Dynamically updating Matplotlib figures in Jupyter notebooks
Sean Lane
Posted on July 6, 2019
Updating matplotlib figures dynamically seems to be a bit of a hassle, but the code below seems to do the trick. This is an example that outputs a figure with multiple subplots, each with multiple plots. Oddly enough, at the time of writing the image will be smaller than the figure until the Jupyter cells stops running, but this can be fixed but generating the figure in one cell, and then updating the image in a subsequent cell 1.
This code is run with the assumption that the following data file can be found in the working directory named data.txt
:
import numpy as np
import time
import matplotlib.pyplot as plt
%matplotlib notebook
def load_data():
data = np.genfromtxt('data.txt', delimiter=',', skip_header=1)
tm = data[:, 0]
Q1 = data[:, 1]
Q2 = data[:, 2]
T1 = data[:, 3]
T2 = data[:, 4]
return (tm, Q1, Q2, T1, T2)
(m_time, Q1s, Q2s, T1s, T2s) = load_data()
n = len(m_time)
labels = [
[r'$T_1$ measured', r'$T_2$ measured'],
# [r'$T_1 set point$', r'$T_2 set point$'],
[r'$Q_1$', r'$Q_2$']
]
colors = [
['r:', 'b-'],
['r:', 'bx']
]
def plot_init(num_subplots=1, x_labels=None, y_labels=None):
if not x_labels:
x_labels = [None] * num_subplots
if not y_labels:
y_labels = [None] * num_subplots
fig = plt.figure(figsize=(12,6), dpi=80)
fig.subplots_adjust(hspace=.5)
axes = []
for i in range(1, num_subplots + 1):
ax = plt.subplot(num_subplots, 1, i)
ax.grid()
if x_labels[i-1]:
ax.set_xlabel(x_labels[i-1])
if y_labels[i-1]:
ax.set_ylabel(y_labels[i-1])
axes.append(ax)
return fig, axes
def plot_update(fig, axes, xs, ys, colors, labels):
for i in range(len(axes)):
for j in range(len(ys[i])):
axes[i].plot(xs, ys[i][j], colors[i][j])
axes[i].legend(labels=labels[i])
fig.canvas.draw()
fig, axes = plot_init(
num_subplots=2,
x_labels=[None, 'Time (sec)'],
y_labels=['Temps (C)', 'Heaters']
)
for i in range(1, n):
try:
ys = [
[T1s[:i], T2s[:i]],
[Q1s[:i], Q2s[:i]]
]
plot_update(fig, axes, m_time[:i], ys, colors, labels)
time.sleep(0.2)
except KeyboardInterrupt:
break
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Sean Lane
Posted on July 6, 2019
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