Question

我有gzip压缩器，出于这个问题的目的可以看作是一个简单的缓冲区。

压缩器写入NIO2 AsynchronousFileChannel.write()。一旦有足够的压缩字节，就刷新缓冲区并启动异步AsynchronousFileChannel调用。

一旦所有写入完成，压缩器可能仍包含数据。所以我需要：

flush＆amp;关闭压缩机
执行最后一次写作
关闭AsynchronousFileChannel.close()本身。

close()的文档声明了

此通道上的任何未完成的异步操作都将完成AsynchronousCloseException异常

这可能包括在调用AsynchronousFileChannel channel = ... ... //Is there a chance that this write will throw an AsynchronousCloseException? channel.write(...); channel.close();之前由同一个线程提交的最后一次写入吗？

换句话说，以下是否安全？

import numpy as np
import scipy
import scipy.optimize as op
import scipy.stats
import scipy.integrate



def fit_model(model_pmt, x_list, y_list, PMT_parra, PMT_bounds=None, tolerance=10**-1, PMT_start_gues=None):
    result = op.minimize(chi_squared, PMT_start_gues, args=(x_list, y_list, model_pmt, PMT_parra[0], PMT_parra[1], PMT_parra[2]),
                     bounds=PMT_bounds, method='SLSQP', options={"ftol": tolerance})
    print result



def chi_squared(fit_parm, x, y_val, model, *non_fit_parm):
    parm = np.concatenate((fit_parm, non_fit_parm))
    y_mod = model(x, *parm)
    X2 = sum(pow(y_val - y_mod, 2))
    return X2



def basic_model(cb_list, max_intesity, sigma_e, noise, N, centre1, centre2, sigma_eb, min_dist=10**-5):
        """
        plateau function consisting of two gaussian CDF functions.
        """
        def get_distance(x, r):
            dist = abs(x - r)
            if dist < min_dist:
                dist = min_dist
            return dist

        def amount_of_material(x):
            A = scipy.stats.norm.cdf((x - centre1) / sigma_e)
            B = (1 - scipy.stats.norm.cdf((x - centre2) / sigma_e))
            cube =  A * B
            return cube

        def amount_of_field_INTEGRAL(x, cb):
        """Integral that is part of my sum"""
            result = scipy.integrate.quad(lambda r: scipy.stats.norm.pdf((r - cb) / sigma_b) / pow(get_distance(x, r), N),
                                          start, end, epsabs=10 ** -1)[0]
            return result



        # Set some constants, not important
        sigma_b = (sigma_eb**2-sigma_e**2)**0.5
        start, end = centre1 - 3 * sigma_e, centre2 + 3 * sigma_e
        integration_range = np.linspace(start, end, int(end - start) / 20)
        intensity_list = []

        # Doing a riemann sum, this is what takes the most time.
        for i, cb_point in enumerate(cb_list):
            intensity = sum([amount_of_material(x) * amount_of_field_INTEGRAL(x, cb_point) for x in integration_range])
            intensity *= (integration_range[1] - integration_range[0])
            intensity_list.append(intensity)


        model_values = np.array(intensity_list) / max(intensity_list)* max_intesity + noise
        return model_values


def get_dummy_data():
"""Can be ignored, produces something resembling my data with noise"""
    # X is just a range
    x_list = np.linspace(0, 300, 300)

    # Y is some sort of step function with noise
    A = scipy.stats.norm.cdf((x_list - 100) / 15.8)
    B = (1 - scipy.stats.norm.cdf((x_list - 200) / 15.8))
    y_list = A * B * .8 + .1 + np.random.normal(0, 0.05, 300)

    return x_list, y_list


if __name__=="__main__":
    # Set some variables
    start_pmt = [0.7, 8, 0.15, 0.6]
    pmt_bounds = [(.5, 1.3), (4, 15), (0.05, 0.3), (0.5, 3)]
    pmt_par = [110, 160, 15]
    x_list, y_list = get_dummy_data()

    fit_model(basic_model, x_list, y_list,  pmt_par, PMT_start_gues=start_pmt, PMT_bounds=pmt_bounds, tolerance=0.1)

或者我需要阻止直到最后一次写回来？

Answer 1

这可能包括在调用close（）之前由同一个线程提交的最后一次写入吗？

是的，当然。

换句话说，以下是否安全？

没有

或者我需要阻止直到最后一次写回来？

您需要在最后一次写入的完成处理程序中执行close操作，或者在成功获取其Future之后执行close操作。

AsynchronousFileChannel - 确保在写入完成后发生close（）

1 个答案: