@Sven的答案是简单的方法，但对于大型数组来说效率相当低。如果你正在处理一个相对较小的数组，你将不会注意到差异，如果你想要一个大的配置文件（例如> 50 MB），你可能想尝试其他几种方法。但是，你需要在像素坐标中工作，因此还有一层额外的复杂性。

@Sven's answer is the easy way, but it's rather inefficient for large arrays. If you're dealing with a relatively small array, you won't notice the difference, if you're wanting a profile from a large (e.g. >50 MB) you may want to try a couple of other approaches. You'll need to work in "pixel" coordinates for these, though, so there's an extra layer of complexity.

还有两种更节省内存的方法。 1）使用 scipy。 ndimage.map_coordinates 如果您需要双线性或三次插值。 2）如果您只想要最近邻居采样，那么只需直接使用索引。

There are two more memory-efficient ways. 1) use scipy.ndimage.map_coordinates if you need bilinear or cubic interpolation. 2) if you just want nearest neighbor sampling, then just use indexing directly.

作为第一个例子：

import numpy as np
import scipy.ndimage
import matplotlib.pyplot as plt

#-- Generate some data...
x, y = np.mgrid[-5:5:0.1, -5:5:0.1]
z = np.sqrt(x**2 + y**2) + np.sin(x**2 + y**2)

#-- Extract the line...
# Make a line with "num" points...
x0, y0 = 5, 4.5 # These are in _pixel_ coordinates!!
x1, y1 = 60, 75
num = 1000
x, y = np.linspace(x0, x1, num), np.linspace(y0, y1, num)

# Extract the values along the line, using cubic interpolation
zi = scipy.ndimage.map_coordinates(z, np.vstack((x,y)))

#-- Plot...
fig, axes = plt.subplots(nrows=2)
axes[0].imshow(z)
axes[0].plot([x0, x1], [y0, y1], 'ro-')
axes[0].axis('image')

axes[1].plot(zi)

plt.show()

使用最近邻插值的等价物看起来像这样：

The equivalent using nearest-neighbor interpolation would look something like this:

import numpy as np
import matplotlib.pyplot as plt

#-- Generate some data...
x, y = np.mgrid[-5:5:0.1, -5:5:0.1]
z = np.sqrt(x**2 + y**2) + np.sin(x**2 + y**2)

#-- Extract the line...
# Make a line with "num" points...
x0, y0 = 5, 4.5 # These are in _pixel_ coordinates!!
x1, y1 = 60, 75
num = 1000
x, y = np.linspace(x0, x1, num), np.linspace(y0, y1, num)

# Extract the values along the line
zi = z[x.astype(np.int), y.astype(np.int)]

#-- Plot...
fig, axes = plt.subplots(nrows=2)
axes[0].imshow(z)
axes[0].plot([x0, x1], [y0, y1], 'ro-')
axes[0].axis('image')

axes[1].plot(zi)

plt.show()

但是，如果你使用的是最近邻居，你可能只需要每个像素的样本，所以你可能会做更像这样的事情，而不是...

However, if you're using nearest-neighbor, you probably would only want samples at each pixel, so you'd probably do something more like this, instead...

import numpy as np
import matplotlib.pyplot as plt

#-- Generate some data...
x, y = np.mgrid[-5:5:0.1, -5:5:0.1]
z = np.sqrt(x**2 + y**2) + np.sin(x**2 + y**2)

#-- Extract the line...
# Make a line with "num" points...
x0, y0 = 5, 4.5 # These are in _pixel_ coordinates!!
x1, y1 = 60, 75
length = int(np.hypot(x1-x0, y1-y0))
x, y = np.linspace(x0, x1, length), np.linspace(y0, y1, length)

# Extract the values along the line
zi = z[x.astype(np.int), y.astype(np.int)]

#-- Plot...
fig, axes = plt.subplots(nrows=2)
axes[0].imshow(z)
axes[0].plot([x0, x1], [y0, y1], 'ro-')
axes[0].axis('image')

axes[1].plot(zi)

plt.show()