blocks|key|1100340|text|以上所有这些可能都是真的，但是对我来说，数字的“在线更新”只适用于一些后端，特别是wx。你可以试着这样做，例如通过ipython+--pylab=wx启动ipython/pylab！祝好运!|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1100341|entityMap^0|15|2|1L|I|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

All of the above might be true, however for me "online-updating" of figures only works with some backends, specifically <code>wx</code>. You just might try to change to this, e.g. by starting ipython/pylab by <code>ipython --pylab=wx</code>! Good luck!

blocks|key|885604|text|我已经发布了一个名为python-drawnow的包，它提供了让图形更新的功能，这通常是在一个for循环中调用的，类似于Matlab的drawnow。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|885605|示例用法：|885606|from+pylab+import+figure,+plot,+ion,+linspace,+arange,+sin,+pi
def+draw_fig():
++++#+can+be+arbitrarily+complex;+just+to+draw+a+figure
++++#figure()+#+don't+call!
++++plot(t,+x)
++++#show()+#+don't+call!

N+=+1e3
figure()+#+call+here+instead!
ion()++++#+enable+interactivity
t+=+linspace(0,+2*pi,+num=N)
for+i+in+arange(100):
++++x+=+sin(2+*+pi+*+i**2+*+t+/+100.0)
++++drawnow(draw_fig)|code-block|syntax|javascript|885607|这个包可以处理任何matplotlib图形，并提供在每次图形更新或放入调试器后等待的选项。|885608|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/scottsievert/python-drawnow^0|1V|7|A|E|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|6|7|12|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|13|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|14|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

I have released a package called <a href="https://github.com/scottsievert/python-drawnow" rel="noreferrer">python-drawnow</a> that provides functionality to let a figure update, typically called within a for loop, similar to Matlab's <code>drawnow</code>.

An example usage:

<pre><code>from pylab import figure, plot, ion, linspace, arange, sin, pi
def draw_fig():
 # can be arbitrarily complex; just to draw a figure
 #figure() # don't call!
 plot(t, x)
 #show() # don't call!

N = 1e3
figure() # call here instead!
ion() # enable interactivity
t = linspace(0, 2*pi, num=N)
for i in arange(100):
 x = sin(2 * pi * i**2 * t / 100.0)
 drawnow(draw_fig)
</code></pre>

This package works with any matplotlib figure and provides options to wait after each figure update or drop into the debugger.

blocks|key|1100453|text|这对我很有效。每次都会重复调用更新图形的函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1100454|import+matplotlib.pyplot+as+plt
import+matplotlib.animation+as+anim

def+plot_cont(fun,+xmax):
++++y+=+[]
++++fig+=+plt.figure()
++++ax+=+fig.add_subplot(1,1,1)

++++def+update(i):
++++++++yi+=+fun()
++++++++y.append(yi)
++++++++x+=+range(len(y))
++++++++ax.clear()
++++++++ax.plot(x,+y)
++++++++print+i,+':+',+yi

++++a+=+anim.FuncAnimation(fig,+update,+frames=xmax,+repeat=False)
++++plt.show()|code-block|syntax|javascript|1100455|"fun“是一个返回整数的函数。FuncAnimation将重复调用"update"，它将执行"xmax“次。|1100456|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This worked for me. Repeatedly calls a function updating the graph every time.

<pre><code>import matplotlib.pyplot as plt
import matplotlib.animation as anim

def plot_cont(fun, xmax):
 y = []
 fig = plt.figure()
 ax = fig.add_subplot(1,1,1)

 def update(i):
 yi = fun()
 y.append(yi)
 x = range(len(y))
 ax.clear()
 ax.plot(x, y)
 print i, ': ', yi

 a = anim.FuncAnimation(fig, update, frames=xmax, repeat=False)
 plt.show()
</code></pre>

"fun" is a function that returns an integer.
FuncAnimation will repeatedly call "update", it will do that "xmax" times.

blocks|key|885663|text|您还可以像下面这样做:这将在for循环的50个循环中绘制一个10x1随机矩阵数据。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|885664|import+matplotlib.pyplot+as+plt
import+numpy+as+np

plt.ion()
for+i+in+range(50):
++++y+=+np.random.random([10,1])
++++plt.plot(y)
++++plt.draw()
++++plt.pause(0.0001)
++++plt.clf()|code-block|syntax|javascript|885665|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can also do like the following:
This will draw a 10x1 random matrix data on the plot for 50 cycles of the for loop.

<pre><code>import matplotlib.pyplot as plt
import numpy as np

plt.ion()
for i in range(50):
 y = np.random.random([10,1])
 plt.plot(y)
 plt.draw()
 plt.pause(0.0001)
 plt.clf()
</code></pre>

blocks|key|885670|text|这对我很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|885671|from+matplotlib+import+pyplot+as+plt
from+IPython.display+import+clear_output
import+numpy+as+np
for+i+in+range(50):
++++clear_output(wait=True)
++++y+=+np.random.random([10,1])
++++plt.plot(y)
++++plt.show()|code-block|syntax|javascript|885672|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This worked for me:

<pre><code>from matplotlib import pyplot as plt
from IPython.display import clear_output
import numpy as np
for i in range(50):
 clear_output(wait=True)
 y = np.random.random([10,1])
 plt.plot(y)
 plt.show()
</code></pre>

blocks|key|885715|text|根据其他答案，我将图的更新包装在python装饰器中，以将图的更新机制与实际图分开。这样，更新任何图就容易多了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|885716|def+plotlive(func):
++++plt.ion()

++++@functools.wraps(func)
++++def+new_func(*args,+**kwargs):

++++++++#+Clear+all+axes+in+the+current+figure.
++++++++axes+=+plt.gcf().get_axes()
++++++++for+axis+in+axes:
++++++++++++axis.cla()

++++++++#+Call+func+to+plot+something
++++++++result+=+func(*args,+**kwargs)

++++++++#+Draw+the+plot
++++++++plt.draw()
++++++++plt.pause(0.01)

++++++++return+result

++++return+new_func+|code-block|syntax|javascript|885717|使用示例|885718|然后你就可以像使用其他装饰器一样使用它了。|885719|@plotlive
def+plot_something_live(ax,+x,+y):
++++ax.plot(x,+y)
++++ax.set_ylim([0,+100])|885720|唯一的限制是您必须在循环之前创建图形：|885721|fig,+ax+=+plt.subplots()
for+i+in+range(100):
++++x+=+np.arange(100)
++++y+=+np.full([100],+fill_value=i)
++++plot_something_live(ax,+x,+y)|885722|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Based on the other answers, I wrapped the figure's update in a python decorator to separate the plot's update mechanism from the actual plot. This way, it is much easier to update any plot.
<pre><code>def plotlive(func):
 plt.ion()

 @functools.wraps(func)
 def new_func(*args, **kwargs):

 # Clear all axes in the current figure.
 axes = plt.gcf().get_axes()
 for axis in axes:
 axis.cla()

 # Call func to plot something
 result = func(*args, **kwargs)

 # Draw the plot
 plt.draw()
 plt.pause(0.01)

 return result

 return new_func 
</code></pre>
<h2>Usage example</h2>
And then you can use it like any other decorator.
<pre><code>@plotlive
def plot_something_live(ax, x, y):
 ax.plot(x, y)
 ax.set_ylim([0, 100])
</code></pre>
The only constraint is that you have to create the figure before the loop:
<pre><code>fig, ax = plt.subplots()
for i in range(100):
 x = np.arange(100)
 y = np.full([100], fill_value=i)
 plot_something_live(ax, x, y)
</code></pre>

I'm having issues with redrawing the figure here. I allow the user to specify the units in the time scale (x-axis) and then I recalculate and call this function <code>plots()</code>. I want the plot to simply update, not append another plot to the figure.

<pre><code>def plots():
 global vlgaBuffSorted
 cntr()

 result = collections.defaultdict(list)
 for d in vlgaBuffSorted:
 result[d['event']].append(d)

 result_list = result.values()

 f = Figure()
 graph1 = f.add_subplot(211)
 graph2 = f.add_subplot(212,sharex=graph1)

 for item in result_list:
 tL = []
 vgsL = []
 vdsL = []
 isubL = []
 for dict in item:
 tL.append(dict['time'])
 vgsL.append(dict['vgs'])
 vdsL.append(dict['vds'])
 isubL.append(dict['isub'])
 graph1.plot(tL,vdsL,'bo',label='a')
 graph1.plot(tL,vgsL,'rp',label='b')
 graph2.plot(tL,isubL,'b-',label='c')

 plotCanvas = FigureCanvasTkAgg(f, pltFrame)
 toolbar = NavigationToolbar2TkAgg(plotCanvas, pltFrame)
 toolbar.pack(side=BOTTOM)
 plotCanvas.get_tk_widget().pack(side=TOP)
</code></pre>

How to update a plot in matplotlib?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我在这里重新绘制图形时遇到了问题。我允许用户指定时间刻度(x轴)中的单位，然后重新计算并调用此函数plots()。我想简单地更新绘图，而不是将另一个绘图附加到图中。def plots():    global vlgaBuffSorted    cntr()    result = collections.defaul...

问如何在matplotlib中更新绘图？
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问如何在matplotlib中更新绘图？
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