在Python中计算PCA的欧几里得距离

Question

我有一个三维的 numpy array，它是用来做主成分分析（PCA）的，内容如下：

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

在这个数组中，每一行代表一个点。我从上面的 PCA 结果中随机选了两行作为一个聚类，内容如下：

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

这样就得到了一个包含两行的 numpy array。

接下来，我需要计算出 out_list 中每一行与 pcar 中每一行（点）之间的欧几里得距离，然后把 pcar 中最近的点添加到 out_list 的聚类中。

Answer 1

编辑

好的，我下载并安装了numpy，还自学了一下。这里是一个numpy版本的代码。

旧答案

我知道你想要一个numpy的答案。我的numpy有点生疏，不过因为没有其他答案，我想给你一个Matlab的版本。转换成numpy应该很简单。我假设你关心的是概念，而不是代码。

请注意，有很多方法可以解决这个问题，我只是提供其中一种。

可用的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, :]

可用的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

在Scipy中，有一个非常快速的实现：

 from scipy.spatial.distance import cdist, pdist

cdist这个函数可以接收两个向量，就像你的pchar那样，然后计算这两个点之间的距离。

而pdist则只会给你这个距离矩阵的上三角部分。

因为它们的背后是用C或Fortran语言实现的，所以运行起来非常高效。