figure of imshow() is too small

Question

I\'m trying to visualize a numpy array using imshow() since it\'s similar to imagesc() in Matlab.

imshow(random.rand(8, 90), interpolation=\'nearest\')


        

Answer 1

If you don't give an aspect argument to imshow, it will use the value for image.aspect in your matplotlibrc. The default for this value in a new matplotlibrc is equal. So imshow will plot your array with equal aspect ratio.

If you don't need an equal aspect you can set aspect to auto

imshow(random.rand(8, 90), interpolation='nearest', aspect='auto')

which gives the following figure

If you want an equal aspect ratio you have to adapt your figsize according to the aspect

fig, ax = subplots(figsize=(18, 2))
ax.imshow(random.rand(8, 90), interpolation='nearest')
tight_layout()

which gives you:

Answer 2

That's strange, it definitely works for me:

from matplotlib import pyplot as plt

plt.figure(figsize = (20,2))
plt.imshow(random.rand(8, 90), interpolation='nearest')

I am using the "MacOSX" backend, btw.

Answer 3

I'm new to python too. Here is something that looks like will do what you want to

axes([0.08, 0.08, 0.94-0.08, 0.94-0.08]) #[left, bottom, width, height]
axis('scaled')`

I believe this decides the size of the canvas.

Answer 4

Update 2020

as requested by @baxxx, here is an update because random.rand is deprecated meanwhile.

This works with matplotlip 3.2.1:

from matplotlib import pyplot as plt
import random
import numpy as np

random = np.random.random ([8,90])

plt.figure(figsize = (20,2))
plt.imshow(random, interpolation='nearest')

This plots:

To change the random number, you can experiment with np.random.normal(0,1,(8,90)) (here mean = 0, standard deviation = 1).