我嘗試使用 Numpy 函數(shù)或向量而不是 for 循環(huán)來(lái)加速此代碼：sommes = []for j in range(vertices.shape[0]): terme = new_vertices[j] - new_vertices[vertex_neighbors[j]] somme_j = np.sum(terme) sommes.append(somme_j)E_int = np.sum(sommes)（它是迭代算法的一部分，并且有很多“頂點(diǎn)”，所以我認(rèn)為 for 循環(huán)花費(fèi)的時(shí)間太長(zhǎng)。）例如，要計(jì)算 j = 0 時(shí)的“terme”，我有：In: new_vertices[0]Out: array([ 10.2533888 , -42.32279717, 68.27230793])In: vertex_neighbors[0]Out: [1280, 2, 1511, 511, 1727, 1887, 759, 509, 1023]In: new_verties[vertex_neighbors[0]]Out: array([[ 10.47121043, -42.00123956, 68.218715 ], [ 10.2533888 , -43.26905874, 62.59473849], [ 10.69773735, -41.26464083, 68.09594854], [ 10.37030712, -42.16729601, 68.24639107], [ 10.12158146, -42.46624547, 68.29621598], [ 9.81850836, -42.71158695, 68.33710623], [ 9.97615447, -42.59625943, 68.31788497], [ 10.37030712, -43.11676015, 62.54960623], [ 10.55512696, -41.82622703, 68.18954624]])In: new_vertices[0] - new_vertices[vertex_neighbors[0]]Out: array([[-0.21782162, -0.32155761, 0.05359293], [ 0. , 0.94626157, 5.67756944], [-0.44434855, -1.05815634, 0.17635939], [-0.11691832, -0.15550116, 0.02591686], [ 0.13180734, 0.1434483 , -0.02390805], [ 0.43488044, 0.38878979, -0.0647983 ], [ 0.27723434, 0.27346227, -0.04557704], [-0.11691832, 0.79396298, 5.7227017 ], [-0.30173816, -0.49657014, 0.08276169]])問(wèn)題是 new_vertices[vertex_neighbors[j]] 并不總是具有相同的大小。例如，當(dāng) j = 7 時(shí)：In: new_vertices[7]Out: array([ 10.74106112, -63.88592276, -70.15593947])In: vertex_neighbors[7]Out: [1546, 655, 306, 1879, 920, 925]沒(méi)有for循環(huán)可以嗎？我的想法已經(jīng)用完了，所以任何幫助將不勝感激！
1 回答
holdtom
TA貢獻(xiàn)1805條經(jīng)驗(yàn) 獲得超10個(gè)贊
是的，這是可能的。這個(gè)想法是用來(lái)np.repeat創(chuàng)建一個(gè)向量，其中的項(xiàng)目重復(fù)可變次數(shù)。這是代碼：
# The two following lines can be done only once if the indices are constant between iterations (precomputation)
counts = np.array([len(e) for e in vertex_neighbors])
flatten_indices = np.concatenate(vertex_neighbors)
E_int = np.sum(np.repeat(new_vertices, counts, axis=0) - new_vertices[flatten_indices])
這是一個(gè)基準(zhǔn)：
import numpy as np
from time import *
n = 32768
vertices = np.random.rand(n, 3)
indices = []
count = np.random.randint(1, 10, size=n)
for i in range(n):
indices.append(np.random.randint(0, n, size=count[i]))
def initial_version(vertices, vertex_neighbors):
sommes = []
for j in range(vertices.shape[0]):
terme = vertices[j] - vertices[vertex_neighbors[j]]
somme_j = np.sum(terme)
sommes.append(somme_j)
return np.sum(sommes)
def optimized_version(vertices, vertex_neighbors):
# The two following lines can be precomputed
counts = np.array([len(e) for e in indices])
flatten_indices = np.concatenate(indices)
return np.sum(np.repeat(vertices, counts, axis=0) - vertices[flatten_indices])
def more_optimized_version(vertices, vertex_neighbors, counts, flatten_indices):
return np.sum(np.repeat(vertices, counts, axis=0) - vertices[flatten_indices])
timesteps = 20
a = time()
for t in range(timesteps):
res = initial_version(vertices, indices)
b = time()
print("V1: time:", b - a)
print("V1: result", res)
a = time()
for t in range(timesteps):
res = optimized_version(vertices, indices)
b = time()
print("V2: time:", b - a)
print("V2: result", res)
a = time()
counts = np.array([len(e) for e in indices])
flatten_indices = np.concatenate(indices)
for t in range(timesteps):
res = more_optimized_version(vertices, indices, counts, flatten_indices)
b = time()
print("V3: time:", b - a)
print("V3: result", res)
這是我機(jī)器上的基準(zhǔn)測(cè)試結(jié)果：
V1: time: 3.656714916229248
V1: result -395.8416223057596
V2: time: 0.19800186157226562
V2: result -395.8416223057595
V3: time: 0.07983255386352539
V3: result -395.8416223057595
正如您所看到的，這一優(yōu)化版本比參考實(shí)現(xiàn)快 18 倍，而預(yù)先計(jì)算索引的版本比參考實(shí)現(xiàn)快 46 倍。
請(qǐng)注意，優(yōu)化版本應(yīng)該需要更多 RAM（特別是如果每個(gè)頂點(diǎn)的鄰居數(shù)量很大）。
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