从多维numpy数组行中选择随机窗口

Question

我有一个很大的数组，其中每一行都是一个时间序列，因此需要保持秩序。

我想为每行选择一个给定大小的随机窗口。

>>>import numpy as np
>>>arr = np.array(range(42)).reshape(6,7)
>>>arr
array([[ 0,  1,  2,  3,  4,  5,  6],
       [ 7,  8,  9, 10, 11, 12, 13],
       [14, 15, 16, 17, 18, 19, 20],
       [21, 22, 23, 24, 25, 26, 27],
       [28, 29, 30, 31, 32, 33, 34],
       [35, 36, 37, 38, 39, 40, 41]])
>>># What I want to do:
>>>select_random_windows(arr, window_size=3)
array([[ 1,  2,  3],
       [11, 12, 13],
       [14, 15, 16],
       [22, 23, 24],
       [38, 39, 40]])

在我看来，理想的解决方案是：

def select_random_windows(arr, window_size):
    offsets = np.random.randint(0, arr.shape[0] - window_size, size = arr.shape[1])
    return arr[:, offsets: offsets + window_size]

但不幸的是，这不起作用

我现在要进行的操作非常慢：

def select_random_windows(arr, wndow_size):
    result = []
    offsets = np.random.randint(0, arr.shape[0]-window_size, size = arr.shape[1])
    for row, offset in enumerate(start_indices):
        result.append(arr[row][offset: offset + window_size])
    return np.array(result)

当然，我可以对列表进行理解（并获得最小的速度提升），但是我想知道是否有某种超级聪明的numpy向量化方​​法可以做到这一点。

Answer 1

这是一种杠杆作用np.lib.stride_tricks.as_strided-

def random_windows_per_row_strided(arr, W=3):
    idx = np.random.randint(0,arr.shape[1]-W+1, arr.shape[0])
    strided = np.lib.stride_tricks.as_strided 
    m,n = arr.shape
    s0,s1 = arr.strides
    windows = strided(arr, shape=(m,n-W+1,W), strides=(s0,s1,s1))
    return windows[np.arange(len(idx)), idx]

在具有10,000行的更大数组上进行运行时测试-

In [469]: arr = np.random.rand(100000,100)

# @Psidom's soln
In [470]: %timeit select_random_windows(arr, window_size=3)
100 loops, best of 3: 7.41 ms per loop

In [471]: %timeit random_windows_per_row_strided(arr, W=3)
100 loops, best of 3: 6.84 ms per loop

# @Psidom's soln
In [472]: %timeit select_random_windows(arr, window_size=30)
10 loops, best of 3: 26.8 ms per loop

In [473]: %timeit random_windows_per_row_strided(arr, W=30)
100 loops, best of 3: 9.65 ms per loop

# @Psidom's soln
In [474]: %timeit select_random_windows(arr, window_size=50)
10 loops, best of 3: 41.8 ms per loop

In [475]: %timeit random_windows_per_row_strided(arr, W=50)
100 loops, best of 3: 10 ms per loop