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的数组中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法是使用for循环。

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

在Numpy中有更简洁的方法来实现吗？

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：使用更高效的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))

示例运行 -

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 Numpy 1.20 开始，我们可以利用新的 sliding_window_view 来滑动/滚动元素窗口。

再加上一个步进[::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]])

滑动的中间结果是：

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]])

Answer 3

@Divakar代码的修改版本，并进行检查以确保内存是连续的并且返回的数组无法修改。（我的DSP应用程序的变量名称已更改。）

def frame(a, framelen, frameadv):
"""frame - Frame a 1D array
a - 1D array
framelen - Samples per frame
frameadv - Samples between starts of consecutive frames
   Set to framelen for non-overlaping consecutive frames

Modified from Divakar's 10/17/16 11:20 solution:
https://stackoverflow.com/questions/40084931/taking-subarrays-from-numpy-array-with-given-stride-stepsize

CAVEATS:
Assumes array is contiguous
Output is not writable as there are multiple views on the same memory

"""

if not isinstance(a, np.ndarray) or \
   not (a.flags['C_CONTIGUOUS'] or a.flags['F_CONTIGUOUS']):
    raise ValueError("Input array a must be a contiguous numpy array")

# Output
nrows = ((a.size-framelen)//frameadv)+1
oshape = (nrows, framelen)

# Size of each element in a
n = a.strides[0]

# Indexing in the new object will advance by frameadv * element size
ostrides = (frameadv*n, n)
return np.lib.stride_tricks.as_strided(a, shape=oshape,
                                       strides=ostrides, writeable=False)

从给定步幅/步长的numpy数组中取出子阵列

3 个答案: