如何将对角矩阵沿轴之一分成相等数量的项目? [英] How to split diagonal matrix into equal number of items each along one of axis?
问题描述
我有一个非常大的对角矩阵,我需要将其拆分以进行并行计算.由于数据局部性问题,遍历矩阵并在 n 个线程之间拆分每个 n 次计算是没有意义的.目前,我正在按以下方式划分 k x k 对角矩阵,但它在计算数量方面产生不相等的分区(最小的部分计算比最大的).
I have a very large diagonal matrix that I need to split for parallel computation. Due to data locality issues it makes no sense to iterate through the matrix and split every n-th calculation between n threads. Currently, I am dividing k x k diagonal matrix in the following way but it yields unequal partitions in terms of the number of the calculations (smallest piece calculates a few times longer than the largest).
def split_matrix(k, n):
split_points = [round(i * k / n) for i in range(n + 1)]
split_ranges = [(split_points[i], split_points[i + 1],) for i in range(len(split_points) - 1)]
return split_ranges
import numpy as np
k = 100
arr = np.zeros((k,k,))
idx = 0
for i in range(k):
for j in range(i + 1, k):
arr[i, j] = idx
idx += 1
def parallel_calc(array, k, si, endi):
for i in range(si, endi):
for j in range(k):
# do some expensive calculations
for start_i, stop_i in split_matrix(k, cpu_cnt):
parallel_calc(arr, k, start_i, stop_i)
您对实现或库函数有什么建议吗?
Do you have any suggestions as to the implementation or library function?
推荐答案
在一侧进行了多次几何计算后,我得出了以下分区,该分区在每个垂直(或水平(如果需要)分区.
After a number of geometrical calculations on a side I arrived at the following partitioning that gives roughly the same number of points of the matrix in each of the vertical (or horizontal, if one wants) partitions.
def offsets_for_equal_no_elems_diag_matrix(matrix_dims, num_of_partitions):
if 2 == len(matrix_dims) and matrix_dims[0] == matrix_dims[1]: # square
k = matrix_dims[0]
# equilateral right angle triangles have area of side**2/2 and from this area == 1/num_of_partitions * 1/2 * matrix_dim[0]**2 comes the below
# the k - ... comes from the change in the axis (for the calc it is easier to start from the smallest triangle piece)
div_points = [0, ] + [round(k * math.sqrt((i + 1)/num_of_partitions)) for i in range(num_of_partitions)]
pairs = [(k - div_points[i + 1], k - div_points[i], ) for i in range(num_of_partitions - 1, -1, -1)]
return pairs
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