如何有效地在二维上合并两个 3d 数组？
假设我有两个形状为 (1.000.000, ???, 50) 的 3 维数组 (a & b)，(??? = 见下文)。
如何合并它们，使结果为 (1.000.000, {a 的形状 + b 的第二个维度} , 50)？
以下是示例，如下所示：（np.arrays 也是可能的）
编辑：添加了可用代码，请滚动^^
[ #a
[
],
[
[1 2 3]
],
[
[0 2 7]
[1 Nan 3]
],
[
[10 0 3]
[NaN 9 9]
[10 NaN 3]
],
[
[8 2 0]
[2 2 3]
[8 1 3]
[1 2 3]
],
[
[0 2 3]
[1 2 9]
[1 2 3]
[1 0 3]
[1 2 3]
]
]
[#b
[
[7 2 3]
[1 2 9]
[1 2 3]
[8 0 3]
[1 7 3]
]
[
[3 9 0]
[2 2 3]
[8 1 3]
[0 2 3]
],
[
[10 0 3]
[0 NaN 9]
[10 NaN 3]
],
[
[0 2 NaN]
[1 Nan 3]
],
[
[1 2 NaN]
],
[
]
]
a = [ [ ],
[ [1, 2, 3] ],
[ [0, 2, 7], [1,np.nan,3] ],
[
[10,0,3], [np.nan,9,9], [10,np.nan,3]
],
[
[8,2,0], [2,2,3], [8,1,3], [1,2,3]
],
[
[0,2,3], [1,2,9], [1,2,3], [1,0,3], [1,2,3]
]
]
b = [
[
[7,2,3], [1,2,9], [1,2,3], [8,0,3], [1,7,3]
],
[
[3,9,0], [2,2,3], [8,1,3], [0,2,3]
],
[
[10,0,3], [0,np.nan,9], [10,np.nan,3]
],
[
[0,2,np.nan], [1,np.nan,3]
],
[
[1,2,np.nan]
],
[
]
]
您知道一种有效的方法吗？
慕后森2回答
-
哈士奇WWW
您不能拥有一个维度长度可变的数组。a和b很可能是列表的列表，而不是数组。您可以将列表理解与 zip 一起使用：np.array([x+y for x,y in zip(a,b)]) 编辑：或根据提供的注释，如果a和b是数组列表：np.array([np.vstack((x,y)) for x,y in zip(a,b)])您的示例的输出如下所示：[[[ 7. 2. 3.] [ 1. 2. 9.] [ 1. 2. 3.] [ 8. 0. 3.] [ 1. 7. 3.]] [[ 1. 2. 3.] [ 3. 9. 0.] [ 2. 2. 3.] [ 8. 1. 3.] [ 0. 2. 3.]] [[ 0. 2. 7.] [ 1. nan 3.] [10. 0. 3.] [ 0. nan 9.] [10. nan 3.]] [[10. 0. 3.] [nan 9. 9.] [10. nan 3.] [ 0. 2. nan] [ 1. nan 3.]] [[ 8. 2. 0.] [ 2. 2. 3.] [ 8. 1. 3.] [ 1. 2. 3.] [ 1. 2. nan]] [[ 0. 2. 3.] [ 1. 2. 9.] [ 1. 2. 3.] [ 1. 0. 3.] [ 1. 2. 3.]]] -
MYYA
要执行串联，请运行：result = np.concatenate([a, b], axis=1)为了测试这段代码，我创建了a和b：a = np.stack([ np.full((2, 3), i) for i in range(1, 6)], axis=1)b = np.stack([ np.full((2, 3), i + 10) for i in range(1, 4)], axis=1)所以它们包含：array([[[1, 1, 1], array([[[11, 11, 11], [2, 2, 2], [12, 12, 12], [3, 3, 3], [13, 13, 13]], [4, 4, 4], [5, 5, 5]], [[11, 11, 11], [12, 12, 12], [[1, 1, 1], [13, 13, 13]]]) [2, 2, 2], [3, 3, 3], [4, 4, 4], [5, 5, 5]]])它们的形状是：(2, 5, 3)和(2, 3, 3)我的串联结果是：array([[[ 1, 1, 1], [ 2, 2, 2], [ 3, 3, 3], [ 4, 4, 4], [ 5, 5, 5], [11, 11, 11], [12, 12, 12], [13, 13, 13]], [[ 1, 1, 1], [ 2, 2, 2], [ 3, 3, 3], [ 4, 4, 4], [ 5, 5, 5], [11, 11, 11], [12, 12, 12], [13, 13, 13]]])形状是(2, 8, 3)，正如它应该的那样。我尝试了您评论中的代码。执行后a = list((map(lambda i: a[:i], range(1,a.shape[0]+1))))，结果是：[array([[1, 1]], dtype=int64), array([[1, 1], [1, 1]], dtype=int64), array([[1, 1], [1, 1], [1, 1]], dtype=int64), array([[1, 1], [1, 1], [1, 1], [1, 1]], dtype=int64), array([[1, 1], [1, 1], [1, 1], [1, 1], [1, 1]], dtype=int64),...所以a是不同大小的数组的列表。您构建数据的方式有问题。首先检查您的两个数组是否都是3-D的，并且它们的形状仅在轴 1 上有所不同。只有这样您才能在它们上运行我的代码。现在a和b都是普通的pythonic 列表，而不是Numpy数组！
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