matplotlib.pyplot/pylab在使用ipython -pylab进行isinteractive()时不更新图形

问题描述

关于matplotlib，pylab，pyplot，ipython的问题很多，所以很抱歉如果您厌倦了看到这个问题.我将尽我所能，因为我一直在研究人们的问题并查看pyplot和pylab的文档，但我仍然不确定自己做错了什么.上的代码:

There are a lot of questions about matplotlib, pylab, pyplot, ipython, so I'm sorry if you're sick of seeing this asked. I'll try to be as specific as I can, because I've been looking through people's questions and looking at documentation for pyplot and pylab, and I still am not sure what I'm doing wrong. On with the code:

目标:每0.5秒绘制一个图形，并在调用plot命令后立即更新图形.

Goal: plot a figure every .5 seconds, and update the figure as soon as the plot command is called.

我尝试对此进行编码(在ipython -pylab上运行):

My attempt at coding this follows (running on ipython -pylab):

import time
ion()
x=linspace(-1,1,51)
plot(sin(x))
for i in range(10):
    plot([sin(i+j) for j in x])
    #see **
    print i
    time.sleep(1)
print 'Done'

它可以正确绘制每条线，但是要等到退出for循环后才能绘制.我尝试通过将draw()放在**位置来强制重绘，但这似乎也不起作用.理想情况下，我想让它简单地添加每一行，而不是进行完整的重绘.但是，如果需要重新绘制，就可以了.

It correctly plots each line, but not until it has exited the for loop. I have tried forcing a redraw by putting draw() where ** is, but that doesn't seem to work either. Ideally, I'd like to have it simply add each line, instead of doing a full redraw. If redrawing is required however, that's fine.

其他尝试解决的问题:
在ion()之后，尝试将hold(True)添加到无效.
对于踢球，尝试对**进行踢球show()
我找到的最接近答案的答案是绘图线不会阻止执行，但是show()没有做任何事情.

Additional attempts at solving:
just after ion(), tried adding hold(True) to no avail.
for kicks tried show() for **
The closest answer I've found to what I'm trying to do was at plotting lines without blocking execution, but show() isn't doing anything.

如果这是一个简单的要求，我深表歉意，而我正在寻找一些显而易见的东西.对于它的价值，这是在我尝试将Matlab代码从类转换为某些python供我自己使用时出现的.我一直在尝试转换的原始matlab(删除了初始化)如下:

I apologize if this is a straightforward request, and I'm looking past something so obvious. For what it's worth, this came up while I was trying to convert matlab code from class to some python for my own use. The original matlab (initializations removed) which I have been trying to convert follows:

for i=1:time
    plot(u)
    hold on
    pause(.01)
    for j=2:n-1
        v(j)=u(j)-2*u(j-1)
    end
    v(1)= pi
    u=v
end

任何帮助，即使只是查找this_method"也将是很好的，因此，我至少可以将工作范围缩小到弄清楚如何使用该方法.如果还有其他有用的信息，请告诉我.

Any help, even if it's just "look up this_method" would be excellent, so I can at least narrow my efforts to figuring out how to use that method. If there's any more information that would be useful, let me know.

推荐答案

所链接问题的第二个答案提供了答案:在每个plot()之后调用draw()使其立即显示；例如:

The second answer to the question you linked provides the answer: call draw() after every plot() to make it appear immediately; for example:

import time
ion()
x = linspace(-1,1,51)
plot(sin(x))
for i in range(10):
    plot([sin(i+j) for j in x])
    # make it appear immediately
    draw()
    time.sleep(1)

如果这不起作用，请尝试在此页面上执行以下操作: http://www.scipy.org/Cookbook/Matplotlib/Animations

If that doesn't work... try what they do on this page: http://www.scipy.org/Cookbook/Matplotlib/Animations

import time

ion()

tstart = time.time()               # for profiling
x = arange(0,2*pi,0.01)            # x-array
line, = plot(x,sin(x))
for i in arange(1,200):
    line.set_ydata(sin(x+i/10.0))  # update the data
    draw()                         # redraw the canvas

print 'FPS:' , 200/(time.time()-tstart)

页面提到line.set_ydata()功能是关键部分.

The page mentions that the line.set_ydata() function is the key part.

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