Question

如果使用--pylab=...或--gui=...等选项调用IPython，是否存在在解释器内部进行检测的规范方法？

原因：我想在单独的过程中进行一些异步绘图，如示例脚本tst_process.py中所示：

#!/usr/bin/env python
# -*- coding: utf-8 -*-
""" File tst_process.py """
# For better Python 3 compatibility:
from __future__ import absolute_import, print_function, unicode_literals, \
                       division

import matplotlib.pylab as plt
from multiprocessing import Process
import numpy as np

def tst_plot(fgoff=0):
    """ Make a test plot """
    print("Drawing figure {}".format(1+fgoff))
    x = np.linspace(0, 5, 500)
    fg = plt.figure(1+fgoff)
    fg.clf()
    ax = fg.add_subplot(1, 1, 1)
    ax.plot(x, np.sin(x))
    ax.set_title("This is a Test-Plot")
    fg.canvas.draw()
    plt.show()


if __name__ == "__main__":

    print("Doing testplot in new process ...")
    pprc1 = Process(target=tst_plot)
    pprc1.start()
    print("Doing testplot in own process ...")
    tst_plot(10)

当我使用命令

运行它时

ipython --i tst_process.py

一切都按预期工作。这样做的：

ipython --pylab=qt --i tst_process.py

给出：

Python 2.7.9 (default, Dec 11 2014, 08:58:12) 
Type "copyright", "credits" or "license" for more information.

IPython 2.3.0 -- An enhanced Interactive Python.
?         -> Introduction and overview of IPython's features.
%quickref -> Quick reference.
help      -> Python's own help system.
object?   -> Details about 'object', use 'object??' for extra details.
Doing testplot in new process ...
Doing testplot in own process ...
Drawing figure 11
Drawing figure 1
 : Fatal IO error: client killed
X Error: BadIDChoice (invalid resource ID chosen for this connection) 14
  Major opcode: 1 (X_CreateWindow)
  Resource id:  0x6a00003
X Error: BadIDChoice (invalid resource ID chosen for this connection) 14
  Extension:    139 (RENDER)
  Minor opcode: 4 (RenderCreatePicture)
  Resource id:  0x6a00004
X Error: BadIDChoice (invalid resource ID chosen for this connection) 14
  Major opcode: 1 (X_CreateWindow)
  Resource id:  0x6a00005
[xcb] Unknown sequence number while processing queue
[xcb] Most likely this is a multi-threaded client and XInitThreads has not been called
[xcb] Aborting, sorry about that.
python: ../../src/xcb_io.c:274: poll_for_event: Zusicherung »!xcb_xlib_threads_sequence_lost« nicht erfüllt.
Abgebrochen

除wx之外的其他后端也不起作用。

对我来说，检测事件循环的存在就足够了。然后我可以使用相同的脚本从命令行和Spyder内部运行。

Answer 1

回答原始问题：Is there a canonical way of detecting inside the interpreter if IPython was called with options like--pylab=... or --gui=...?

是的，有。最简单的方法是检查命令行参数：

import sys
print sys.argv # returns the commandline arguments
# ['ipython', '--pylab', 'inline']

更好的方法是使用内置模块optparse。

但是，这只会让您根据命令行参数查看它运行的模式 - 这是您的主要问题。它不会帮助您解决gui事件外观+多处理问题，如评论中提到的@tcaswell。

Answer 2

您可以使用

检查qt事件循环是否正在运行

import PyQt4.QtCore
if PyQt4.QtCore.QCoreApplication.instance():
    print("Event loop detected")

PS：为了更精细地检查事件循环何时开始，您可以使用应用程序startingUp方法。

检测IPython Pylab GUI事件循环是否处于活动状态

2 个答案: