这是一种方法
NumPy strides，该方法主要使用剩余元素填充，然后
strides帮助我们非常有效地创建转换后的版本-
def strided_method(ar): a = np.concatenate(( ar, ar[:-1] )) L = len(ar) n = a.strides[0] return np.lib.stride_tricks.as_strided(a[L-1:], (L,L), (-n,n))
样品运行-
In [42]: ar = np.array([1, 2, 3, 4])In [43]: strided_method(ar)Out[43]: array([[4, 1, 2, 3], [3, 4, 1, 2], [2, 3, 4, 1], [1, 2, 3, 4]])In [44]: ar = np.array([4,9,3,6,1,2])In [45]: strided_method(ar)Out[45]: array([[2, 4, 9, 3, 6, 1], [1, 2, 4, 9, 3, 6], [6, 1, 2, 4, 9, 3], [3, 6, 1, 2, 4, 9], [9, 3, 6, 1, 2, 4], [4, 9, 3, 6, 1, 2]])
运行时测试-
In [5]: a = np.random.randint(0,9,(1000))# @Eric's solnIn [6]: %timeit roll_matrix(a)100 loops, best of 3: 3.39 ms per loop# @Warren Weckesser's solnIn [8]: %timeit circulant(a[::-1])100 loops, best of 3: 2.03 ms per loop# Strides methodIn [18]: %timeit strided_method(a)100000 loops, best of 3: 6.7 µs per loop
制作副本（如果您想进行更改，而不仅仅是用作只读数组），对于该
strides方法不会对我们造成太大的伤害-
In [19]: %timeit strided_method(a).copy()1000 loops, best of 3: 381 µs per loop
