我正在使用numpy einsum来计算形状(3,N)的列向量pts数组的点积,其自身形成矩阵dotps,形状(N,N),所有点产品。这是我使用的代码:
dotps = np.einsum('ij,ik->jk', pts, pts)
这样可行,但我只需要主对角线以上的值。即。没有对角线的结果的上三角部分。是否可以使用einsum仅计算这些值?或以比使用einsum计算整个矩阵更快的任何其他方式?
我的pts数组可能非常大,所以如果我只计算我需要的值,那么我的计算速度就会翻倍。
答案 0 :(得分:4)
您可以对相关列进行切片,然后使用np.einsum
-
R,C = np.triu_indices(N,1)
out = np.einsum('ij,ij->j',pts[:,R],pts[:,C])
示例运行 -
In [109]: N = 5
...: pts = np.random.rand(3,N)
...: dotps = np.einsum('ij,ik->jk', pts, pts)
...:
In [110]: dotps
Out[110]:
array([[ 0.26529103, 0.30626052, 0.18373867, 0.13602931, 0.51162729],
[ 0.30626052, 0.56132272, 0.5938057 , 0.28750708, 0.9876753 ],
[ 0.18373867, 0.5938057 , 0.84699103, 0.35788749, 1.04483158],
[ 0.13602931, 0.28750708, 0.35788749, 0.18274288, 0.4612556 ],
[ 0.51162729, 0.9876753 , 1.04483158, 0.4612556 , 1.82723949]])
In [111]: R,C = np.triu_indices(N,1)
...: out = np.einsum('ij,ij->j',pts[:,R],pts[:,C])
...:
In [112]: out
Out[112]:
array([ 0.30626052, 0.18373867, 0.13602931, 0.51162729, 0.5938057 ,
0.28750708, 0.9876753 , 0.35788749, 1.04483158, 0.4612556 ])
进一步优化 -
让我们来看看我们的方法,看看是否有任何改善效果的范围。
In [126]: N = 5000
In [127]: pts = np.random.rand(3,N)
In [128]: %timeit np.triu_indices(N,1)
1 loops, best of 3: 413 ms per loop
In [129]: R,C = np.triu_indices(N,1)
In [130]: %timeit np.einsum('ij,ij->j',pts[:,R],pts[:,C])
1 loops, best of 3: 1.47 s per loop
保持内存限制,看起来我们在优化np.einsum
方面做得不够。因此,让我们将重点转移到np.triu_indices
。
对于N = 4
,我们有:
In [131]: N = 4
In [132]: np.triu_indices(N,1)
Out[132]: (array([0, 0, 0, 1, 1, 2]), array([1, 2, 3, 2, 3, 3]))
它似乎创造了一个规则的模式,有点像移动的模式。这可以使用在3
和5
位置发生变化的累计和来编写。一般来说,我们最终会编写类似这样的代码 -
def triu_indices_cumsum(N):
# Length of R and C index arrays
L = (N*(N-1))/2
# Positions along the R and C arrays that indicate
# shifting to the next row of the full array
shifts_idx = np.arange(2,N)[::-1].cumsum()
# Initialize "shift" arrays for finally leading to R and C
shifts1_arr = np.zeros(L,dtype=int)
shifts2_arr = np.ones(L,dtype=int)
# At shift positions along the shifts array set appropriate values,
# such that when cumulative summed would lead to desired R and C arrays.
shifts1_arr[shifts_idx] = 1
shifts2_arr[shifts_idx] = -np.arange(N-2)[::-1]
# Finall cumsum to give R, C
R_arr = shifts1_arr.cumsum()
C_arr = shifts2_arr.cumsum()
return R_arr, C_arr
让它为各种N's
!
In [133]: N = 100
In [134]: %timeit np.triu_indices(N,1)
10000 loops, best of 3: 122 µs per loop
In [135]: %timeit triu_indices_cumsum(N)
10000 loops, best of 3: 61.7 µs per loop
In [136]: N = 1000
In [137]: %timeit np.triu_indices(N,1)
100 loops, best of 3: 17 ms per loop
In [138]: %timeit triu_indices_cumsum(N)
100 loops, best of 3: 16.3 ms per loop
因此,对于体面的N's
来说,基于triu_indices
的自定义cumsum可能值得一看!