在numpy中查找对角线总和(更快) [英] Find diagonals sums in numpy (faster)
问题描述
我有一些像这样的board
numpy数组:
I have some board
numpy arrays like that:
array([[0, 0, 0, 1, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 1, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 1],
[0, 1, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 1, 0],
[1, 0, 0, 0, 0, 1, 0, 0]])
我正在使用以下代码在电路板的-7至8(及其镜像版本)的第n个对角线上找到元素的总和.
And I'm using the following code to find the sum of elements on each nth diagonal from -7 to 8 of the board (and the mirrored version of it).
n = 8
rate = [b.diagonal(i).sum()
for b in (board, board[::-1])
for i in range(-n+1, n)]
经过分析后,此操作大约花费了总运行时间的2/3,这似乎是由于以下两个因素造成的:
After some profiling, this operation is taking about 2/3 of overall running time and it seems to be because of 2 factors:
-
.diagonal
方法将构建一个新数组,而不是一个视图(看起来numpy 1.7将具有一个新的.diag
方法来解决该问题) - 迭代是在列表理解内的python中完成的
- The
.diagonal
method builds a new array instead of a view (looks like numpy 1.7 will have a new.diag
method to solve that) - The iteration is done in python inside the list comprehension
那么,有什么方法可以更快地找到这些总和(可能在numpy的C层中)?
So, there are any methods to find these sums faster (possibly in the C layer of numpy)?
经过更多测试后,通过缓存此操作,我可以减少7.5倍的总时间...也许我在寻找错误的瓶颈?
After some more tests, I could reduce 7.5x the total time by caching this operation... Maybe I was looking for the wrong bottleneck?
还有一件事:
只是找到了.trace
方法来代替diagonal(i).sum()
东西,并且...性能没有太大提高(大约2%到4%).
Just found the .trace
method that replaces the diagonal(i).sum()
thing and... There wasn't much improvement in performance (about 2 to 4%).
所以问题应该出在理解上.有什么想法吗?
So the problem should be the comprehension. Any ideas?
推荐答案
有可能使用stride_tricks
的解决方案.这部分是基于此问题的答案中可用的大量信息,但是问题恰恰不同,我认为,不应将其视为重复项.这是应用于方阵的基本思想,有关实现更通用解决方案的函数,请参见下文.
There's a possible solution using stride_tricks
. This is based in part on the plethora of information available in the answers to this question, but the problem is just different enough, I think, not to count as a duplicate. Here's the basic idea, applied to a square matrix -- see below for a function implementing the more general solution.
>>> cols = 8
>>> a = numpy.arange(cols * cols).reshape((cols, cols))
>>> fill = numpy.zeros((cols - 1) * cols, dtype='i8').reshape((cols - 1, cols))
>>> stacked = numpy.vstack((a, fill, a))
>>> major_stride, minor_stride = stacked.strides
>>> strides = major_stride, minor_stride * (cols + 1)
>>> shape = (cols * 2 - 1, cols)
>>> numpy.lib.stride_tricks.as_strided(stacked, shape, strides)
array([[ 0, 9, 18, 27, 36, 45, 54, 63],
[ 8, 17, 26, 35, 44, 53, 62, 0],
[16, 25, 34, 43, 52, 61, 0, 0],
[24, 33, 42, 51, 60, 0, 0, 0],
[32, 41, 50, 59, 0, 0, 0, 0],
[40, 49, 58, 0, 0, 0, 0, 0],
[48, 57, 0, 0, 0, 0, 0, 0],
[56, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 7],
[ 0, 0, 0, 0, 0, 0, 6, 15],
[ 0, 0, 0, 0, 0, 5, 14, 23],
[ 0, 0, 0, 0, 4, 13, 22, 31],
[ 0, 0, 0, 3, 12, 21, 30, 39],
[ 0, 0, 2, 11, 20, 29, 38, 47],
[ 0, 1, 10, 19, 28, 37, 46, 55]])
>>> diags = numpy.lib.stride_tricks.as_strided(stacked, shape, strides)
>>> diags.sum(axis=1)
array([252, 245, 231, 210, 182, 147, 105, 56, 7, 21, 42, 70, 105,
147, 196])
当然,我不知道这实际上有多快.但是我敢打赌它会比Python列表理解要快.
Of course, I have no idea how fast this will actually be. But I bet it will be faster than a Python list comprehension.
为方便起见,这是一个完全通用的diagonals
函数.假设您想沿最长轴移动对角线:
For convenience, here's a fully general diagonals
function. It assumes you want to move the diagonal along the longest axis:
def diagonals(a):
rows, cols = a.shape
if cols > rows:
a = a.T
rows, cols = a.shape
fill = numpy.zeros(((cols - 1), cols), dtype=a.dtype)
stacked = numpy.vstack((a, fill, a))
major_stride, minor_stride = stacked.strides
strides = major_stride, minor_stride * (cols + 1)
shape = (rows + cols - 1, cols)
return numpy.lib.stride_tricks.as_strided(stacked, shape, strides)
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