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I am plotting a 2D data array with imshow in matplotlib. I have a problem trying to scale the resulting plot. The size of the array is 30x1295 points, but the extent in units are: extent = [-130,130,0,77] If I plot the array without the extent, I get the right plot, but if I use extent, I get this plot with the wrong aspect. It is a pretty beginner question, but there is always a first time: How I can control the aspect and the size of the plot at the same time? Thanks,

Alex

P.D. The code is, for the right case: imshow(np.log10(psirhoz+1e-5),origin='lower')

and for the wrong one: imshow(np.log10(psirhoz+1e-5),origin='lower', extent =[z_ax.min(),z_ax.max(),rho_ax.min(),rho_ax.max()])

I hope this clarify a bit things.

AlexNoir
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2 Answers2

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I'm guessing that you're wanting "square" pixels in the final plot?

For example, if we plot random data similar to yours:

import numpy as np
import matplotlib.pyplot as plt

data = np.random.random((30, 1295))

fig, ax = plt.subplots()
ax.imshow(data, extent=[-130,130,0,77])
plt.show()

We'll get an image with "stretched" pixels:

enter image description here

So, first off, "aspect" in matplotlib refers to the aspect in data coordinates. This means we have to jump through a couple of hoops to get what you want.

import numpy as np
import matplotlib.pyplot as plt

def main():
    shape = (30, 1295)
    extent = [-130,130,0,77]

    data = np.random.random(shape)

    fig, ax = plt.subplots()
    ax.imshow(data, extent=extent, aspect=calculate_aspect(shape, extent))
    plt.show()

def calculate_aspect(shape, extent):
    dx = (extent[1] - extent[0]) / float(shape[1])
    dy = (extent[3] - extent[2]) / float(shape[0])
    return dx / dy

main()
Joe Kington
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    Thanks very much! The key point that make me understand is that the aspect refers to the data rather than the axes. – AlexNoir Sep 13 '13 at 12:22
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In this case, pyplot.matshow() might also be useful:

from matplotlib import pyplot as plt
import numpy as np
dat = np.array(range(9)).reshape(3,3)
plt.matshow(dat)
plt.show()

result:

enter image description here

Keith Hughitt
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