How to update a plot in matplotlib


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

    result = collections.defaultdict(list)
    for d in vlgaBuffSorted:

    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:

    plotCanvas = FigureCanvasTkAgg(f, pltFrame)
    toolbar = NavigationToolbar2TkAgg(plotCanvas, pltFrame)
Asked By: thenickname



You essentially have two options:

  1. 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.

  2. 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...

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))
Answered By: Joe Kington

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!

Answered By: jkeyser

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?


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,
    for k in range(nBins):
        frame   = frames[k]
        tempCS1 = plt.imshow(frame,
        del tempCS1
        #time.sleep(1e-2) #unnecessary, but useful

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):
    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.

Answered By: CavemanPhD

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)

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

Answered By: Scott

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()
        x = range(len(y))
        ax.plot(x, y)
        print i, ': ', yi

    a = anim.FuncAnimation(fig, update, frames=xmax, repeat=False)

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

Answered By: Victor Basso

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

for i in range(50):
    y = np.random.random([10,1])
Answered By: Arindam

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):
    y = np.random.random([10,1])
Answered By: Julian

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.

def plotlive(func):

    def new_func(*args, **kwargs):

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

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

        # Draw the plot

        return result

    return new_func 

Usage example

And then you can use it like any other decorator.

def plot_something_live(ax, x, y):
    ax.plot(x, y)
    ax.set_ylim([0, 100])

The only constraint is that you have to create the figure before the loop:

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)
Answered By: Ladislav Ondris
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