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如何广播这个欧氏距离?

[英]How to broadcast this euclidean distance?

So, I have this code, I tried to calculate euclidean distance on each element on list1 it throws an error if list1 has 2 elements, any idea on this?所以,我有这段代码,我试图计算 list1 上每个元素的欧氏距离,如果 list1 有 2 个元素,它会抛出错误,对此有什么想法吗?

import numpy as np
from scipy.spatial import distance
list1 =[(10.2,20.2),(5.3,9.2)]
list2 = [(2.2,3.3)]
list1 =np.array(list1)
dist1= distance.euclidean(list1,list2)
print("distance",dist1)

prints:印刷:

ValueError: Input vector should be 1-D.

You can directly manipulate numpy arrays in order to find euclidean distances here.您可以直接操作numpy arrays 以在此处找到欧氏距离。

I am assuming either list1 or list2 contains 1 element and distances are to be calculated between each element of the other list and the single element.我假设list1list2包含1元素,并且要计算另一个列表的每个元素与单个元素之间的距离。 Rest is taken care of by numpy broadcasting . Rest 由numpy 广播处理

import numpy as np
list1 =[(10.2,20.2),(5.3,9.2)]
list2 = [(2.2,3.3)]

a = np.array(list1)
b = np.array(list2)

dist = np.sqrt(((b - a)**2).sum(axis = 1))

Output: dist dist :距离

array([18.69786084,  6.66483308])

where dist[0] gives distance(list1[0], list2[0]) and dist[1] gives distance(list1[1], list2[0]) .其中dist[0]给出distance(list1[0], list2[0])并且dist[1]给出distance(list1[1], list2[0])

It generalizes even when list1 has arbitrary number of points, the only constraint is the other list should have only one point.即使list1具有任意数量的点,它也可以概括,唯一的限制是另一个列表应该只有一个点。

As you say a for loop is probably easiest here.正如您所说,for 循环在这里可能是最简单的。 I have changed your lists to be lists-of-lists rather than lists-of-tuples although I'm not sure that's actually necessary.我已将您的列表更改为列表列表而不是元组列表,尽管我不确定这是否真的有必要。 I don't have scipy installed to check.我没有安装 scipy 来检查。

from scipy.spatial import distance
list1 =[[10.2,20.2],[5.3,9.2]]
list2 = [2.2,3.3]
for point in list1:
    dist1= distance.euclidean(point,list2)
print("distance",dist1)

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