8 acquired
16.0 acquiring
16.0 acquired
19.2 acquiring
19.2 acquired
22.4 acquiring
22.4 acquired
< hangs >

等等,release打印在哪里?似乎信号量没有被触摸,或者正在为每个呼叫复制,并且总是重新初始化。嗯。

因此,让我们寻找really_predict = True并交换其值:predict()永远不会以这种方式到达。

....这很好,很好!!!因此,问题不能完全解决multiprocessing,而是解决keras预测和multiprocessing池之间的奇怪链接。有什么建议吗?

0 个答案:

没有答案