提取 2d numpy 数组的随机 2d windows
Extract random 2d windows of a 2d numpy array
import numpy as np
arr = np.array(range(60)).reshape(6,10)
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, 42, 43, 44, 45, 46, 47, 48, 49],
> [50, 51, 52, 53, 54, 55, 56, 57, 58, 59]])
我需要的:
select_random_windows(arr, number_of windows= 3, window_size=3)
> array([[[ 1, 2, 3],
> [11, 12, 13],
> [21, 22, 23]],
>
> [37, 38, 39],
> [47, 48, 49],
> [57, 58, 59]],
>
> [31, 32, 33],
> [41, 42, 43],
> [51, 52, 53]]])
在这个假设的情况下,我在主数组 (arr) 中选择 3 个 windows of 3x3。
我的实际阵列是一个栅格,我基本上需要一堆(数千个)小 3x3 windows。
我们将不胜感激任何帮助甚至提示。
实际上我还没有找到任何实用的解决方案...因为很多很多小时
谢谢!
我们可以利用 np.lib.stride_tricks.as_strided
based scikit-image's view_as_windows
to get sliding windows. .
from skimage.util.shape import view_as_windows
def select_random_windows(arr, number_of_windows, window_size):
# Get sliding windows
w = view_as_windows(arr,window_size)
# Store shape info
m,n = w.shape[:2]
# Get random row, col indices for indexing into windows array
lidx = np.random.choice(m*n,number_of_windows,replace=False)
r,c = np.unravel_index(lidx,(m,n))
# If duplicate windows are allowed, use replace=True or np.random.randint
# Finally index into windows and return output
return w[r,c]
样本运行-
In [209]: arr
Out[209]:
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, 42, 43, 44, 45, 46, 47, 48, 49],
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59]])
In [210]: np.random.seed(0)
In [211]: select_random_windows(arr, number_of_windows=3, window_size=(2,4))
Out[211]:
array([[[41, 42, 43, 44],
[51, 52, 53, 54]],
[[26, 27, 28, 29],
[36, 37, 38, 39]],
[[22, 23, 24, 25],
[32, 33, 34, 35]]])
你可以试试[numpy.random.choice()][1]
。它采用一维或 ndarray,并通过从给定的 ndarray 中采样元素来创建单个元素或 ndarray。您还可以选择提供您想要的数组大小作为输出。
import numpy as np
arr = np.array(range(60)).reshape(6,10)
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, 42, 43, 44, 45, 46, 47, 48, 49],
> [50, 51, 52, 53, 54, 55, 56, 57, 58, 59]])
我需要的:
select_random_windows(arr, number_of windows= 3, window_size=3)
> array([[[ 1, 2, 3],
> [11, 12, 13],
> [21, 22, 23]],
>
> [37, 38, 39],
> [47, 48, 49],
> [57, 58, 59]],
>
> [31, 32, 33],
> [41, 42, 43],
> [51, 52, 53]]])
在这个假设的情况下,我在主数组 (arr) 中选择 3 个 windows of 3x3。
我的实际阵列是一个栅格,我基本上需要一堆(数千个)小 3x3 windows。
我们将不胜感激任何帮助甚至提示。
实际上我还没有找到任何实用的解决方案...因为很多很多小时
谢谢!
我们可以利用 np.lib.stride_tricks.as_strided
based scikit-image's view_as_windows
to get sliding windows.
from skimage.util.shape import view_as_windows
def select_random_windows(arr, number_of_windows, window_size):
# Get sliding windows
w = view_as_windows(arr,window_size)
# Store shape info
m,n = w.shape[:2]
# Get random row, col indices for indexing into windows array
lidx = np.random.choice(m*n,number_of_windows,replace=False)
r,c = np.unravel_index(lidx,(m,n))
# If duplicate windows are allowed, use replace=True or np.random.randint
# Finally index into windows and return output
return w[r,c]
样本运行-
In [209]: arr
Out[209]:
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, 42, 43, 44, 45, 46, 47, 48, 49],
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59]])
In [210]: np.random.seed(0)
In [211]: select_random_windows(arr, number_of_windows=3, window_size=(2,4))
Out[211]:
array([[[41, 42, 43, 44],
[51, 52, 53, 54]],
[[26, 27, 28, 29],
[36, 37, 38, 39]],
[[22, 23, 24, 25],
[32, 33, 34, 35]]])
你可以试试[numpy.random.choice()][1]
。它采用一维或 ndarray,并通过从给定的 ndarray 中采样元素来创建单个元素或 ndarray。您还可以选择提供您想要的数组大小作为输出。