data_dict = {'x': {(0, 0): 3760.448435678077,
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根据上面的数据，我想制作一个动画群图matplotlib and moviepy。然而，通过每帧的以下代码，我得到了额外的点，但保留了旧的点：
import numpy as np
import pandas as pd
from scipy.stats import gaussian_kde
from matplotlib import pyplot as plt
from moviepy.editor import VideoClip
from moviepy.video.io.bindings import mplfig_to_npimage
fps = 10
df = pd.DataFrame(data_dict)
fig, ax = plt.subplots(1, 1)
def swarm_plot(x):
kde = gaussian_kde(x)
density = kde(x) # estimate the local density at each datapoint
# ax.clear()
jitter = np.random.rand(*x.shape) - .5
# scale the jitter by the KDE estimate and add it to the centre x-coordinate
y = 1 + (density * jitter * 1000 * 2)
ax.scatter(x, y, s = 30, c = 'g')
# plt.axis('off')
return fig
def draw_swarmplot(t):
f = int(t * fps)
fig, ax = plt.subplots(1, 1)
dff = df.loc[f]
return mplfig_to_npimage(swarm_plot(dff['x']))
anim = VideoClip(lambda x: draw_swarmplot(x), duration=2)
anim.to_videofile('swarmplot.mp4', fps=fps)
结果，所有点都在动画中累积。我相信这是因为matplotlib fig and ax对象使用不当。然而，在draw_swarmplot功能我重置fig and ax每次迭代后的对象。尽管如此，我仍然需要初始化fig and ax在两个函数之外都不会出现错误ax目的。因此，我的问题是两者如何fig and ax应该引用，我缺少什么导致我的代码无法按预期工作？
您的范围fig and ax变量受变量范围 and 跨越界限的部分变量和范围文档。具体相关，
当我们在函数内部使用赋值运算符 (=) 时，它的默认行为是创建一个新的局部变量 - 除非局部作用域中已经定义了同名变量。
请注意，警告“除非已经定义了同名变量“实际上仅限于local变量。正如在example,
a = 0
def my_function():
a = 3
print(a)
my_function()
print(a)
这将输出
3
0
这是因为
默认情况下，赋值语句在局部范围内创建变量。所以函数内部的赋值不会修改全局变量[...]
如果要从函数内修改全局变量，请使用关键字global，作为答案@iliar says.
However不建议这样做 -
请注意，从函数内部访问全局变量通常是非常糟糕的做法，修改它们的做法更糟糕。这使得我们很难将我们的程序安排成逻辑上封装的部分，而这些部分不会以意想不到的方式相互影响。如果函数需要访问某些外部值，我们应该将该值作为参数传递给函数。 [...]
两种选择是
class fig and ax into draw_swarmplot().前者
class SwarmPlot:
def __init__(self):
self.fig, self.ax = plt.subplots(1, 1)
anim = VideoClip(lambda x: self.draw_swarmplot(x, self.fig, self.ax), duration=2)
anim.to_videofile('swarmplot.mp4', fps=fps)
def swarm_plot(self, x):
kde = gaussian_kde(x)
density = kde(x) # estimate the local density at each datapoint
jitter = np.random.rand(*x.shape) - .5
y = 1 + (density * jitter * 1000 * 2)
self.ax.scatter(x, y, s = 30, c = 'g')
return self.fig
def draw_swarmplot(self, t, fig, ax):
self.fig, self.ax = plt.subplots(1, 1)
f = int(t * fps)
dff = df.loc[f]
return mplfig_to_npimage(self.swarm_plot(dff['x']))
S = SwarmPlot()
后者
def draw_swarmplot(t, fig, ax):
fig, ax = plt.subplots(1, 1)
f = int(t * fps)
dff = df.loc[f]
return mplfig_to_npimage(swarm_plot(dff['x']))
anim = VideoClip(lambda x: draw_swarmplot(x, fig, ax), duration=2)
对于像这样的简单情况，我可能会偏向后者，但在更复杂的情况下，前者可能更可取。两者似乎都能正确生成所需的输出：
当然，如果你不覆盖的话，这一切都是可以避免的figure and axis通过使用清除函数之一来代替每次迭代中的实例：
fig.clear()清除图形fig（相当于plt.clf() if fig是当前的数字）ax（相当于plt.cla() if ax是个
当前轴）ax.clear() or plt.cla()在这种情况下可能是最合适的，并且可以按如下方式使用
fig, ax = plt.subplots(1, 1)
def swarm_plot(x):
kde = gaussian_kde(x)
density = kde(x) # estimate the local density at each datapoint
jitter = np.random.rand(*x.shape) - .5
y = 1 + (density * jitter * 1000 * 2)
ax.clear()
ax.scatter(x, y, s = 30, c = 'g')
return fig
def draw_swarmplot(t):
f = int(t * fps)
dff = df.loc[f]
return mplfig_to_npimage(swarm_plot(dff['x']))
这也将产生上面所示的输出。