我正在使用嵌套数组。我需要对该数组的每个元素应用相当简单但成本高昂的算术运算。
下面是MWE
,其中“第二个 block ”是占用大部分时间的 block (它将运行数千次)。
考虑到获取a、b、c的真实方式,第一个和第二个 block 需要分开,因为第一个 block 仅处理一次
时间成本非常高。
我不确定如何提高应用于嵌套数组的每个元素的操作的性能。当然,numpy
会更快地完成此操作,但我不太熟悉数组上的广播操作。
import numpy as np
import time
# Generate some random data.
N = 100
x, y, z = [np.random.uniform(0., 10., N) for _ in range(3)]
# Grid of values in 2 dimensions.
M = 200
p_lst, q_lst = np.linspace(0., 50., M), np.linspace(0., 25., M)
# Define empty nested list to be filled below.
# The shape is given by the length of the lists defined above.
abc_lst = [[[] for _ in p_lst] for _ in q_lst]
# First block. This needs to be separated from the block below.
# Fill nested list with values.
for i, p in enumerate(p_lst):
for j, q in enumerate(q_lst):
# a,b,c are obtained via some complicated function of p,q.
# This is just for the purpose of this example.
a, b, c = 1.*p, 1.*q, p+q
# Store in nested list.
abc_lst[i][j] = [a, b, c]
# Second block <-- THIS IS THE BOTTLENECK
tik = time.time()
# Apply operation on nested list.
lst = []
for i in range(len(p_lst)):
for j in range(len(q_lst)):
# Extract a,b,c values from nested list.
a, b, c = abc_lst[i][j]
# Apply operation. This is the *actual* operation
# I need to apply.
d = sum(abs(a*x + y*b + c*z))
# Store value.
lst.append(d)
print time.time() - tik
最佳答案
我在 this question 中找到了答案,使用np.outer()功能。
只需要对第一个 block 进行一些重新排列,第二个 block 的运行速度就会快很多倍。
# First block. Store a,b,c separately.
a_lst, b_lst, c_lst = [], [], []
for i, p in enumerate(p_lst):
for j, q in enumerate(q_lst):
# a,b,c are obtained via some complicated function of p,q.
# This is just for the purpose of this example.
a_lst.append(1.*p)
b_lst.append(1.*q)
c_lst.append(p+q)
# As arrays.
a_lst, b_lst, c_lst = np.asarray(a_lst), np.asarray(b_lst), np.asarray(c_lst)
# Second block.
# Apply operation on nested list using np.outer.
lst = np.sum(abs(np.outer(a_lst, x) + np.outer(b_lst, y) + np.outer(c_lst, z)), axis=1)
关于python - 加速嵌套 python 数组的操作,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34903246/