问题:如何在matplotlib中更新图?
我在这里重新绘制图形时遇到问题。我允许用户在时间刻度(x轴)中指定单位,然后重新计算并调用此函数plots()
。我希望该图可以简单地更新,而不是将另一个图添加到该图上。
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)
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 plots()
. I want the plot to simply update, not append another plot to the figure.
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)
回答 0
您基本上有两个选择:
完全执行当前操作,但是在重新配置数据之前先调用graph1.clear()
和graph2.clear()
。这是最慢,但最简单和最可靠的选择。
除了重新绘制外,您还可以更新绘图对象的数据。您需要在代码中进行一些更改,但这应该比每次重新绘制都快得多。但是,您要绘制的数据的形状无法更改,并且如果数据范围正在更改,则需要手动重置x和y轴限制。
举一个第二种选择的例子:
import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(0, 6*np.pi, 100)
y = np.sin(x)
# You probably won't need this if you're embedding things in a tkinter plot...
plt.ion()
fig = plt.figure()
ax = fig.add_subplot(111)
line1, = ax.plot(x, y, 'r-') # Returns a tuple of line objects, thus the comma
for phase in np.linspace(0, 10*np.pi, 500):
line1.set_ydata(np.sin(x + phase))
fig.canvas.draw()
fig.canvas.flush_events()
You essentially have two options:
Do exactly what you’re currently doing, but call graph1.clear()
and graph2.clear()
before replotting the data. This is the slowest, but most simplest and most robust option.
Instead of replotting, you can just update the data of the plot objects. You’ll need to make some changes in your code, but this should be much, much faster than replotting things every time. However, the shape of the data that you’re plotting can’t change, and if the range of your data is changing, you’ll need to manually reset the x and y axis limits.
To give an example of the second option:
import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(0, 6*np.pi, 100)
y = np.sin(x)
# You probably won't need this if you're embedding things in a tkinter plot...
plt.ion()
fig = plt.figure()
ax = fig.add_subplot(111)
line1, = ax.plot(x, y, 'r-') # Returns a tuple of line objects, thus the comma
for phase in np.linspace(0, 10*np.pi, 500):
line1.set_ydata(np.sin(x + phase))
fig.canvas.draw()
fig.canvas.flush_events()
回答 1
您还可以执行以下操作:这将在for循环的50个循环的绘图上绘制10×1随机矩阵数据。
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()
You can also do like the following:
This will draw a 10×1 random matrix data on the plot for 50 cycles of the for loop.
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()
回答 2
这对我有用。每次重复调用更新图形的函数。
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()
“ fun”是一个返回整数的函数。FuncAnimation将反复调用“更新”,它将执行“ xmax”次。
This worked for me. Repeatedly calls a function updating the graph every time.
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()
“fun” is a function that returns an integer.
FuncAnimation will repeatedly call “update”, it will do that “xmax” times.
回答 3
如果有人碰到本文来寻找我想要的东西,我在
如何使用Matplotlib可视化标量2D数据?
和
http://mri.brechmos.org/2009/07/automatically-update-a-figure-in-a-loop(在web.archive.org上)
然后对其进行修改,以将imshow与输入的帧堆栈一起使用,而不是动态生成和使用轮廓。
从3D形状的图像阵列(nBins,nBins,nBins)开始,称为frames
。
def animate_frames(frames):
nBins = frames.shape[0]
frame = frames[0]
tempCS1 = plt.imshow(frame, cmap=plt.cm.gray)
for k in range(nBins):
frame = frames[k]
tempCS1 = plt.imshow(frame, cmap=plt.cm.gray)
del tempCS1
fig.canvas.draw()
#time.sleep(1e-2) #unnecessary, but useful
fig.clf()
fig = plt.figure()
ax = fig.add_subplot(111)
win = fig.canvas.manager.window
fig.canvas.manager.window.after(100, animate_frames, frames)
我还发现了进行此整个过程的简单得多的方法,尽管不够健壮:
fig = plt.figure()
for k in range(nBins):
plt.clf()
plt.imshow(frames[k],cmap=plt.cm.gray)
fig.canvas.draw()
time.sleep(1e-6) #unnecessary, but useful
请注意,这两个似乎只适用于ipython --pylab=tk
,backend = TkAgg
感谢您提供的所有帮助。
In case anyone comes across this article looking for what I was looking for, I found examples at
How to visualize scalar 2D data with Matplotlib?
and
http://mri.brechmos.org/2009/07/automatically-update-a-figure-in-a-loop
(on web.archive.org)
then modified them to use imshow with an input stack of frames, instead of generating and using contours on the fly.
Starting with a 3D array of images of shape (nBins, nBins, nBins), called frames
.
def animate_frames(frames):
nBins = frames.shape[0]
frame = frames[0]
tempCS1 = plt.imshow(frame, cmap=plt.cm.gray)
for k in range(nBins):
frame = frames[k]
tempCS1 = plt.imshow(frame, cmap=plt.cm.gray)
del tempCS1
fig.canvas.draw()
#time.sleep(1e-2) #unnecessary, but useful
fig.clf()
fig = plt.figure()
ax = fig.add_subplot(111)
win = fig.canvas.manager.window
fig.canvas.manager.window.after(100, animate_frames, frames)
I also found a much simpler way to go about this whole process, albeit less robust:
fig = plt.figure()
for k in range(nBins):
plt.clf()
plt.imshow(frames[k],cmap=plt.cm.gray)
fig.canvas.draw()
time.sleep(1e-6) #unnecessary, but useful
Note that both of these only seem to work with ipython --pylab=tk
, a.k.a.backend = TkAgg
Thank you for the help with everything.
回答 4
我发布了一个名为python-drawnow的软件包,该软件包提供了使图形更新的功能(通常在for循环内调用),类似于Matlab的drawnow
。
用法示例:
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)
该软件包适用于任何matplotlib图形,并提供了在每个图形更新或放入调试器后等待的选项。
I have released a package called python-drawnow that provides functionality to let a figure update, typically called within a for loop, similar to Matlab’s drawnow
.
An example usage:
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)
This package works with any matplotlib figure and provides options to wait after each figure update or drop into the debugger.
回答 5
以上所有内容可能都是正确的,但是对我来说,“在线更新”数字仅适用于某些后端wx
。您可能只是想尝试更改为该值,例如通过启动ipython / pylab ipython --pylab=wx
!祝好运!
All of the above might be true, however for me “online-updating” of figures only works with some backends, specifically wx
. You just might try to change to this, e.g. by starting ipython/pylab by ipython --pylab=wx
! Good luck!
回答 6
这对我有用:
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()
This worked for me:
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()