如何减少python中多处理的时间

时间:2016-11-22 17:06:03

标签: python multiprocessing overhead-minimization

我正在尝试在python中构建多处理以降低计算速度,但似乎在多处理之后,整体计算速度显着下降。我创建了4个不同的进程,并将dataFrame拆分为4个不同的数据帧,这些数据帧将作为每个进程的输入。在对每个流程进行计时后,似乎开销成本很高,并且想知道是否有办法降低这些管理费用。

我使用的是windows7,python 3.5,我的机器有8个核心。

def doSomething(args, dataPassed,):

    processing data, and calculating outputs

def parallelize_dataframe(df, nestedApply):
    df_split = np.array_split(df, 4)
    pool = multiprocessing.Pool(4)
    df = pool.map(nestedApply, df_split)
    print ('finished with Simulation')
    time = float((dt.datetime.now() - startTime).total_seconds())

    pool.close()
    pool.join()

def nestedApply(df):

    func2 = partial(doSomething, args=())
    res = df.apply(func2, axis=1)
    res =  [output Tables]
    return res

if __name__ == '__main__':

data = pd.read_sql_query(query, conn)

parallelize_dataframe(data, nestedApply)

1 个答案:

答案 0 :(得分:1)

我建议使用队列而不是将DataFrame作为块提供。你需要大量的资源来复制每个块,这需要相当长的时间才能完成。如果你的DataFrame非常大,你可能会耗尽内存。使用队列可以从pandas中的快速迭代器中受益。 这是我的方法。开销随着工人的复杂性而降低。不幸的是,我的工作人员很难真正证明这一点,但sleep模拟了复杂性。

import pandas as pd
import multiprocessing as mp
import numpy as np
import time


def worker(in_queue, out_queue):
    for row in iter(in_queue.get, 'STOP'):
        value = (row[1] * row[2] / row[3]) + row[4]
        time.sleep(0.1)
        out_queue.put((row[0], value))

if __name__ == "__main__":
    # fill a DataFrame
    df = pd.DataFrame(np.random.randn(1e5, 4), columns=list('ABCD'))

    in_queue = mp.Queue()
    out_queue = mp.Queue()

    # setup workers
    numProc = 2
    process = [mp.Process(target=worker,
                          args=(in_queue, out_queue)) for x in range(numProc)]

    # run processes
    for p in process:
        p.start()

    # iterator over rows
    it = df.itertuples()

    # fill queue and get data
    # code fills the queue until a new element is available in the output
    # fill blocks if no slot is available in the in_queue
    for i in range(len(df)):
        while out_queue.empty():
            # fill the queue
            try:
                row = next(it)
                in_queue.put((row[0], row[1], row[2], row[3], row[4]), block=True)  # row = (index, A, B, C, D) tuple
            except StopIteration:
                break
        row_data = out_queue.get()
        df.loc[row_data[0], "Result"] = row_data[1]

    # signals for processes stop
    for p in process:
        in_queue.put('STOP')

    # wait for processes to finish
    for p in process:
        p.join()

使用numProc = 2每个循环需要50秒,numProc = 4速度是其两倍。