吳恩達機器學習 - 評估假設

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				版權宣告：如果感覺寫的不錯，轉載標明出處連結哦~blog.csdn.net/wyg1997					https://blog.csdn.net/wyg1997/article/details/80778511				</div>
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						<p>題目連結：<a href="https://s3.amazonaws.com/spark-public/ml/exercises/on-demand/machine-learning-ex5.zip" rel="nofollow" target="_blank">點選開啟連結</a></p>
 

正則線性迴歸：

視覺化資料：

Code:

load ('ex5data1.mat');
plot(X, y, 'rx', 'MarkerSize', 10, 'LineWidth', 1.5);
  

   
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結果為：

代價函式：

公式（並正則化）：

Code(填充在linearRegCostFunction.m中):

t = X*theta-y;
J = t'*t/(2.0*m) + lambda/(2.0*m)*theta(2:end)'*theta(2:end);
  

   
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求正則線性迴歸梯度

公式：

Code：

grad(1) = (X(:,1)'*t)/m;
grad(2:end) = (X(:,2:end)'*t./m) + (lambda/m).*theta(2:end);
  

   
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訓練樣本

效果如下：

偏差和方差

樣本數與訓練誤差、交叉驗證誤差的關係

Code（learningCurve.m）:

function [error_train, error_val] = ...
    learningCurve(X, y, Xval, yval, lambda)
%LEARNINGCURVE Generates the train and cross validation set errors needed 
 

%to plot a learning curve
%   [error_train, error_val] = ...
%       LEARNINGCURVE(X, y, Xval, yval, lambda) returns the train and
%       cross validation set errors for a learning curve. In particular, 
%       it returns two vectors of the same length - error_train and 
%       error_val. Then, error_train(i) contains the training error for
%       i examples (and similarly for error_val(i)).
%
%   In this function, you will compute the train and test errors for
%   dataset sizes from 1 up to m. In practice, when working with larger
%   datasets, you might want to do this in larger intervals.
%

% Number of training examples
m = size(X, 1);

% You need to return these values correctly
error_train = zeros(m, 1);
error_val   = zeros(m, 1);

% ====================== YOUR CODE HERE ======================
% Instructions: Fill in this function to return training errors in 
%               error_train and the cross validation errors in error_val. 
%               i.e., error_train(i) and 
%               error_val(i) should give you the errors
%               obtained after training on i examples.
%
% Note: You should evaluate the training error on the first i training
%       examples (i.e., X(1:i, :) and y(1:i)).
%
%       For the cross-validation error, you should instead evaluate on
%       the _entire_ cross validation set (Xval and yval).
%
% Note: If you are using your cost function (linearRegCostFunction)
%       to compute the training and cross validation error, you should 
%       call the function with the lambda argument set to 0. 
%       Do note that you will still need to use lambda when running
%       the training to obtain the theta parameters.
%
% Hint: You can loop over the examples with the following:
%
%       for i = 1:m
%           % Compute train/cross validation errors using training examples 
%           % X(1:i, :) and y(1:i), storing the result in 
%           % error_train(i) and error_val(i)
%           ....
%           
%       end
%

% ---------------------- Sample Solution ----------------------

for i = 1:m
    theta = trainLinearReg(X(1:i,:),y(1:i),lambda);
    [error_train(i), ~] = linearRegCostFunction(X(1:i,:),y(1:i),theta,0);

    [error_val(i), ~] = linearRegCostFunction(Xval,yval,theta,0);
end


% -------------------------------------------------------------

% =========================================================================

end
  

   
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圖示（高偏差）：

多項式迴歸

特徵擴張（將一次線性的特徵擴張到p次）

Code（polyFeatures.m）：

function [X_poly] = polyFeatures(X, p)
%POLYFEATURES Maps X (1D vector) into the p-th power
%   [X_poly] = POLYFEATURES(X, p) takes a data matrix X (size m x 1) and
%   maps each example into its polynomial features where
%   X_poly(i, :) = [X(i) X(i).^2 X(i).^3 ...  X(i).^p];
%


% You need to return the following variables correctly.
X_poly = zeros(numel(X), p);

% ====================== YOUR CODE HERE ======================
% Instructions: Given a vector X, return a matrix X_poly where the p-th 
%               column of X contains the values of X to the p-th power.
%
% 

for i = 1:p
    X_poly(:,i) = X.^i;
end

% =========================================================================

end
  

   
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效果檢視

這裡說明一下接下來執行ex5的效果。因為我們把特徵擴張到了8次方（本例中），所以最後一個特徵的數值特別特別大，所以我們這裡需要特徵歸一化。

程式是這麼呼叫的（實際上我們可以自己寫這個部分，也不是很麻煩）（featureNormalize.m）：

function [X_norm, mu, sigma] = featureNormalize(X)
%FEATURENORMALIZE Normalizes the features in X 
%   FEATURENORMALIZE(X) returns a normalized version of X where
%   the mean value of each feature is 0 and the standard deviation
%   is 1. This is often a good preprocessing step to do when
%   working with learning algorithms.

mu = mean(X);
X_norm = bsxfun(@minus, X, mu);

sigma = std(X_norm);
X_norm = bsxfun(@rdivide, X_norm, sigma);


% ============================================================

end
  

   
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然後就是利用我們之前寫好的函式計算代價啦：

效果圖如下：

從圖上看來是高方差了（過擬合）

然後我們看一下不同的λ對結果有什麼影響吧（這一步不計分，只是幫助我們來理解）~

Code（直接在控制檯執行）（只需對第一行進行修改）：

lambda = 1;
[theta] = trainLinearReg(X_poly, y, lambda);

% Plot training data and fit
figure(1);
plot(X, y, 'rx', 'MarkerSize', 10, 'LineWidth', 1.5);
plotFit(min(X), max(X), mu, sigma, theta, p);
xlabel('Change in water level (x)');
ylabel('Water flowing out of the dam (y)');
title (sprintf('Polynomial Regression Fit (lambda = %f)', lambda));

figure(2);
[error_train, error_val] = ...
    learningCurve(X_poly, y, X_poly_val, yval, lambda);
plot(1:m, error_train, 1:m, error_val);

title(sprintf('Polynomial Regression Learning Curve (lambda = %f)', lambda));
xlabel('Number of training examples')
ylabel('Error')
axis([0 13 0 100])
legend('Train', 'Cross Validation')
  

   
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λ=1時（擬合的不錯）

λ=32時（欠擬合啦）

那麼我們改小一點：λ=0.1時（懲罰的力度不夠，還是有點過擬合）

使用交叉驗證集選擇合適的λ（畫出λ-Error曲線）

效果圖：

發現λ大概在3的位置上比較好

我們來看看λ=3的曲線：