Question

一直在玩Python的绘图库，并且遇到了matplotlib，它似乎已经过战斗测试和验证。但是，我在线程中创建一个简单的绘图时遇到了一个问题。

在下面的示例中， Dummy 类 plotme 方法在一个线程中连续运行两次，但在第二次迭代中会卡住/冻结。很可能是明显的并且与线程本身有关，但到目前为止我没有发现它。

import matplotlib.pyplot as plt
from numpy import arange, sin, pi
import threading

class Dummy():

    def plotme(self, iteration = 1):

        print "%ix plotting... " % iteration,
        t = arange(0.0, 2.0, 0.01)
        s = sin(2*pi*t)
        plt.plot(t, s)
        plt.xlabel('time (s)')
        plt.ylabel('voltage (mV)')
        plt.title('About as simple as it gets, folks')
        #savefig("test.png") # irrelevant here
        plt.close()

    def threadme(self, iteration = 1):

        thread_plot = threading.Thread(target=self.plotme,
                                      args=(iteration,))
        thread_plot.start()
        thread_plot.join()

dummy = Dummy()
dummy.threadme(1)
dummy.threadme(2)

Answer 1

首先，请注意pyplot - 接口不是线程安全的。

然后：使用“Agg”-backend进行非交互式创建多个图像。

一个工作示例（由于线程可能导致的问题）是：

import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt
from numpy import arange, sin, pi
import threading

class Dummy():

    def plotme(self, iteration = 1):

        print "%ix plotting... " % iteration,
        t = arange(0.0, 2.0, 0.01)
        s = sin(2*pi*t)
        plt.plot(t, s)
        plt.xlabel('time (s)')
        plt.ylabel('voltage (mV)')
        plt.title('About as simple as it gets, folks')
        plt.savefig("19110942_%i_test.png" % iteration) # irrelevant here
        plt.clf()

    def threadme(self, iteration = 1):

        thread_plot = threading.Thread(target=self.plotme,
                                      args=(iteration,))
        thread_plot.start()
        thread_plot.join()

dummy = Dummy()
dummy.threadme(1)
dummy.threadme(2)

线程安全版本如下所示：

import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt
from numpy import arange, sin, pi
import threading

class Dummy():

    def plotme(self, iteration = 1):

        print "%ix plotting... " % iteration,
        t = arange(0.0, 2.0, 0.01)
        s = sin(2*pi*t)

        fig, ax = plt.subplots()
        ax.plot(t, s)
        ax.set_xlabel('time (s)')
        ax.set_ylabel('voltage (mV)')
        ax.set_title('About as simple as it gets, folks (%i)' % iteration)
        fig.savefig("19110942_%i_test.png" % iteration)

    def threadme(self, iteration = 1):

        thread_plot = threading.Thread(target=self.plotme,
                                      args=(iteration,))
        thread_plot.start()
        thread_plot.join()

dummy = Dummy()
dummy.threadme(1)
dummy.threadme(2)

matplotlib在简单的线程中绘制冻结

1 个答案: