线性指标上三角矩阵 [英] Linear index upper triangular matrix
问题描述
如果我有一个矩阵的上三角部分,上面的对角线偏移,存储为线性阵列,如何矩阵可以在(I,J)
指数元件从阵列的线性索引提取
例如,线性阵列 [A0,A1,A2,A3,A4,A5,A6,A7,A8,A9
是矩阵存储器对>
0 A0 A1 A2 A3
0 0 A4 A5 A6
0 0 0 A7 A8
0 0 0 0 A9
0 0 0 0 0
我们想知道(I,J),对应的线性矩阵偏移数组中的索引,不递归。
有一个合适的结果, k2ij(INT K,INT N) - GT&; (INT,INT)
将满足,例如:
k2ij(K = 0,N = 5)=(0,1)
k2ij(K = 1,N = 5)=(0,2)
k2ij(K = 2,N = 5)=(0,3)
k2ij(K = 3,N = 5)=(0,4)
k2ij(K = 4,N = 5)=(1,2)
k2ij(K = 5,N = 5)=(1,3)
[等等]
方程从线性指标将(I,J)
指数均
I = N - 2 - 楼(开方(-8 * K + 4 * N *(N-1)-7)/2.0 - 0.5)
J = K + I + 1 - N *(N-1)/ 2 +(N-1)*((N-1)-1)/ 2
逆操作,从(I,J)
指数线性指数
K =(N *(N-1)/ 2) - (N-1)*((N-1)-1)/ 2 + J - I - 1
验证与
从numpy的进口triu_indices,开方
N = 10
对于在范围K(N *(N-1)/ 2):
I = N - 2 - INT(SQRT(-8 * K + 4 * N *(N-1)-7)/2.0 - 0.5)
J = K + I + 1 - N *(N-1)/ 2 +(N-1)*((N-1)-1)/ 2
断言np.triu_indices(N,K = 1)[0] [K] ==我
断言np.triu_indices(N,K = 1)[1] [k]的==Ĵ因为我在范围(N):
对于在范围Ĵ第(i + 1,n)的:
K =(N *(N-1)/ 2) - (N-1)*((N-1)-1)/ 2 + J - I - 1
断言triu_indices(N,K = 1)[0] [K] ==我
断言triu_indices(N,K = 1)[1] [k]的==Ĵ
If I have the upper triangular portion of a matrix, offset above the diagonal, stored as a linear array, how can the (i,j)
indices of a matrix element be extracted from the linear index of the array?
For example, the linear array [a0, a1, a2, a3, a4, a5, a6, a7, a8, a9
is storage for the matrix
0 a0 a1 a2 a3
0 0 a4 a5 a6
0 0 0 a7 a8
0 0 0 0 a9
0 0 0 0 0
And we want to know the (i,j) index in the array corresponding to an offset in the linear matrix, without recursion.
A suitable result, k2ij(int k, int n) -> (int, int)
would satisfy, for example
k2ij(k=0, n=5) = (0, 1)
k2ij(k=1, n=5) = (0, 2)
k2ij(k=2, n=5) = (0, 3)
k2ij(k=3, n=5) = (0, 4)
k2ij(k=4, n=5) = (1, 2)
k2ij(k=5, n=5) = (1, 3)
[etc]
The equations going from linear index to (i,j)
index are
i = n - 2 - floor(sqrt(-8*k + 4*n*(n-1)-7)/2.0 - 0.5)
j = k + i + 1 - n*(n-1)/2 + (n-i)*((n-i)-1)/2
The inverse operation, from (i,j)
index to linear index is
k = (n*(n-1)/2) - (n-i)*((n-i)-1)/2 + j - i - 1
Verify with:
from numpy import triu_indices, sqrt
n = 10
for k in range(n*(n-1)/2):
i = n - 2 - int(sqrt(-8*k + 4*n*(n-1)-7)/2.0 - 0.5)
j = k + i + 1 - n*(n-1)/2 + (n-i)*((n-i)-1)/2
assert np.triu_indices(n, k=1)[0][k] == i
assert np.triu_indices(n, k=1)[1][k] == j
for i in range(n):
for j in range(i+1, n):
k = (n*(n-1)/2) - (n-i)*((n-i)-1)/2 + j - i - 1
assert triu_indices(n, k=1)[0][k] == i
assert triu_indices(n, k=1)[1][k] == j
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