Question

我希望通过更改滑块d_in来更改对象（椭圆形）的x位置（= d_in）。这就是我得到的：

from numpy import pi, sin
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.widgets import Slider, Button, RadioButtons
from matplotlib.patches import Ellipse
from scipy.optimize import fsolve
import pylab


axis_color = 'lightgoldenrodyellow'
#variable
d_in=80

fig = plt.figure(figsize=(12,6))
ax = fig.add_subplot(111)
fig.subplots_adjust(left=0.25, bottom=0.35)
x = np.arange(0.0, 300, 0.01)


# object ellipse
Spiegel = Ellipse(xy=(d_in, 0), width=2, height=73.2,
                        edgecolor='black', fc='#808080', lw=1)
ax.add_patch(Spiegel)


#Draw d_in slider
d_in_slider_ax = fig.add_axes([0.25, 0.1, 0.65, 0.03], axisbg=axis_color)
d_in_slider = Slider(d_in_slider_ax, 'd_in', 1, 150, valinit=d_in)


#axis range
ax.set_xlim([0, 300])
ax.set_ylim([-40, 40])

plt.show()

如何告诉滑块更改椭圆的位置？

谢谢

Answer 1

您需要在滑块值更改时注册一个回调，

slider.on_changed(callback)

并在该回调函数中，更新椭圆的位置

ellipse.center = new_center

在这里，它可能如下所示

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.widgets import Slider
from matplotlib.patches import Ellipse

#variable
d_in=80

fig = plt.figure(figsize=(12,6))
ax = fig.add_subplot(111)
fig.subplots_adjust(left=0.25, bottom=0.35)
x = np.arange(0.0, 300, 0.01)


# object ellipse
Spiegel = Ellipse(xy=(d_in, 0), width=2, height=73.2,
                  edgecolor='black', fc='#808080', lw=1)
ax.add_patch(Spiegel)


#Draw d_in slider
d_in_slider_ax = fig.add_axes([0.25, 0.1, 0.65, 0.03])
d_in_slider = Slider(d_in_slider_ax, 'd_in', 1, 150, valinit=d_in)

def update(val):
    Spiegel.center = (val,0)

d_in_slider.on_changed(update)


#axis range
ax.set_xlim([0, 300])
ax.set_ylim([-40, 40])

plt.show()

对象上的matplotlib小部件滑块（椭圆）

1 个答案: