<p>对于全Python解决方案(使用numpy/scipy array ops,这将比任何显式的每像素循环都快),它:</p>
<pre><code>#!/usr/bin/env python
import math
import numpy as np
import scipy
import scipy.misc
import scipy.ndimage.interpolation
import subprocess
src=scipy.misc.imread("ji80w.png")
size=256
frames=50
for frame in xrange(0,frames):
# Image pixel co-ordinates
px=np.arange(-1.0,1.0,2.0/size)+1.0/size
py=np.arange(-1.0,1.0,2.0/size)+1.0/size
hx,hy=scipy.meshgrid(px,py)
# Compute z of sphere hit position, if pixel's ray hits
r2=hx*hx+hy*hy
hit=(r2<=1.0)
hz=np.where(
hit,
-np.sqrt(1.0-np.where(hit,r2,0.0)),
np.NaN
)
# Some spin and tilt to make things interesting
spin=2.0*np.pi*(frame+0.5)/frames
cs=math.cos(spin)
ss=math.sin(spin)
ms=np.array([[cs,0.0,ss],[0.0,1.0,0.0],[-ss,0.0,cs]])
tilt=0.125*np.pi*math.sin(2.0*spin)
ct=math.cos(tilt)
st=math.sin(tilt)
mt=np.array([[1.0,0.0,0.0],[0.0,ct,st],[0.0,-st,ct]])
# Rotate the hit points
xyz=np.dstack([hx,hy,hz])
xyz=np.tensordot(xyz,mt,axes=([2],[1]))
xyz=np.tensordot(xyz,ms,axes=([2],[1]))
x=xyz[:,:,0]
y=xyz[:,:,1]
z=xyz[:,:,2]
# Compute map position of hit
latitude =np.where(hit,(0.5+np.arcsin(y)/np.pi)*src.shape[0],0.0)
longitude=np.where(hit,(1.0+np.arctan2(z,x)/np.pi)*0.5*src.shape[1],0.0)
latlong=np.array([latitude,longitude])
# Resample, and zap non-hit pixels
dst=np.zeros((size,size,3))
for channel in [0,1,2]:
dst[:,:,channel]=np.where(
hit,
scipy.ndimage.interpolation.map_coordinates(
src[:,:,channel],
latlong,
order=1
),
0.0
)
# Save to f0000.png, f0001.png, ...
scipy.misc.imsave('f{:04}.png'.format(frame),dst)
# Use imagemagick to make an animated gif
subprocess.call('convert -delay 10 f????.png anim.gif',shell=True)
</code></pre>
<p>会得到你的</p>
<p><img src="https://i.stack.imgur.com/m1iYx.gif" alt="animated thing"/>。在</p>
<p>尽管如此,OpenGL确实是一个进行这种像素争夺的地方,尤其是对于任何交互式的东西。在</p>