Question

我正在尝试“有效地”计算两个迭代器的乘积。他们每个人花一点时间来产生每个结果，并且有很多产生的结果。既然unsigned int ix = get_global_id(0); unsigned int iy = get_global_id(1); //PROBLEM LIES IN THESE 3 LINES: varfloat y = yaw - fov/2.0 + fov*ix/width; //get yaw using horizontal fov varfloat f = (fov * height) / width; //calculate vertical fov from horizontal fov scaled proportionally based on width and height varfloat p = pitch - f/2.0 + f*iy/height; //get pitch using vertical fov varfloat3 direction = {sin(y)*cos(p), cos(y)*cos(p), sin(p)}; //unit vector for ray direction似乎首先计算了所有项目，那么要获得第一对就需要很多时间。

MCVE是：

itertools.product

输出为：

import time
from itertools import product

def costlygen(n):
    for i in range(n):
        time.sleep(1)
        yield i

g1 = costlygen(5)
g2 = costlygen(5)

now = time.time()
g = product(g1,g2)

for x in g:
    print(x)
    print(time.time()-now)

从结果很明显，(0, 0) 10.027392148971558 (0, 1) 10.027477979660034 (0, 2) 10.027528285980225 ... (4, 3) 10.028220176696777 (4, 4) 10.028250217437744计算出每个生成器生成的所有项，因此第一个结果仅在10秒后产生，而可能在2秒后才产生。 / p>

一旦产生结果，有什么方法可以得到结果吗？

Answer 1

有一种可能的解决方案，它通过gone列表使用缓存：

import time
from itertools import product

def costlygen(n):
    for i in range(n):
        time.sleep(1)
        yield i

def simple_product(it1, it2):
    gone = []
    x = next(it1)
    for y in it2:
        gone.append(y)
        yield x, y
    for x in it1:
        for y in gone:
            yield x, y

def complex_product(*iterables):
    if len(iterables) == 2:
        yield from simple_product(*iterables)
        return
    it1, *rest = iterables
    gone = []
    x = next(it1)
    for t in complex_product(*rest):
        gone.append(t)
        yield (x,) + t
    for x in it1:
        for t in gone:
            yield (x,) + t

g1 = costlygen(5)
g2 = costlygen(5)
g3 = costlygen(5)

now = time.time()
g = complex_product(g1,g2,g3)

for x in g:
    print(x)
    print(time.time()-now)

时间：

(0, 0, 0)
3.002698898315429  # as soon as possible
(0, 0, 1)
4.003920316696167  # after one second
(0, 0, 2)
5.005135536193848
(0, 0, 3)
6.006361484527588
(0, 0, 4)
7.006711721420288
(0, 1, 0)
8.007975101470947
(0, 1, 1)
8.008066892623901  # third gen was already gone, so (*, *, 1) will be produced instantly after (*, *, 0)
(0, 1, 2)
8.008140802383423
(0, 1, 3)
8.00821304321289
(0, 1, 4)
8.008255004882812
(0, 2, 0)
9.009203910827637

有没有一种方法可以有效地计算两个（或多个）迭代器的乘积？

1 个答案: