ZhankunLuo_TakeHomeExamProblem2

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#pattern recognition #homework

本文探讨了糖尿病预测中的特征选择方法，通过Relief算法确定了四个关键特征，并利用KNN、SSE和LDA三种分类器进行对比分析，展示了不同分类器在预测精度上的差异。

Take Home Exam Problem2

Zhankun Luo

PUID: 0031195279

Email: luo333@pnw.edu

Fall-2018-ECE-59500-009

Instructor: Toma Hentea

文章目录

Function

plot_point

function plot_point(X,y)
%this function can handle up to 6 different classes
[l,N]=size(X); %N=no. of data vectors, l=dimensionality
if(l~=2)
    fprintf('NO PLOT CAN BE GENERATED\n')
    return
else
    pale=['ro';'g+';'b.';'y.';'m.';'c.']; 
    %Plot of the data vectors
    hold on
    for i=1:N
       plot(X(1,i),X(2,i),pale(y(i),:))
    end    
    hold off
end

Step 1: find the best combination of 4 features.

With Relief( X, y ), get weight of each feature X( i, ? to outcome y.
Relief
diff

Relief

function weight = Relief( X, y )
%weight = Relief( X, y )
% for 2 classes
%OUTPUT: weight of different features(dimesions) to y
[l, N] = size(X);
max_X = max(X, [], 2) * ones(1, N); % return max of each row/dimension
min_X = min(X, [], 2) * ones(1, N);
X_process = (X - min_X)./ (max_X - min_X); % 0 =< X_process =< 1
X_process_class1 = X_process(:, find(y == 1)); N1 = size(X_process_class1, 2);
X_process_class2 = X_process(:, find(y == 0)); N2 = size(X_process_class2, 2);
weight = zeros(l, 1);
for i = 1:N1
    dist_1 = X_process_class1 - X_process_class1(:, i) * ones(1, N1);
    [Dist_1, Index_1] = sort(sum(dist_1.^2));
    diff_1 = Dist_1(2); index_1 = Index_1(2);
    dist_2 = X_process_class2 - X_process_class1(:, i) * ones(1, N2);
    [Dist_2, Index_2] = sort(sum(dist_2.^2));
    diff_2 = Dist_2(1); index_2 = Index_2(1);
    weight = weight - abs(dist_1(:, index_1)) + abs(dist_2(:, index_2));
end
for i = 1:N2
    dist_2 = X_process_class2 - X_process_class2(:, i) * ones(1, N2);
    [Dist_2, Index_2] = sort(sum(dist_2.^2));
    diff_2 = Dist_2(2); index_2 = Index_2(2);
    dist_1 = X_process_class1 - X_process_class2(:, i) * ones(1, N1);
    [Dist_1, Index_1] = sort(sum(dist_1.^2));
    diff_1 = Dist_1(1); index_1 = Index_1(1);
    weight = weight - abs(dist_2(:, index_2)) + abs(dist_1(:, index_1));
end
weight = weight / N;
end

Step 2: Design a classifier using the training set (X1R, Y1) , use it to predict the labels Y2 for the test set X2R.

(0) normalize data

normalize

function [X_norm, Xtest_norm] = normalize( X, Xtest )
%X_norm = normalize( X )
[l, N] = size(X);
Ntest = size(Xtest, 2);
max_X = max(X, [], 2) * ones(1, N); % return max of each row/dimension
min_X = min(X, [], 2) * ones(1, N);
X_norm = (X - min_X)./ (max_X - min_X); % 0 =< X_norm =< 1
max_Xtest = max(X, [], 2) * ones(1, Ntest); % return max of each row/dimension
min_Xtest = min(X, [], 2) * ones(1, Ntest);
Xtest_norm = (Xtest - min_Xtest)./ (max_Xtest - min_Xtest); % 0 =< Xtest_norm =< 1
end

(1) KNN algorithm

k_nn_classifier

function z = k_nn_classifier(Z, v, k, X)
[l, N1] = size(Z); %in Z we have the training data
[l, N] = size(X); %in X we have the points to be classified
c = max(v); %The number of classes
%in v we have the classes to which the vectors in Z belong

%Computation of the (squared) Euclidean distance of a point in X from each
%reference vector
for i = 1:N
    dist = sum((X(:, i) * ones(1, N1) - Z).^2);
    %sorting the above distances in ascending order
    [sorted, nearest]=sort(dist); % MODE=‘ASCEND’
    %counting the class occurrences among the k-closest reference vectors
    %Z(:,i)
    refe = zeros(1, c); %Counting the reference vectors per class
    for q = 1:k
        class = v(nearest(q));
        refe(class) = refe(class) + 1;
    end
    [val,z(i)] = max(refe); % maximizes the occurrences among the k-closest reference vectors
end

Adjust according to Number of Classes.

k_nn_classifier_adjust

function z = k_nn_classifier_adjust(Z, v, k, X)
[l, N1] = size(Z); %in Z we have the training data
[l, N] = size(X); %in X we have the points to be classified
c = max(v); %The number of classes
%in v we have the classes to which the vectors in Z belong
N_class = zeros(1, c);
for i = 1:c
    N_class(i) = length(find(v == i));
end
%Computation of the (squared) Euclidean distance of a point in X from each
%reference vector
for i = 1:N
    dist = sum((X(:, i) * ones(1, N1) - Z).^2);
    %sorting the above distances in ascending order
    [sorted, nearest]=sort(dist); % MODE=‘ASCEND’
    %counting the class occurrences among the k-closest reference vectors
    %Z(:,i)
    refe = zeros(1, c); %Counting the reference vectors per class
    for q = 1:k
        class = v(nearest(q));
        refe(class) = refe(class) + 1;
    end
    [val,z(i)] = max(refe./ N_class); % maximizes the occurrences among the k-closest reference vectors
end

(2) SSE classifier

SSE

function [w, cost_func, mis_clas] = SSE(X, y)
% FUNCTION
%  [w, cost_func, mis_clas] = SSE(X, y)
% INPUT ARGUMENTS:
%  X:       lxN matrix whose columns are the data vectors to
%           be classfied.
%  y:       N-dimensional vector whose i-th  component contains the
%           label of the class where the i-th data vector belongs (+1 or
%           -1).
% OUTPUT ARGUMENTS:
%  w:       the final estimate of the parameter vector.
%  cost_func: value of cost function = 0.5 * @sum(y - w'*X)^2
%  mis_clas: number of misclassified data vectors.
w = (X*X') \ (X*y');
[l,N] = size(X);
cost_func = 0.5 * (y - w'*X) * (y - w'*X)';  % calculate cost function
mis_clas = 0;  % calculate number of misclassified vectors
for i = 1:N
    if((X(:,i)' * w) * y(i) < 0)
        mis_clas = mis_clas + 1;
    end
end

(3) LDA classifier

FDR

function [Lambda, FDR, w ] = FDR( X, y , D_y)
%function [ FDR, w ] = FDR( X, y , D_y)
% Fisher's Discriminant Ratio  
% INPUT: 
%   X: points
%   y: y==i ==> belong to Class i
%   D_y: dimension of w, how many features Z_i = w_i'* X need to classify
% OUTPUT:
%   FDR: trace((w * S_w * w') \ (w * S_b * w'))
%   w: use w ==> make Z = w'* X => calculate tr(S_w \ S_b) of Z
%      <=> maximize FDR of X
[ S_w, S_b, S_m ] = Calc_SwSbSm( X, y );
[ Vector, Diag ] = eig( S_w \ S_b );
[Lambda, Index]= sort(diag(Diag), 'descend'); % make highest eig show first
w = Vector(:, Index(1:D_y)); % select D_y vectors corresponding to D_y highest eig values
FDR = trace((w'* S_w * w) \ (w'* S_b * w));
end

Calc_SwSbSm

function [ S_w, S_b, S_m ] = Calc_SwSbSm( X, y )
% [ S_w, S_b, S_m ] = Calc_SwSbSm( X, y )
%   Calculate S_w, S_b, S_m
% OUTPUT:
%   S_w: the within-class
%   S_b: the between-class
%   S_m: the mixture Sm = Sw + Sb
c = max(y); % number of classes
[l, N] = size(X); % N: number of vectors, l: dimensions
mu = zeros(l, c);
S_w = zeros(l, l); S_b = zeros(l, l); mu_0 = zeros(l, 1);
P = zeros(1, c);
for i = 1:c
    index_class_i = find(y == i);
    Mu = sum(X(:, index_class_i), 2) / length(index_class_i);
    mu(:, i) = Mu; mu_0 = mu_0 + sum(X(:, index_class_i), 2) / N;
    P(i) = length(index_class_i) / N;
    X_relative = X(:, index_class_i) - repmat(Mu, 1, length(index_class_i));
    S_wi = zeros(l, l);
    for j = 1:length(index_class_i)
        S_wi = S_wi + X_relative(:, j) * X_relative(:, j)';
    end
    S_w = S_w + S_wi / N;    
end
for i = 1:c
    S_b = S_b + P(i) * (mu(:, i) - mu_0) * (mu(:, i) - mu_0)';
end
S_m = S_w + S_b;

Problem2_TakeHomeExam

%% Take Home Problem 2
%  @Author:     Zhankun Luo
%  @Date:       10/21/2018
%  @Instructor: Toma Hentea
clear all; close all; clc;
load diabet2.mat
fprintf('weight of 8 features:\n');
[l, N] = size(X1);
weight = Relief(X1, Y1)  % get weight of 8 features
[weight_sort, weight_index] = sort(weight,'descend') 
X1R = X1(weight_index(1:4), :); % remain 4 most important weight
X2R = X2(weight_index(1:4), :);
[X1R_norm, X2R_norm] = normalize(X1R, X2R);
%% KNN (k = 9)
Y1_knn = Y1; Y1_knn(Y1_knn == 0) = 2;
Y2_knn = k_nn_classifier(X1R_norm, Y1_knn, 9, X2R_norm);
Y2_knn(Y2_knn == 2) = 0;
%% SSE
fprintf('SSE:\n');
X1_SSE = [X1R_norm; ones(1, N)];
X2_SSE = [X2R_norm; ones(1, N)];
Y1_SSE = Y1; Y1_SSE(Y1_SSE == 0) = -1;
[w, cost_func, mis_clas] = SSE(X1_SSE, Y1_SSE)
err_rate = mis_clas / N
Y2_SSE = sign(w'* X2_SSE); Y2_SSE(Y2_SSE == -1) = 0;
%% LDA
fprintf('LDA:\n');
[Lambda, FDR1, w] = FDR(X1R_norm, Y1_knn, 1)
LDA_X1 = w'* X1R_norm;
[LDA_X1_sort, index_LDA_X1] = sort(LDA_X1);
err1_best = N; LDA_threshold = LDA_X1_sort(1);
for i = 1:N-1
    y1_pred = [ones(1, i), zeros(1, N-i)];
    err1_pred = sum(xor(y1_pred, Y1(index_LDA_X1)));
    if (err1_best > err1_pred)
        err1_best = err1_pred;
        LDA_threshold = LDA_X1_sort(i);
    end
end
err1_best, LDA_threshold
LDA_X2 = w'* X2R_norm;
Y2_LDA = heaviside(LDA_threshold - LDA_X2);
%% plot figures
Y1_plot = Y1; Y1_plot(Y1_plot == 0) = 2;
plot_point(X1R([1,2], :), Y1_plot);
figure 
plot(LDA_X1, Y1, 'g+')
hold on; plot(LDA_threshold * ones(1,100), linspace(0,1,100), 'r-');
ylabel('HAVE diabetes or NOT'); xlabel('w * normalized X1R');
text(-0.78, 0.5, 'Threshold');
%% save predicted Y2
save('Y2', 'Y2_knn', 'Y2_SSE', 'Y2_LDA');

Result

Select Features: Weight of different Features to Y

weight of 8 features:

weight =
    0.0149
    0.0120
    0.0082
    0.0034
    0.0004
    0.0133
    0.0109
    0.0024
weight_sort =
    0.0149
    0.0133
    0.0120
    0.0109
    0.0082
    0.0034
    0.0024
    0.0004
weight_index =  % sorted descend weight 
     1
     6
     2
     7
     3
     4
     8
     5

So, choose Feature: 1, 6, 2, 7

Number of times pregnant
Body mass index (weight in kg/(height in m)^2)
undefinedPlasma glucose concentration
undefinedDiabetes pedigree function

As the 4 Features chosen.

(1) KNN (k = 9)

sum(Y2_knn)
ans =   102

Number of (For X2R)

Diabet 1: 102

Healthy 0: 282

Y2_knn =
     0     0     0     0     0     0     0     1     0     0     1     1     0     0     0
     1     0     0     0     0     1     0     0     0     1     1     0     0     1     0
     0     1     0     1     0     0     1     0     0     0     1     1     0     1     0
     0     0     0     0     0     0     0     1     0     0     0     1     0     0     1
     0     1     0     0     0     0     0     0     0     0     0     1     0     0     1
     0     0     0     0     0     1     0     0     0     1     1     0     0     0     0
     0     0     0     1     1     0     0     0     0     0     1     1     0     1     0
     1     0     0     0     1     0     1     0     0     1     1     0     0     1     0
     0     0     1     0     0     1     0     0     0     0     0     0     1     0     0
     0     0     0     1     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     1     1     0     0     0     0     0     1     1     0     1
     1     0     0     1     0     0     1     0     1     0     0     0     1     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     1     0     0     0     0     0     0     0     0     1     0     1     0     0     1
     1     1     0     0     1     0     0     0     0     1     1     1     1     0     0
     0     0     1     1     0     0     0     1     0     1     0     0     1     1     0
     0     0     0     0     0     0     1     0     0     0     1     0     0     0     1
     0     0     0     0     0     0     0     0     1     0     0     0     0     0     0
     0     0     0     1     0     0     1     1     1     0     0     0     0     0     0
     1     1     0     0     1     0     1     1     0     0     0     0     1     0     0
     0     1     0     0     0     0     0     1     1     0     0     0     0     0     0
     1     0     0     1     0     0     0     0     0     1     0     0     0     0     0
     0     1     1     0     0     0     0     0     1     0     0     0     0     0     0
     0     0     1     1     0     0     0     0     0     0     0     1     0     0     1
     1     0     1     0     1     0     1     0     0     1     1     1     1     0     0
     1     0     1     0     0     0     0     0     0

(2) SSE

SSE:   % training w of decision line

w =
    0.7887
    1.4265
    1.7585
    0.8521
   -1.9029
cost_func =  126.4948
mis_clas =    95  % misclassified number for X1
err_rate =	0.2474  % error rate for X1

When w * X > 0

X => Y = 1 : Diabet

X => Y = 0 : Healthy

sum(Y2_SSE)
ans =   112

Number of (For X2R)

Diabet 1: 98

Healthy 0: 286

Y2_SSE =

     0     0     0     0     1     0     0     1     0     0     1     0     0     0     0
     1     0     0     0     0     1     0     0     0     1     1     0     0     1     0
     0     1     0     1     0     0     0     0     0     0     1     1     0     1     0
     0     0     0     0     0     0     0     1     0     0     0     1     0     0     0
     0     1     0     0     0     0     0     0     0     0     0     1     0     0     1
     0     0     0     0     0     0     0     0     0     0     1     1     0     0     0
     0     0     0     0     0     0     1     0     0     0     1     0     1     1     0
     1     0     0     0     0     0     1     0     0     1     1     0     0     1     0
     0     0     1     0     0     0     0     0     0     0     0     1     1     1     0
     0     0     0     0     1     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     1     0     0     0     0     0     1     1     0     1
     1     0     0     0     0     0     0     0     0     1     0     0     1     0     
     0     0     0     0     1     0     0     0     0     0     0     0     0     0     0
     1     1     0     0     0     0     0     1     0     1     0     1     0     0     0
     0     1     0     0     1     0     0     0     0     1     1     0     1     0     1
     0     0     1     1     0     1     0     0     0     1     0     0     0     1     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     1     0     1     1     0     0     0     0     0
     0     1     0     1     1     0     1     1     1     1     0     0     0     0     0
     1     1     0     0     1     0     1     1     0     0     0     0     1     0     0
     0     0     0     0     0     1     0     1     0     1     0     0     1     0     0
     1     0     0     1     0     0     0     0     0     1     0     1     0     1     0
     0     1     1     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     1     0     0     0     0     0     0     0     1     0     0     1
     1     0     1     0     1     0     1     0     0     1     1     1     1     0     0
     1     0     1     0     0     0     0     0     0

(3) LDA

Linear Discriminant Analysis:

LDA:

Lambda =
   0.4269 + 0.0000i
   0.0000 + 0.0000i
   0.0000 - 0.0000i
   0.0000 + 0.0000i
FDR1 =    0.4269
w =
   -0.3099
   -0.5606
   -0.6911
   -0.3348
err1_best =    90  % misclassified number for X1
LDA_threshold =   -0.8031

LDA_threshold

When w * normalized X > LDA_threshold

X => Y = 0: Healthy

When w * normalized X < LDA_threshold

X => Y = 1: Diabet

sum(Y2_LDA)
ans =   73

Number of (For X2R)

Diabet 1: 73

Healthy 0: 311

Y2_LDA =
     0     0     0     0     0     0     0     1     0     0     1     0     0     0     0
     1     0     0     0     0     1     0     0     0     1     1     0     0     1     0
     0     1     0     1     0     0     0     0     0     0     1     1     0     1     0
     0     0     0     0     0     0     0     1     0     0     0     1     0     0     0
     0     1     0     0     0     0     0     0     0     0     0     1     0     0     1
     0     0     0     0     0     0     0     0     0     0     1     1     0     0     0
     0     0     0     0     0     0     1     0     0     0     1     0     0     1     0
     1     0     0     0     0     0     0     0     0     1     1     0     0     0     0
     0     0     1     0     0     0     0     0     0     0     0     0     1     0     0
     0     0     0     0     1     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     1     1     0     0
     1     0     0     0     0     0     0     0     0     0     0     0     1     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     1     0     0     0     0     0     0     0     0     1     0     1     0     0     0
     0     1     0     0     1     0     0     0     0     1     1     0     1     0     0
     0     0     1     1     0     1     0     0     0     0     0     0     0     1     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     1     0     0     0     0     0     0
     0     0     0     0     1     0     1     1     1     1     0     0     0     0     0
     1     1     0     0     1     0     1     0     0     0     0     0     1     0     0
     0     0     0     0     0     1     0     1     0     0     0     0     0     0     0
     1     0     0     1     0     0     0     0     0     1     0     0     0     1     0
     0     1     1     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     1     0     0     0     0     0     0     0     1     0     0     1
     1     0     1     0     1     0     0     0     0     1     1     0     0     0     0
     1     0     1     0     0     0     0     0     0

Compare 3 different Y2

sum(xor(Y2_knn,Y2_SSE)) % Compare Y2_knn & Y2_SSE
ans =	50
sum(xor(Y2_knn,Y2_LDA)) % Compare Y2_knn & Y2_LDA
ans =    45
sum(xor(Y2_LDA,Y2_SSE)) % Compare Y2_LDA & Y2_SSE
ans =    25

These result are different for which Feature:

find(xor(Y2_knn,Y2_SSE) == 1) % Y2_knn & Y2_SSE
ans =
     5    12    37    60    81    85    87    94    95    97   102   103   110   126   132
   134   139   140   155   169   172   174   175   185   197   203   210   211   222   225
   231   233   238   247   251   255   262   265   272   275   280   302   306   309   310
   313   327   329   339   348

find(xor(Y2_knn,Y2_LDA) == 1) % Y2_knn & Y2_LDA
ans =
    12    37    60    81    85    87    94    95    97   102   110   112   119   126   139
   140   155   156   165   169   172   174   210   211   222   231   233   235   238   247
   251   255   274   275   280   293   302   306   309   329   339   348   367   372   373
   
find(xor(Y2_SSE,Y2_LDA) == 1) % Y2_SSE & Y2_LDA
ans =
     5   103   112   119   132   134   156   165   175   185   197   203   225   235   262
   265   272   274   293   310   313   327   367   372   373
解决cannot modify header information - headers already sent by (......)
luo2010的专栏
10-01 433
这两天被这个报错困扰了很久，今天终于找到了解决办法，主要是把utf8-bom  转为utf8，所以一定要注意编码方式，其他的方式网上有很多，以下转载一下供参考。      错误信息经常提示：cannot modify header information - headers already sent by (......)。其实已经实现需要的效果了，就是这个错误信息看着不爽，网上找了很多办法...
Java web项目报错：HTTP Status 404 – Not Found Type Status Report Message /WebProject/ Description The ori
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Take Home Exam Problem1 Zhankun Luo PUID: 0031195279 Email: luo333@pnw.edu Fall-2018-ECE-59500-009 Instructor: Toma Hentea 文章目录Take Home Exam Problem1Functionplot_pointk_nn_classifierperceptron_ANDW2\...
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Zhankun Luo PUID: 0031195279 Email: luo333@pnw.edu Fall-2018-ECE-59500-009 Instructor: Toma Hentea #Homework 2 文章目录Problem 2.12(a) Bayesian classifier that minimizes the error probability(b) Bayesian ...
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Zhankun Luo PUID: 0031195279 Email: luo333@pnw.edu Fall-2018-ECE-59500-009 Instructor: Toma Hentea 文章目录Homework 5Functionplot_point.mCalc_SwSbSm.mJ3.mFDR.mProblemProblem 5.1Proof of $E[z] = \lVert \m...
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编译环境： vs2022 win10 x64 anaconda3+python3.10 torch==2.3.1+cu118 cuda11.8.0+cudnn8.9.7 triton==2.1.0 causal_conv1d==1.4.0 mamba==2.2.2 RTX2070显卡 注意编译的whl是不能用于RTX50显卡的，可以用于RTX20-RTX40系列显卡，安装时候尽量和模块一致
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toplus1s_calculator_32040_1765656584703.zip
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基于可靠性评估序贯蒙特卡洛模拟法的配电网可靠性评估研究（Matlab代码实现）内容概要：本文围绕“基于可靠性评估序贯蒙特卡洛模拟法的配电网可靠性评估研究”，介绍了利用Matlab代码实现配电网可靠性的仿真分析方法。重点采用序贯蒙特卡洛模拟法对配电网进行长时间段的状态抽样与统计，通过模拟系统元件的故障与修复过程，评估配电网的关键可靠性指标，如系统停电频率、停电持续时间、负荷点可靠性等。该方法能够有效处理复杂网络结构与设备时序特性，提升评估精度，适用于含分布式电源、电动汽车等新型负荷接入的现代配电网。文中提供了完整的Matlab实现代码与案例分析，便于复现和扩展应用。; 适合人群：具备电力系统基础知识和Matlab编程能力的高校研究生、科研人员及电力行业技术人员，尤其适合从事配电网规划、运行与可靠性分析相关工作的人员； 使用场景及目标：①掌握序贯蒙特卡洛模拟法在电力系统可靠性评估中的基本原理与实现流程；②学习如何通过Matlab构建配电网仿真模型并进行状态转移模拟；③应用于含新能源接入的复杂配电网可靠性定量评估与优化设计； 阅读建议：建议结合文中提供的Matlab代码逐段调试运行，理解状态抽样、故障判断、修复逻辑及指标统计的具体实现方式，同时可扩展至不同网络结构或加入更多不确定性因素进行深化研究。
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