我有 3D 的 PCA numpy array as
pcar =[[xa ya za]
[xb yb zb]
[xc yc zc]
.
.
[xn yn zn]]
其中每一行都是一个点，我从PCA 上方选择了任意两个随机行作为集群
out_list=pcar[numpy.random.randint(0,pcar.shape[0],2)]
给出 2 行的 numpy 数组。
我必须从 out_list 的每一行与 pcar 中的每一行（点）找到欧几里得距离，并将该 pcar 点添加到 out_list 集群中最近的点。
【问题讨论】：
标签： python numpy pca euclidean-distance
Scipy 中有一个非常快速的实现：
from scipy.spatial.distance import cdist, pdist
cdist 采用两个向量，例如您的 pchar 一个，并计算每个点之间的距离。 pdist 只会为您提供该矩阵的上三角形。
由于它们是在幕后用 C 或 Fortran 实现的，因此它们的性能非常好。
【讨论】：
编辑 好的，我下载、安装并自学了 numpy。这是一个numpy版本
旧答案
我知道你想要一个 numpy 的答案。我的 numpy 生锈了，但由于没有其他答案，我想我会在 Matlab 中给你一个。转换应该很简单。我假设问题是概念，而不是代码。
请注意，给这只猫剥皮的方法有很多种，我只提供一种。
工作 Numpy 版本
import numpy as np
pcar = np.random.rand(10,3)
out_list=pcar[np.random.randint(0,pcar.shape[0],2)]
ol_1 = out_list[0,:]
ol_2 = out_list[1,:]
## Get the individual distances
## The trick here is to pre-multiply the 1x3 ol vector with a row of
## ones of size 10x1 to get a 10x3 array with ol replicated, so that it
## can simply be subtracted
d1 = pcar - ones( size(pcar,1))*ol_1
d2 = pcar - ones( size(pcar,1))*ol_2
##% Square them using an element-wise square
d1s = np.square(d1)
d2s = np.square(d2)
##% Sum across the rows, not down columns
d1ss = np.sum(d1s, axis=1)
d2ss = np.sum(d2s, axis=1)
##% Square root using an element-wise square-root
e1 = np.sqrt(d1ss)
e2 = np.sqrt(d2ss)
##% Assign to class one or class two
##% Start by assigning one to everything, then select all those where ol_2
##% is closer and assign them the number 2
assign = ones(size(e1,0));
assign[e2<e1] = 2
##% Separate
pcar1 = pcar[ assign==1, :]
pcar2 = pcar[ assign==2, :]
工作 Matlab 版本
close all
clear all
% Create 10 records each with 3 attributes
pcar = rand(10, 3)
% Pick two (normally at random of course)
out_list = pcar(1:2, :)
% Hard-coding this separately, though this can be done iteratively
ol_1 = out_list(1,:)
ol_2 = out_list(2,:)
% Get the individual distances
% The trick here is to pre-multiply the 1x3 ol vector with a row of
% ones of size 10x1 to get a 10x3 array with ol replicated, so that it
% can simply be subtracted
d1 = pcar - ones( size(pcar,1), 1)*ol_1
d2 = pcar - ones( size(pcar,1), 1)*ol_2
% Square them using an element-wise square
d1s = d1.^2
d2s = d2.^2
% Sum across the rows, not down columns
d1ss = sum(d1s, 2)
d2ss = sum(d2s, 2)
% Square root using an element-wise square-root
e1 = sqrt(d1ss)
e2 = sqrt(d2ss)
% Assign to class one or class two
% Start by assigning one to everything, then select all those where ol_2
% is closer and assign them the number 2
assign = ones(length(e1),1);
assign(e2<e1)=2
% Separate
pcar1 = pcar( assign==1, :)
pcar2 = pcar( assign==2, :)
% Plot
plot3(pcar1(:,1), pcar1(:,2), pcar1(:,3), 'g+')
hold on
plot3(pcar2(:,1), pcar2(:,2), pcar2(:,3), 'r+')
plot3(ol_1(1), ol_1(2), ol_1(3), 'go')
plot3(ol_2(1), ol_2(2), ol_2(3), 'ro')
【讨论】：