Python 将二维数组转换为重复现有值的三维数组

我有一个形状数组（360480），包含从1到11的值

 Array([[ 1,  1,  1, ...,  1,  1,  1],
   [ 1,  1,  1, ...,  1,  1,  1],
   [ 1,  1,  1, ...,  1,  1,  1],
   ...,
   [ 4,  4,  4, ..., 11, 11, 11],
   [ 4,  4,  4, ..., 11, 11, 11],
   [ 4,  4,  4, ..., 11, 11, 11]])

我怎样才能将这个数组重塑成一个数组的形状（360480,3）呢

np.all(array[:,:,0]==array[:,:,1])

及

都是真的吗？ 预期结果应该是：

array([[[ 1,  1,  1],
    [ 1,  1,  1],
    [ 1,  1,  1],
    ...,
    [ 1,  1,  1],
    [ 1,  1,  1],
    [ 1,  1,  1]],

   [[ 4,  4,  4],
    [ 4,  4,  4],
    [ 4,  4,  4],
    ...,
    [11, 11, 11],
    [11, 11, 11],
    [11, 11, 11]],

   [[ 4,  4,  4],
    [ 4,  4,  4],
    [ 4,  4,  4],
    ...,
    [11, 11, 11],
    [11, 11, 11],
    [11, 11, 11]]])

---

您可以使用numpy.repeat函数

---

这正是我要找的！它按我的要求工作！

array([[[ 1,  1,  1],
    [ 1,  1,  1],
    [ 1,  1,  1],
    ...,
    [ 1,  1,  1],
    [ 1,  1,  1],
    [ 1,  1,  1]],

   [[ 4,  4,  4],
    [ 4,  4,  4],
    [ 4,  4,  4],
    ...,
    [11, 11, 11],
    [11, 11, 11],
    [11, 11, 11]],

   [[ 4,  4,  4],
    [ 4,  4,  4],
    [ 4,  4,  4],
    ...,
    [11, 11, 11],
    [11, 11, 11],
    [11, 11, 11]]])

array3d = np.repeat(array2d[:, :, None], repeats=3, axis=2)