Question

我有3D numpy array的PCA

pcar =[[xa ya za]
       [xb yb zb]
       [xc yc zc]
       .
       .
       [xn yn zn]]

其中每一行都是一个点，我从上面PCA选择任意两个随机行作为一个群集

out_list=pcar[numpy.random.randint(0,pcar.shape[0],2)]

给numpy数组2行。

我必须找到每个out_list行的欧几里德距离与pcar中的每一行（点），并将该pcar点添加到out_list集群中的最近点。

Answer 1

修改好的，我下载，安装和自学numpy。这是一个numpy版本

旧答案

我意识到你想要一个愚蠢的答案。我的numpy生锈了，但由于没有其他答案，我想我会在Matlab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，有许多方法可以给这只猫留下皮肤，我只是给它一只。

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, :]

使用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')

Answer 2

Scipy中有一个非常快速的实现：

 from scipy.spatial.distance import cdist, pdist

cdist接受两个向量，例如你的pchar，并计算每个点之间的距离。 pdist只会给你那个矩阵的上三角。

由于它们是在幕后的C或Fortran中实现的，因此它们非常高效。

计算python中PCA的欧氏距离

2 个答案: