由不规则(开始、停止)对分割的 numpy 数组

Question

我有一个带有x和y点值的numpy数组。我有另一个数组，其中包含开始和结束索引对。最初这个数据是熊猫DataFrame，但由于超过6000万个项目，loc算法非常慢。有什么快速的方法来拆分它吗？

import numpy as np
xy_array = np.arange(100).reshape(2,-1)
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, 60, 61, 62, 63, 64, 65, 66,
67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83,
84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]])
split_paris = [[0, 10], [10, 13], [13, 17], [20, 22]]
expected_result = [
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [50, 51, 52, 53, 54, 55, 56, 57, 58, 59]],
[[10, 11, 12], [60, 61, 62]],
[[13, 14, 15, 16], [63, 64, 65, 66]],
[[20, 21], [70, 71]]
]

更新：并非总是如此，下一对将从前一对的末尾开始。

Answer 1

这将做到这一点：

import numpy as np
xy_array = np.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, 60, 61, 62, 63, 64, 65, 66,
67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83,
84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]])
split_paris = [[0, 10], [10, 13], [13, 17]]
expected_result = [xy_array[:, x:y] for x, y in split_paris]
expected_result
#[array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9],
#       [50, 51, 52, 53, 54, 55, 56, 57, 58, 59]]), array([[10, 11, 12],
#       [60, 61, 62]]), array([[13, 14, 15, 16],
#       [63, 64, 65, 66]])]

它使用索引切片基本上在某种意义上工作array[rows, columns]让:获取所有行，x:y将列从xy获取。

Answer 2

您可以随时使用 numpy 提供的np.array_split函数。 并使用您想要的范围

x = np.arange(8.0)
>>> np.array_split(x, 3)
[array([ 0.,  1.,  2.]), array([ 3.,  4.,  5.]), array([ 6.,  7.])]