从给定步幅/步长的numpy数组中取出子阵列

Question

假设我有一个Python Numpy数组a.

a = numpy.array([1,2,3,4,5,6,7,8,9,10,11])

我想从这个长度为5的数组创建一个子序列矩阵,步长为3.结果矩阵因此如下所示:

numpy.array([[1,2,3,4,5],[4,5,6,7,8],[7,8,9,10,11]])

实现这一点的一种可能方式是使用for循环.

result_matrix = np.zeros((3, 5))
for i in range(0, len(a), 3):
  result_matrix[i] = a[i:i+5]

有没有更简洁的方法来实现这个Numpy？

Answer 1

方法#1: 使用broadcasting-

def broadcasting_app(a, L, S ):  # Window len = L, Stride len/stepsize = S
    nrows = ((a.size-L)//S)+1
    return a[S*np.arange(nrows)[:,None] + np.arange(L)]

方法#2:使用更高效NumPy strides-

def strided_app(a, L, S ):  # Window len = L, Stride len/stepsize = S
    nrows = ((a.size-L)//S)+1
    n = a.strides[0]
    return np.lib.stride_tricks.as_strided(a, shape=(nrows,L), strides=(S*n,n))

样品运行 -

In [143]: a
Out[143]: array([ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11])

In [144]: broadcasting_app(a, L = 5, S = 3)
Out[144]: 
array([[ 1,  2,  3,  4,  5],
       [ 4,  5,  6,  7,  8],
       [ 7,  8,  9, 10, 11]])

In [145]: strided_app(a, L = 5, S = 3)
Out[145]: 
array([[ 1,  2,  3,  4,  5],
       [ 4,  5,  6,  7,  8],
       [ 7,  8,  9, 10, 11]])

Answer 2

从 开始Numpy 1.20，我们可以使用新功能sliding_window_view来滑动/滚动元素窗口。

再加上stepping [::3]，它就变成了：

from numpy.lib.stride_tricks import sliding_window_view

# values = np.array([1,2,3,4,5,6,7,8,9,10,11])
sliding_window_view(values, window_shape = 5)[::3]
# array([[ 1,  2,  3,  4,  5],
#        [ 4,  5,  6,  7,  8],
#        [ 7,  8,  9, 10, 11]])

其中滑动的中间结果为：

sliding_window_view(values, window_shape = 5)
# array([[ 1,  2,  3,  4,  5],
#        [ 2,  3,  4,  5,  6],
#        [ 3,  4,  5,  6,  7],
#        [ 4,  5,  6,  7,  8],
#        [ 5,  6,  7,  8,  9],
#        [ 6,  7,  8,  9, 10],
#        [ 7,  8,  9, 10, 11]])