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Matlab

Question to Matlab solve-function

/ 2024-09-30

Hey guys I have a simple question:
I am using Matlab R2010a and trying to solve an equation with 1 variable like this:

solve(‘4-t^2=0’)

ans =

-2
2

No problem as you can see 😉
My question is, when I try to replace the *4* with a variable (lets call it A), it wont work how I want it to work:

solve(‘A-t^2=0’)

ans =

A^(1/2)
-A^(1/2)

All fine, solution found but A is not a variable, it has a defined value of 4, so that the solution would be -2,2.
Can anyone help me to find the solution to this problem if its possible?

Thanks a lot guys!Hey guys I have a simple question:
I am using Matlab R2010a and trying to solve an equation with 1 variable like this:

solve(‘4-t^2=0’)

ans =

-2
2

No problem as you can see 😉
My question is, when I try to replace the *4* with a variable (lets call it A), it wont work how I want it to work:

solve(‘A-t^2=0’)

ans =

A^(1/2)
-A^(1/2)

All fine, solution found but A is not a variable, it has a defined value of 4, so that the solution would be -2,2.
Can anyone help me to find the solution to this problem if its possible?

Thanks a lot guys! Hey guys I have a simple question:
I am using Matlab R2010a and trying to solve an equation with 1 variable like this:

solve(‘4-t^2=0’)

ans =

-2
2

No problem as you can see 😉
My question is, when I try to replace the *4* with a variable (lets call it A), it wont work how I want it to work:

solve(‘A-t^2=0’)

ans =

A^(1/2)
-A^(1/2)

All fine, solution found but A is not a variable, it has a defined value of 4, so that the solution would be -2,2.
Can anyone help me to find the solution to this problem if its possible?

Thanks a lot guys! solve, variable MATLAB Answers — New Questions

Matlab

The power trendline does not work correctly.

/ 2024-09-30

Power Trend line equation has the form y=ax^b, but in MATLAB Online it always appear a=b.Power Trend line equation has the form y=ax^b, but in MATLAB Online it always appear a=b. Power Trend line equation has the form y=ax^b, but in MATLAB Online it always appear a=b. power trend line MATLAB Answers — New Questions

Matlab

Fitting data to a Gaussian when data is a little bit skewed (but not too much)

/ 2024-09-30

Hello, I have some data below that should approx to a Gaussian.
1.00 NaN
2.00 4.00
3.00 3.00
4.00 6.00
5.00 9.00
6.00 7.00
7.00 9.00
8.00 57.00
9.00 141.00
10.00 205.00
11.00 204.00
12.00 190.00
13.00 86.00
14.00 23.00
15.00 5.00
16.00 4.00
17.00 7.00
18.00 5.00
19.00 7.00
20.00 4.00
21.00 5.00
I understand its not perfect, but when I give my gaussian fitting the parameters represented by the orange curve, it fails to do any kind of fit.

This is my Routine
b0 = 11.00
c0 = 1.73 % This is actually sqrt (fwhm guess = 3)

function [fwhm,xdataFine,fitGaus,xpeak,ypeak,rsquared]=myGaussianFit2(xdata,ydata, b0,c0)

%——————–Gaussian Fit—————————————-
%Input is xdata, ydata and must be in column vector format
%only two initial guesses required as the following are determine by ydata

a0 = max(ydata(:)); %-min(ydata(:));
d0 = min(ydata(:));

%b0 is the guess at x-location of peak
%c0 is guess at width of peak

%Output gives the fitted parameters as well as xpeak location and its
%yvalue (i.e. xpeak & ypeak
%————————————————————————-

%Define Gauss Equation (remember the . notation
gaussEqn =’a*exp(-0.5*((x-b)/(c^2)).^2)+d’;

%Use startpoint to define guess values (otherwise fit doesn’t perform well)
try
%f = fit(xdata,ydata,gaussEqn,’Normalize’,’off’, ‘StartPoint’,[a0,b0,sqrt(c0),d0]); %use c^2 instred of c to enforce postive sigma/fwhm
[f,gof]=fit(xdata,ydata,gaussEqn,’Normalize’,’off’, ‘StartPoint’,[a0,b0,sqrt(c0),d0]); %use c^2 instead of c to enforce postive sigma/fwhm
coeffs=coeffvalues(f);
a=coeffs(1); b=coeffs(2);
c=coeffs(3); %need to square it as used c^2 in fitting equation to enforce +ve values
c=c^2;
d=coeffs(4);
rsquared=gof.rsquare;
fwhm=c*sqrt(log(256));
% %Increase resolution of x data (by 30)
xdataFine=(linspace(xdata(1),xdata(end),100))’;
% Create high res Gauss function
fitGaus = a*exp(-0.5*((xdataFine-b)/c).^2)+d;

%Find max value and its x location
ypeak=max(fitGaus);
xpeak=b;
xpeak=mean(xpeak(:)); %Take mean incase more than one peak found

catch
a=0;b=0;c=0;d=0;xpeak=0;ypeak=0;
end

Is there anyhting I can do here to make the fit work, although I understand its not great, but I still think it should be able to fit something!Hello, I have some data below that should approx to a Gaussian.
1.00 NaN
2.00 4.00
3.00 3.00
4.00 6.00
5.00 9.00
6.00 7.00
7.00 9.00
8.00 57.00
9.00 141.00
10.00 205.00
11.00 204.00
12.00 190.00
13.00 86.00
14.00 23.00
15.00 5.00
16.00 4.00
17.00 7.00
18.00 5.00
19.00 7.00
20.00 4.00
21.00 5.00
I understand its not perfect, but when I give my gaussian fitting the parameters represented by the orange curve, it fails to do any kind of fit.

This is my Routine
b0 = 11.00
c0 = 1.73 % This is actually sqrt (fwhm guess = 3)

function [fwhm,xdataFine,fitGaus,xpeak,ypeak,rsquared]=myGaussianFit2(xdata,ydata, b0,c0)

a0 = max(ydata(:)); %-min(ydata(:));
d0 = min(ydata(:));

%b0 is the guess at x-location of peak
%c0 is guess at width of peak

%Output gives the fitted parameters as well as xpeak location and its
%yvalue (i.e. xpeak & ypeak
%————————————————————————-

%Define Gauss Equation (remember the . notation
gaussEqn =’a*exp(-0.5*((x-b)/(c^2)).^2)+d’;

%Find max value and its x location
ypeak=max(fitGaus);
xpeak=b;
xpeak=mean(xpeak(:)); %Take mean incase more than one peak found

catch
a=0;b=0;c=0;d=0;xpeak=0;ypeak=0;
end

Is there anyhting I can do here to make the fit work, although I understand its not great, but I still think it should be able to fit something! Hello, I have some data below that should approx to a Gaussian.
1.00 NaN
2.00 4.00
3.00 3.00
4.00 6.00
5.00 9.00
6.00 7.00
7.00 9.00
8.00 57.00
9.00 141.00
10.00 205.00
11.00 204.00
12.00 190.00
13.00 86.00
14.00 23.00
15.00 5.00
16.00 4.00
17.00 7.00
18.00 5.00
19.00 7.00
20.00 4.00
21.00 5.00
I understand its not perfect, but when I give my gaussian fitting the parameters represented by the orange curve, it fails to do any kind of fit.

This is my Routine
b0 = 11.00
c0 = 1.73 % This is actually sqrt (fwhm guess = 3)

function [fwhm,xdataFine,fitGaus,xpeak,ypeak,rsquared]=myGaussianFit2(xdata,ydata, b0,c0)

a0 = max(ydata(:)); %-min(ydata(:));
d0 = min(ydata(:));

%b0 is the guess at x-location of peak
%c0 is guess at width of peak

%Output gives the fitted parameters as well as xpeak location and its
%yvalue (i.e. xpeak & ypeak
%————————————————————————-

%Define Gauss Equation (remember the . notation
gaussEqn =’a*exp(-0.5*((x-b)/(c^2)).^2)+d’;

%Find max value and its x location
ypeak=max(fitGaus);
xpeak=b;
xpeak=mean(xpeak(:)); %Take mean incase more than one peak found

catch
a=0;b=0;c=0;d=0;xpeak=0;ypeak=0;
end

Is there anyhting I can do here to make the fit work, although I understand its not great, but I still think it should be able to fit something! fit, gaussian MATLAB Answers — New Questions

Matlab

About the DeLong test

/ 2024-09-30

Dear Professors,
I apologize for bothering you during your busy schedules. We have conducted the program described below, reading three files and performing ROC analysis through machine learning to calculate three AUCs. We have attempted to create a program to demonstrate the statistical significance of the differences among these AUCs, but despite multiple efforts, we have encountered errors that prevent progress. Could you please help us revise the program? I am attaching a program below. I would greatly appreciate it if you could add the necessary modifications to it. I am having a lot of trouble implementing the DeLong test in the program. Thank you very much for your assistance.Thank you very much for your assistance.
Best regards,
% CSVファイルのパスを指定
filePaths = {‘C:Usersrms56Desktop1-491B.csv’, …
‘C:Usersrms56Desktop1-491C.csv’, …
‘C:Usersrms56Desktop1-491D.csv’};

% X と y を格納するための cell 配列を作成
X_all = cell(3, 1); % 特徴量 X 用の cell 配列
y_all = cell(3, 1); % ラベル y 用の cell 配列

% 3つのファイルを順番に読み込み、X と y に割り当てる
for i = 1:3
% CSVファイルの読み込み
data = readmatrix(filePaths{i});

% 各ファイルに応じて X と y の列を指定
if i == 1 % ‘1-491B.csv’: 3列目までがX、4列目がY
X_all{i} = data(:, 1:3); % 1～3列目を X に設定
y_all{i} = data(:, 4); % 4列目を Y に設定

elseif i == 2 % ‘1-491C.csv’: 6列目までがX、7列目がY
X_all{i} = data(:, 1:6); % 1～6列目を X に設定
y_all{i} = data(:, 7); % 7列目を Y に設定

elseif i == 3 % ‘1-491D.csv’: 3列目までがX、4列目がY
X_all{i} = data(:, 1:3); % 1～3列目を X に設定
y_all{i} = data(:, 4); % 4列目を Y に設定
end
end

% ファイルごとの解析をループで実行
for fileIndex = 1:3
% ファイルに対応するデータを取得
X = X_all{fileIndex}; % 特徴量
y = y_all{fileIndex}; % ラベル

% クロスバリデーションの設定
k = 5; % フォールド数
cv = cvpartition(y, ‘KFold’, k); % クロスバリデーションの分割
accuracy = zeros(k, 1); % 各フォールドの精度を格納する配列

% 各フォールドごとにトレーニングとテストを実行
for i = 1:k
trainIdx = training(cv, i);
testIdx = test(cv, i);

% データの分割
XTrain = X(trainIdx, :);
yTrain = y(trainIdx, :);
XTest = X(testIdx, :);
yTest = y(testIdx, :);

% SVMモデルのトレーニング
model = fitcsvm(XTrain, yTrain, …
‘KernelFunction’, ‘polynomial’, …
‘PolynomialOrder’, 2, …
‘KernelScale’, ‘auto’, …
‘BoxConstraint’, 1, …
‘Standardize’, true);

% モデルを使用してテストセットを予測
[predictions, score] = predict(model, XTest);

% 現在のフォールドの精度を計算
accuracy(i) = sum(predictions == yTest) / length(yTest);
fprintf(‘ファイル %d – Fold %d Accuracy: %.2f%%n’, fileIndex, i, accuracy(i) * 100);
end

% 全フォールドの平均精度を計算
averageAccuracy = mean(accuracy);
fprintf(‘ファイル %d – Average Accuracy: %.2f%%n’, fileIndex, averageAccuracy * 100);

% ROC曲線とAUCの計算
[~, ~, ~, AUC_final] = perfcurve(yTest, score(:, 2), 1);

% ブートストラップ法で信頼区間を計算
nBoot = 1000; % ブートストラップの反復回数
[AUC_bootstrap, CI_final] = bootstrapAUC(yTest, score(:, 2), nBoot);

% 混同行列の計算
confusionMatrix_final = confusionmat(yTest, predictions);
tn = confusionMatrix_final(1, 1);
fp = confusionMatrix_final(1, 2);
fn = confusionMatrix_final(2, 1);
tp = confusionMatrix_final(2, 2);

% 指標の計算
sensitivity_final = tp / (tp + fn);
specificity_final = tn / (tn + fp);
ppv_final = tp / (tp + fp);
npv_final = tn / (tn + fn);
accuracy_final = (tp + tn) / sum(confusionMatrix_final(:));

% 結果の表示
fprintf(‘ファイル %d – 最終的なAUC: %.2fn’, fileIndex, AUC_final);
fprintf(‘ファイル %d – ブートストラップ法による95%%信頼区間: [%.2f, %.2f]n’, fileIndex, CI_final(1), CI_final(2));
fprintf(‘感度: %.2fn’, sensitivity_final);
fprintf(‘特異度: %.2fn’, specificity_final);
fprintf(‘陽性的中率: %.2fn’, ppv_final);
fprintf(‘陰性的中率: %.2fn’, npv_final);
fprintf(‘診断精度: %.2fn’, accuracy_final);

% ROC曲線を描画
figure;
plot(Xroc, Yroc, ‘b-‘, ‘LineWidth’, 2);
xlabel(‘特異度’);
ylabel(‘感度’);
title(sprintf(‘ファイル %d – ROC曲線’, fileIndex));
grid on;

% 混同行列を描画
figure;
confusionchart(confusionMatrix_final, {‘Negative’, ‘Positive’}, ‘RowSummary’, ‘row-normalized’, …
‘ColumnSummary’, ‘column-normalized’);
title(sprintf(‘ファイル %d – 混同行列’, fileIndex));
end
% ブートストラップ法によるAUCと信頼区間を計算する関数

function [AUC, CI] = bootstrapAUC(yTrue, scores, nBoot)
% 初期化
AUC = zeros(nBoot, 1);

for i = 1:nBoot
idx = randi(length(yTrue), [length(yTrue), 1]); % リプレースメントで再サンプリング
yBoot = yTrue(idx);
scoresBoot = scores(idx);
[~, ~, ~, AUC(i)] = perfcurve(yBoot, scoresBoot, 1); % AUC計算
end

% 信頼区間の計算
CI = prctile(AUC, [2.5 97.5]); % 95%信頼区間
endDear Professors,
I apologize for bothering you during your busy schedules. We have conducted the program described below, reading three files and performing ROC analysis through machine learning to calculate three AUCs. We have attempted to create a program to demonstrate the statistical significance of the differences among these AUCs, but despite multiple efforts, we have encountered errors that prevent progress. Could you please help us revise the program? I am attaching a program below. I would greatly appreciate it if you could add the necessary modifications to it. I am having a lot of trouble implementing the DeLong test in the program. Thank you very much for your assistance.Thank you very much for your assistance.
Best regards,
% CSVファイルのパスを指定
filePaths = {‘C:Usersrms56Desktop1-491B.csv’, …
‘C:Usersrms56Desktop1-491C.csv’, …
‘C:Usersrms56Desktop1-491D.csv’};