主页 / coding / 问题 / 79335463
Accepted
user28311778
Asked: 2025-01-07 17:28:28 +0800 CST

如何获取我想要的 NumPy 数组

  • 772

我怎样才能像这样改变数组？

arr = [
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31],
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41],
...,
[130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141],
]

=> [
[ [0, 1], [20, 21], [30, 31], ,.., [130, 131]],
[ [2, 3], [22, 23], [32, 33], .., [132, 133]],
[ [4, 5], [24, 25], [34, 35], .., [134, 135]],
[ [6, 7], [26, 27], [36, 37], .., [136, 137]],
[ [8, 9], [28, 29], [38, 39], .., [138, 139]],
[ [10, 11], [30, 31], [40, 41], .., [140, 141]],
...
]

我的尝试最好修改为以下内容，但这不是我想要的结果。

[ [0, 1], [2, 3], [4, 5], ,.., [10, 11]],
[ [20, 21], [22, 23], [24, 25], .., [30, 31]],
numpy

4 个回答

  1. Best Answer

    您可以使用reshape+ swapaxes：

    n, m = arr.shape
out = arr.reshape(n, m//2, 2).swapaxes(0, 1)

# or
out = arr.reshape(n, -1, 2).swapaxes(0, 1)

    例子：

    # input
arr = np.array([[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
                [20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31],
                [30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41],
                [130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141]])

# output
array([[[  0,   1], [ 20,  21], [ 30,  31], [130, 131]],
       [[  2,   3], [ 22,  23], [ 32,  33], [132, 133]],
       [[  4,   5], [ 24,  25], [ 34,  35], [134, 135]],
       [[  6,   7], [ 26,  27], [ 36,  37], [136, 137]],
       [[  8,   9], [ 28,  29], [ 38,  39], [138, 139]],
       [[ 10,  11], [ 30,  31], [ 40,  41], [140, 141]]])
    • 6

  2. 添加到@Mozway 解决方案，您可以使用 stride 来更高级（且不安全）地使用 numpy：

    n, m = arr.shape
stride_x, stride_y = arr.strides
out = np.lib.stride_tricks.as_strided( arr, shape=(m//2,n,2), strides=(stride_y*2, stride_x, stride_y))

    有关文档，请参阅https://numpy.org/doc/stable/reference/generated/numpy.lib.stride_tricks.as_strided.html

    • 2

  3. 您想要以 2 为步长（对于成对的）进行逐列配对。想象您的arr列表有行和列。您需要成对地遍历原始列表的列，并将所有行中的相应元素分组。

    result = [
    [[arr[row][col], arr[row][col + 1]] for row in range(len(arr))]
    for col in range(0, len(arr[0]), 2)
]
    • 0

  4. 您应该使用双重 for 循环，一个循环以两个步骤遍历元素，第二个 for 循环用于创建两个对。

    output = [
    # Going over the elements with steps of 2.
    [
        # Create pairs of two
        [row[i], row[i + 1]]
        for row in arr
    ]
    for i in range(0, len(arr[0]), 2)
]

    当然，这是假设所有元素的arr长度相同

    • 0

相关问题