Difficulty animating a matplotlib graph with moviepy

Question

I have to make an animation of a large number (~90,000) figures. For context, it\'s a plot of a map for every day from 1700 - 1950, with events of interest marked on relevent da

Answer 1

Same solution as JuniorCompressor, with just one frame kept in memory to avoid RAM issues. This example runs in 30 seconds on my machine and produces a good quality 400-second clip of 6000 frames, weighing 600k.

import numpy as np
import matplotlib.pyplot as plt
from moviepy.video.io.bindings import mplfig_to_npimage
import moviepy.editor as mpy

fig = plt.figure(facecolor="white") # <- ADDED FACECOLOR FOR WHITE BACKGROUND
ax = plt.axes()
x = np.random.randn(10, 1)
y = np.random.randn(10, 1)
p = plt.plot(x, y, 'ko')
time = np.arange(2341973, 2342373)

last_i = None
last_frame = None

def animate(t):
    global last_i, last_frame

    i = int(t)
    if i == last_i:
        return last_frame

    xn = x + np.sin(2 * np.pi * time[i] / 10.0)
    yn = y + np.cos(2 * np.pi * time[i] / 8.0)
    p[0].set_data(xn, yn)

    last_i = i
    last_frame = mplfig_to_npimage(fig)
    return last_frame

duration = len(time)
fps = 15
animation = mpy.VideoClip(animate, duration=duration)
animation.write_videofile("test.mp4", fps=fps)

On a sidenote, there is dedicated class of videoclips called DataVideoClip for precisely this purpose, which looks much more like matplotlib's animate. For the moment it's not really speed-efficient (I didn't include that little memoizing trick above). Here is how it works:

from moviepy.video.VideoClip import DataVideoClip

def data_to_frame(time):
    xn = x + np.sin(2 * np.pi * time / 10.0)
    yn = y + np.cos(2 * np.pi * time / 8.0)
    p[0].set_data(xn, yn)
    return mplfig_to_npimage(fig)

times = np.arange(2341973, 2342373)
clip = DataVideoClip(times, data_to_frame, fps=1) # one plot per second

#final animation is 15 fps, but still displays 1 plot per second
animation.write_videofile("test2.mp4", fps=15) 

Answer 2

Same observations:

• In animate a float number will be passed
• One frame per second may cause playback problems in many players. It's better to use a bigger frame rate like 15 fps.
• Using 15 fps will need many frames. It's better to use caching.

So you can do the following:

import numpy as np
import matplotlib.pyplot as plt
from moviepy.video.io.bindings import mplfig_to_npimage
import moviepy.editor as mpy

fig = plt.figure()
ax = plt.axes()
x = np.random.randn(10, 1)
y = np.random.randn(10, 1)
p = plt.plot(x, y, 'ko')
time = np.arange(2341973, 2342373)
cache = {}

def animate(t):
    i = int(t)
    if i in cache:
        return cache[i]

    xn = x + np.sin(2 * np.pi * time[i] / 10.0)
    yn = y + np.cos(2 * np.pi * time[i] / 8.0)
    p[0].set_data(xn, yn)
    cache.clear()
    cache[i] = mplfig_to_npimage(fig)
    return cache[i]

duration = len(time)
fps = 15
animation = mpy.VideoClip(animate, duration=duration)
animation.write_videofile("test.mp4", fps=fps)