Matplotlib冻结当Spyder中使用input（）时

2 回答

• 3

  经过更多的挖掘后，我得出的结论是，你应该制作一个GUI . 我建议你使用PySide或PyQt . 为了使matplotlib具有图形窗口，它运行一个事件循环 . 任何单击或鼠标移动都会触发一个事件，触发图形部分执行某些操作 . 脚本编写的问题是每个代码都是顶级的;它表明代码是按顺序运行的 .

  当您手动将代码输入ipython控制台时，它可以工作！这是因为ipython已经启动了一个GUI事件循环 . 您调用的每个命令都在事件循环中处理，允许其他事件发生 .

  您应该创建一个GUI并将GUI后端声明为相同的matplotlib后端 . 如果你有一个按钮点击触发anomaly_selection函数，那么该函数在一个单独的线程中运行，并且应该允许你仍然在GUI中进行交互 .

  有很多摆弄和移动你调用fucntions的方式你可以让thread_input函数工作 .

  幸运的是，PySide和PyQt允许您手动调用处理GUI事件 . 我添加了一个方法，要求在单独的线程中输入并循环等待结果 . 在等待时，它告诉GUI处理事件 . 如果安装了PySide（或PyQt）并将其用作matplotlib的后端， return_input 方法将有用 .

  import threading
  
  def _get_input(msg, func):
      """Get input and run the function."""
      value = input(msg)
      if func is not None:
          func(value)
      return value
  # end _get_input
  
  def thread_input(msg="", func=None):
      """Collect input from the user without blocking. Call the given function when the input has been received.
  
      Args:
          msg (str): Message to tell the user.
          func (function): Callback function that will be called when the user gives input.
      """
      th = threading.Thread(target=_get_input, args=(msg, func))
      th.daemon = True
      th.start()
  # end thread_input
  
  def return_input(msg=""):
      """Run the input method in a separate thread, and return the input."""
      results = []
      th = threading.Thread(target=_store_input, args=(msg, results))
      th.daemon = True
      th.start()
      while len(results) == 0:
          QtGui.qApp.processEvents()
          time.sleep(0.1)
  
      return results[0]
  # end return_input
  
  if __name__ == "__main__":
  
      stop = [False]
      def stop_print(value):
          print(repr(value))
          if value == "q":
              stop[0] = True
              return
          thread_input("Enter value:", stop_print)
  
      thread_input("Enter value:", stop_print)
      add = 0
      while True:
          add += 1
          if stop[0]:
              break
  
      print("Total value:", add)
  

  这段代码似乎对我有用 . 虽然它确实给了我一些ipython内核的问题 .

  from matplotlib import pyplot as pl
  
  import threading
  
  
  def anomaly_selection(selected, indexes, fig, ax):
      for i in range(0, len(indexes)):
          index = indexes[i]
          ax.set_xlim(index-100, index+100)
          ax.autoscale_view()
          #fig.canvas.draw_idle() # Do not need because of pause
          print("[%d/%d] Index %d " % (i, len(indexes), index), end="")
          while True:
              response = input("Particle? ")
              if response == "y":
                  selected.append(index)
                  break
              elif response == "x":
                  selected[0] = True
                  return selected
              elif response == "n":
                  break
  
      selected[0] = True
      return selected
  
  
  fig, ax = pl.subplots(2, sharex=True)
  ax[0].plot([1, 2, 3, 4, 5]) # just pretend data
  pl.show(block=False)
  
  sel = [False]
  th = threading.Thread(target=anomaly_selection, args=(sel, [100, 1000, 53000, 4300], fig, ax[0]))
  th.start()
  #sel = anomaly_selection([100, 1000, 53000, 4300], fig, ax[0])
  
  
  while not sel[0]:
      pl.pause(1)
  th.join()
  

  回复于 2024-04-26T07:54:44+08:00
• 1

  知道比我更好的人，如果可能的话请发一个答案 . 我对Python / Scipy / Spyder知之甚少

  这是我编写的一个kludgy模块，可以防止Matplotlib窗口在Spyder下输入（）处于挂起状态时冻结 .

  您必须事先调用 prompt_hack.start() 并在之后调用 prompt_hack.finish() ，并将 input() 替换为 prompt_hack.input()

  prompt_hack.py

  import matplotlib.pyplot
  import time
  import threading
  
  # Super Hacky Way of Getting input() to work in Spyder with Matplotlib open
  # No efforts made towards thread saftey!
  
  prompt = False
  promptText = ""
  done = False
  waiting = False
  response = ""
  
  regular_input = input
  
  def threadfunc():
      global prompt
      global done
      global waiting
      global response
  
      while not done:   
          if prompt:   
              prompt = False
              response = regular_input(promptText)
              waiting = True
          time.sleep(0.1)
  
  def input(text):
      global waiting
      global prompt
      global promptText
  
      promptText = text
      prompt = True
  
      while not waiting:
          matplotlib.pyplot.pause(0.01)
      waiting = False
  
      return response
  
  def start():
      thread = threading.Thread(target = threadfunc)
      thread.start()
  
  def finish():
      global done
      done = True
  

  回复于 2024-04-26T07:54:44+08:00