有时将行向量或列向量“克隆”到矩阵中是有用的。克隆的意思是转换行向量，比如
[1, 2, 3]
变成一个矩阵
[[1, 2, 3],
[1, 2, 3],
[1, 2, 3]]
或者一个列向量，比如
[[1],
[2],
[3]]
into
[[1, 1, 1]
[2, 2, 2]
[3, 3, 3]]
在MATLAB或八度音阶中，这很容易做到:
x = [1, 2, 3]
a = ones(3, 1) * x
a =
1 2 3
1 2 3
1 2 3
b = (x') * ones(1, 3)
b =
1 1 1
2 2 2
3 3 3
我想在numpy中重复这一点，但不成功
In [14]: x = array([1, 2, 3])
In [14]: ones((3, 1)) * x
Out[14]:
array([[ 1., 2., 3.],
[ 1., 2., 3.],
[ 1., 2., 3.]])
# so far so good
In [16]: x.transpose() * ones((1, 3))
Out[16]: array([[ 1., 2., 3.]])
# DAMN
# I end up with
In [17]: (ones((3, 1)) * x).transpose()
Out[17]:
array([[ 1., 1., 1.],
[ 2., 2., 2.],
[ 3., 3., 3.]])
为什么第一种方法(在[16])不工作?有没有一种更优雅的方法在python中实现这个任务?
另一个解决方案
>> x = np.array([1,2,3])
>> y = x[None, :] * np.ones((3,))[:, None]
>> y
array([[ 1., 2., 3.],
[ 1., 2., 3.],
[ 1., 2., 3.]])
为什么?当然，重复和平铺是正确的方法。但是None索引是一个强大的工具，它可以让我多次快速向量化操作(尽管它很快就会非常消耗内存!)
下面是我自己代码中的一个例子:
# trajectory is a sequence of xy coordinates [n_points, 2]
# xy_obstacles is a list of obstacles' xy coordinates [n_obstacles, 2]
# to compute dx, dy distance between every obstacle and every pose in the trajectory
deltas = trajectory[:, None, :2] - xy_obstacles[None, :, :2]
# we can easily convert x-y distance to a norm
distances = np.linalg.norm(deltas, axis=-1)
# distances is now [timesteps, obstacles]. Now we can for example find the closest obstacle at every point in the trajectory by doing
closest_obstacles = np.argmin(distances, axis=1)
# we could also find how safe the trajectory is, by finding the smallest distance over the entire trajectory
danger = np.min(distances)
使用numpy.tile:
>>> tile(array([1,2,3]), (3, 1))
array([[1, 2, 3],
[1, 2, 3],
[1, 2, 3]])
或者对于重复列:
>>> tile(array([[1,2,3]]).transpose(), (1, 3))
array([[1, 1, 1],
[2, 2, 2],
[3, 3, 3]])
