Python 如何在numpy中获得3x3x3数组上的欧氏距离
假设我有一个这样的(3,3,3)数组Python 如何在numpy中获得3x3x3数组上的欧氏距离,python,matrix,numpy,scipy,linear-algebra,Python,Matrix,Numpy,Scipy,Linear Algebra,假设我有一个这样的(3,3,3)数组 array([[[1, 1, 1], [1, 1, 1], [0, 0, 0]], [[2, 2, 2], [2, 2, 2], [2, 2, 2]], [[3, 3, 3], [3, 3, 3], [1, 1, 1]]]) 如何得到对应于3个值的每个向量与第0个值之间的欧几里德距离的9个值 例如做一个numpy.linalg.n
array([[[1, 1, 1],
[1, 1, 1],
[0, 0, 0]],
[[2, 2, 2],
[2, 2, 2],
[2, 2, 2]],
[[3, 3, 3],
[3, 3, 3],
[1, 1, 1]]])
如何得到对应于3个值的每个向量与第0个值之间的欧几里德距离的9个值
例如做一个numpy.linalg.norm([1,1,1]-[1,1,1])
2次,然后做norm([0,0,0]-[0,0,0])
,然后做norm([2,2,2,2]-[1,1,1])
2次,norm([2,2,2]-[0,0])
,然后做norm([3,3,3]-[1,1])
2次,最后做norm([1,0,0])
有什么好方法可以矢量化这个吗?我想把距离存储在(3,3,1)矩阵中
结果将是:
array([[[0. ],
[0. ],
[0. ]],
[[1.73],
[1.73],
[3.46]]
[[3.46],
[3.46],
[1.73]]])
您可能需要考虑,它有效地计算两个输入集合中的点对之间的距离(包括标准欧几里得度量,等等)。下面是示例代码:
import numpy as np
import scipy.spatial.distance as dist
i = np.array([[[1, 1, 1],
[1, 1, 1],
[0, 0, 0]],
[[2, 2, 2],
[2, 2, 2],
[2, 2, 2]],
[[3, 3, 3],
[3, 3, 3],
[1, 1, 1]]])
n,m,o = i.shape
# compute euclidean distances of each vector to the origin
# reshape input array to 2-D, as required by cdist
# only keep diagonal, as cdist computes all pairwise distances
# reshape result, adapting it to input array and required output
d = dist.cdist(i.reshape(n*m,o),i[0]).reshape(n,m,o).diagonal(axis1=2).reshape(n,m,1)
d
适用于:
array([[[ 0. ],
[ 0. ],
[ 0. ]],
[[ 1.73205081],
[ 1.73205081],
[ 3.46410162]],
[[ 3.46410162],
[ 3.46410162],
[ 1.73205081]]])
这种方法的最大警告是,我们正在计算n*m*o
距离,而我们只需要n*m
(这涉及到大量的重塑) keepdims
参数添加到numpy 1.7中,您可以使用它来保持总和轴:
np.sum((x - [1, 1, 1])**2, axis=-1, keepdims=True)**0.5
结果是:
[[[ 0. ]
[ 0. ]
[ 0. ]]
[[ 1.73205081]
[ 1.73205081]
[ 1.73205081]]
[[ 3.46410162]
[ 3.46410162]
[ 0. ]]]
array([[[ 0. ],
[ 0. ],
[ 0. ]],
[[ 1.73205081],
[ 1.73205081],
[ 3.46410162]],
[[ 3.46410162],
[ 3.46410162],
[ 1.73205081]]])
编辑
np.sum((x - x[0])**2, axis=-1, keepdims=True)**0.5
结果是:
[[[ 0. ]
[ 0. ]
[ 0. ]]
[[ 1.73205081]
[ 1.73205081]
[ 1.73205081]]
[[ 3.46410162]
[ 3.46410162]
[ 0. ]]]
array([[[ 0. ],
[ 0. ],
[ 0. ]],
[[ 1.73205081],
[ 1.73205081],
[ 3.46410162]],
[[ 3.46410162],
[ 3.46410162],
[ 1.73205081]]])
我正在做一些类似的事情,即计算视频卷中每对帧的平方距离之和(SSD)。我想这可能对你有帮助
视频卷是单个4d numpy阵列。此数组应具有维度
(时间、行、列、3)和数据类型np.uint8
输出是一个dtype float的正方形2d numpy数组。输出[i,j]应包含
帧i和j之间的SSD
video_volume = video_volume.astype(float)
size_t = video_volume.shape[0]
output = np.zeros((size_t, size_t), dtype = np.float)
for i in range(size_t):
for j in range(size_t):
output[i, j] = np.square(video_volume[i,:,:,:] - video_volume[j,:,:,:]).sum()
是的,很遗憾,norm
不允许轴
arg。我不知道为什么。你可能会在这篇感谢你的帮助的文章中找到你想要的答案。你可能离答案不远了;我编辑了这个问题,以显示我如何需要arr[0]中的第0个值,而不仅仅是arr[0][0]。@chimpsarehungry,我编辑了答案,只需将[1,1,1]
替换为x[0]
,您就会得到结果。谢谢,我的也是一个视频。如果可以的话,我想避免这样的循环。@chimpsarehungry:我修改后的答案应该能解决你们的澄清。