如何获取散景图像以像matplotlib一样在左上角绘制原点？

Question

How can bokeh be used to plot images with the origin in the top left like matplotlib?

I can accomplish this by rotating and transposing the 2d array, but is there a way to do this without modifying the array? (Modifying the array isn't ideal, because then the axis labels don't match the original data.)

from bokeh.plotting import figure, output_file, show
import matplotlib.pyplot as plt
import numpy as np

img = np.zeros((5,5), dtype=np.int8)
img[0][0] = 250 
img[0][1] = 250 
img[0][4] = 500
img[0][3] = 500 
img[1][4] = 500 

output_file("default.html")
plot1 = figure(x_range=(0, img.shape[0]), y_range=(0, img.shape[1]))
plot1.image(image=[img], x=0, y=0, dw=img.shape[0], dh=img.shape[1], palette="Spectral10")
show(plot1)

output_file("rotated_transpose.html")
plot2 = figure(x_range=(0, img.shape[0]), y_range=(0, img.shape[1]))
plot2.image(image=[np.rot90(np.transpose(img))], x=0, y=0, dw=img.shape[0], dh=img.shape[1], palette="Spectral10")
show(plot2)

plt.imshow(img, cmap=plt.get_cmap('Spectral'), interpolation="nearest")
plt.show()

I've tried setting the y range in reverse and specifying the axis locations, but this still doesn't match the matplotlib coordinate system:

plot3 = figure(x_range=(0, img.shape[0]), y_range=(img.shape[1], 0), x_axis_location='above', y_axis_location='left')
plot3.image(image=[img], x=0, y=0, dw=img.shape[0], dh=img.shape[1], palette="Spectral10")
show(plot3)

Answer 1

Same issue with Bokeh image plot.The only thing that worked for me was adding img=np.flipud(img) before plotting img. Hope that helps !