我试图在我的一个程序中复制主要瓶颈。
我想同时获得几个非整数像素值的linearly (or rather bilinearly) interpolated值。 不每个像素坐标以相同方式被扰动的情况。下面是一个完整/最小的脚本以及演示该问题的注释。如何加快result
的计算?
import numpy as np
import time
im = np.random.rand(640,480,3) # my "image"
xx, yy = np.meshgrid(np.arange(im.shape[1]), np.arange(im.shape[0]))
print "Check these are the right indices:",np.sum(im - im[yy,xx,:])
# perturb the indices slightly
# I want to calculate the interpolated
# values of "im" at these locations
xx = xx + np.random.normal(size=im.shape[:2])
yy = yy + np.random.normal(size=im.shape[:2])
# integer value/pixel locations
x_0 = np.int_(np.modf(xx)[1])
y_0 = np.int_(np.modf(yy)[1])
x_1, y_1 = x_0 + 1, y_0 + 1
# the real-valued offsets/coefficients pixels
a = np.modf(xx)[0][:,:,np.newaxis]
b = np.modf(yy)[0][:,:,np.newaxis]
# make sure we don't go out of bounds at edge pixels
np.clip(x_0,0,im.shape[1]-1,out=x_0)
np.clip(x_1,0,im.shape[1]-1,out=x_1)
np.clip(y_0,0,im.shape[0]-1,out=y_0)
np.clip(y_1,0,im.shape[0]-1,out=y_1)
# now perform linear interpolation: THIS IS THE BOTTLENECK!
tic = time.time()
result = ((1-a) * (1-b) * im[y_0, x_0, :] +
a * (1-b) * im[y_1, x_0, :] +
(1-a) * b * im[y_0, x_1, :] +
a * b * im[y_1, x_1, :] )
toc = time.time()
print "interpolation time:",toc-tic
答案 0 :(得分:6)
感谢@JoeKington提出的建议。这是我使用scipy.ndimage.map_coordinates
# rest as before
from scipy import ndimage
tic = time.time()
new_result = np.zeros(im.shape)
coords = np.array([yy,xx,np.zeros(im.shape[:2])])
for d in range(im.shape[2]):
new_result[:,:,d] = ndimage.map_coordinates(im,coords,order=1)
coords[2] += 1
toc = time.time()
print "interpolation time:",toc-tic
更新:添加了评论中建议的调整,并尝试了其他一两件事。这是最快的版本:
tic = time.time()
new_result = np.zeros(im.shape)
coords = np.array([yy,xx])
for d in range(im.shape[2]):
ndimage.map_coordinates(im[:,:,d],
coords,order=1,
prefilter=False,
output=new_result[:,:,d] )
toc = time.time()
print "interpolation time:",toc-tic
示例运行时间:
original version: 0.463063955307
better version: 0.204537153244
best version: 0.121845006943