如何在numpy中合并多个2d数组？

Question

MWE：假设我有12个2D阵列。 我想使用循环将它们合并为某个固定大小。 下面的代码给出了我想要得到X的结果。在这里，我连接了3个2D数组，但是如果我想根据自己的选择使用循环来固定此大小，该如何做呢？

import numpy as np
a1 = np.array([[0, 12, 3], [5, 8, 9]])
a2 = np.array([[2, 13, 3], [5, 9, 9]])
a3 = np.array([[0, 24, 4], [6, 10, 9]])
a4 = np.array([[1, 55, 6], [4, 5, 19]])
a5 = np.array([[1, 56, 6], [4, 01, 9]])
a6 = np.array([[1, 57, 6], [4, 20, 9]])
a7 = np.array([[1, 58, 6], [4, 30, 9]])
a8 = np.array([[1, 59, 6], [4, 40, 9]])
a9 = np.array([[1, 51, 6], [4, 60, 9]])
a10 = np.array([[1, 34, 6], [4, 60, 9]])
a11 = np.array([[1, 51, 62], [4, 30, 9]])
a12 = np.array([[1, 1, 6], [4, 7, 9]]) 

M=[a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12]
d1 = np.concatenate((a1,a2,a3), 1)
d2 = np.concatenate((a4,a5,a6), 1)
d3 = np.concatenate((a7,a8,a9), 1)
d4 = np.concatenate((a10,a11,a12), 1)
X=[d1,d2,d3,d4]

Answer 1

您的问题有点不清楚，我想您想将串联的2D数组（例如，在d1，d2，..中）的数量设为可变，并通过循环执行与在代码中相同的操作。

您可以这样操作：

这将产生与您的代码相同的结果，并且您可以为不同的结果更改变量chosenSize 。

    a1 = np.array([[0, 12, 3], [5, 8, 9]])
    a2 = np.array([[2, 13, 3], [5, 9, 9]])
    a3 = np.array([[0, 24, 4], [6, 10, 9]])
    a4 = np.array([[1, 55, 6], [4, 5, 19]])
    a5 = np.array([[1, 56, 6], [4, 1, 9]])
    a6 = np.array([[1, 57, 6], [4, 20, 9]])
    a7 = np.array([[1, 58, 6], [4, 30, 9]])
    a8 = np.array([[1, 59, 6], [4, 40, 9]])
    a9 = np.array([[1, 51, 6], [4, 60, 9]])
    a10 = np.array([[1, 34, 6], [4, 60, 9]])
    a11 = np.array([[1, 51, 62], [4, 30, 9]])
    a12 = np.array([[1, 1, 6], [4, 7, 9]])

    M = [a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11, a12]
    d1 = []  # will be used to concatenate
    X = []
    chosenSize = 3  # chosen size

    numberOfIterations = M.__len__()//chosenSize

    for i in range(0, numberOfIterations):
        for j in range(0, chosenSize):
            d1.append(M[j + chosenSize * i])
        d1 = np.concatenate(d1, 1)
        X.append(d1)
        d1 = []

    print(X)