R

再現線性判別分析投影圖

  • August 11, 2014

我正在與線性判別分析(LDA)中的投影點作鬥爭。許多關於多元統計方法的書籍都用下圖說明了 LDA 的思想。

圖1

問題描述如下。首先,我們需要繪製決策邊界,添加垂直線,然後在其上繪製數據點的投影。我想知道如何將投影點添加到垂直線。

有什麼建議/指示嗎?

判別軸(點在圖 1 上投影到的軸)由第一個特徵向量給出. 在只有兩個類的情況下,這個特徵向量與, 在哪裡是類質心。對這個向量(或得到的特徵向量)進行歸一化,得到單位軸向量. 這足以繪製軸。

要將(中心)點投影到該軸上,您只需計算. 這裡是一個線性投影儀.

以下是您的保管箱和 LDA 投影中的數據樣本:

LDA 投影

這是生成此圖的 MATLAB 代碼(根據要求):

% # data taken from your example
X = [-0.9437    -0.0433; -2.4165    -0.5211; -2.0249    -1.0120; ...
   -3.7482 0.2826; -3.3314 0.1653; -3.1927 0.0043; -2.2233 -0.8607; ...
   -3.1965 0.7736; -2.5039 0.2960; -4.4509 -0.3555];
G = [1 1 1 1 1 2 2 2 2 2];

% # overall mean
mu = mean(X);

% # loop over groups
for g=1:max(G)
   mus(g,:) = mean(X(G==g,:)); % # class means
   Ng(g) = length(find(G==g)); % # number of points per group
end

Sw = zeros(size(X,2)); % # within-class scatter matrix
Sb = zeros(size(X,2)); % # between-class scatter matrix
for g=1:max(G)
   Xg = bsxfun(@minus, X(G==g,:), mus(g,:)); % # centred group data
   Sw = Sw + transpose(Xg)*Xg;
   Sb = Sb + Ng(g)*(transpose(mus(g,:) - mu)*(mus(g,:) - mu));
end

St = transpose(bsxfun(@minus,X,mu)) * bsxfun(@minus,X,mu); % # total scatter matrix
assert(sum(sum((St-Sw-Sb).^2)) < 1e-10, 'Error: Sw + Sb ~= St')

% # LDA
[U,S] = eig(Sw\Sb);

% # projecting data points onto the first discriminant axis
Xcentred = bsxfun(@minus, X, mu);
Xprojected = Xcentred * U(:,1)*transpose(U(:,1));
Xprojected = bsxfun(@plus, Xprojected, mu);

% # preparing the figure
colors = [1 0 0; 0 0 1];
figure
hold on
axis([-5 0 -2.5 2.5])
axis square

% # plot discriminant axis
plot(mu(1) + U(1,1)*[-2 2], mu(2) + U(2,1)*[-2 2], 'k')
% # plot projection lines for each data point
for i=1:size(X,1)
   plot([X(i,1) Xprojected(i,1)], [X(i,2) Xprojected(i,2)], 'k--')
end
% # plot projected points
scatter(Xprojected(:,1), Xprojected(:,2), [], colors(G, :))
% # plot original points
scatter(X(:,1), X(:,2), [], colors(G, :), 'filled')

引用自:https://stats.stackexchange.com/questions/111421

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