Python imshow 当您绘制数据而不是图像时。方面和范围之间的关系？
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imshow when you are plotting data, not images. Realtion between aspect and extent?
提问by AlexNtheitroad
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,
我正在用 matplotlib 中的 imshow 绘制一个二维数据数组。我在尝试缩放结果图时遇到问题。数组的大小是 30x1295 点，但范围的单位是：
extent = [-130,130,0,77]如果我绘制没有范围的数组，我会得到正确的图，但是如果我使用范围，我会得到这个带有错误方面的图。这是一个非常初学者的问题，但总是有第一次：我如何同时控制情节的方面和大小？谢谢，
Alex
亚历克斯
P.D. The code is, for the right case:
imshow(np.log10(psirhoz+1e-5),origin='lower')
PD 代码是，对于正确的情况：
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()])
对于错误的：
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.
我希望这能澄清一些事情。
采纳答案by Joe Kington
I'm guessing that you're wanting "square" pixels in the final plot?
我猜你想要最终图中的“方形”像素？
For example, if we plot random data similar to yours:
例如，如果我们绘制与您相似的随机数据：
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:
我们将得到一个带有“拉伸”像素的图像：
So, first off, "aspect" in matplotlib refers to the aspect in datacoordinates. This means we have to jump through a couple of hoops to get what you want.
因此，首先，matplotlib 中的“方面”是指数据坐标中的方面。这意味着我们必须跳过几个环节才能得到您想要的东西。
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()
回答by Keith Hughitt
In this case, pyplot.matshow()might also be useful:
在这种情况下，pyplot.matshow()也可能有用：
from matplotlib import pyplot as plt
import numpy as np
dat = np.array(range(9)).reshape(3,3)
plt.matshow(dat)
plt.show()
result:
结果：
