how to divide a data set randomly into training and testing data set?

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chocho
chocho 2018 年 4 月 16 日
Hello guys, I have a dataset of a matrix of size 399*6 type double and I want to divide it randomly into 2 subsets training and testing sets by using the cross-validation.
i have tried this code but did get what i want https://www.mathworks.com/help/stats/cvpartition-class.html
Could anyone help me to do that?
Expected outputs:
training_data: k*6 double
testing_data: l*6 double

採用された回答

KSSV
KSSV 2018 年 4 月 16 日
編集済み: KSSV 2018 年 4 月 16 日
Let A be your data of size 399*6. To divide data into training and testing with given percentage:
[m,n] = size(A) ;
P = 0.70 ;
idx = randperm(m) ;
Training = A(idx(1:round(P*m)),:) ;
Testing = A(idx(round(P*m)+1:end),:) ;
  20 件のコメント
Gentil Andres Collazos Escobar
Gentil Andres Collazos Escobar 2021 年 9 月 10 日
Thank you!!
Abhijit Bhattacharjee
Abhijit Bhattacharjee 2023 年 3 月 4 日
If it hasn't been covered already, you can also use cvpartition to split the dataset. See THIS answer for more details.

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その他の回答 (8 件)

Jeremy Breytenbach
Jeremy Breytenbach 2019 年 5 月 24 日
編集済み: Jeremy Breytenbach 2019 年 5 月 24 日
Hi there.
If you have the Deep Learning toolbox, you can use the function dividerand: https://www.mathworks.com/help/deeplearning/ref/dividerand.html
[trainInd,valInd,testInd] = dividerand(Q,trainRatio,valRatio,testRatio) separates targets into three sets: training, validation, and testing.

ALDO
ALDO 2020 年 2 月 2 日
you can use The helper function 'helperRandomSplit', It performs the random split. helperRandomSplit accepts the desired split percentage for the training data and Data. The helperRandomSplit function outputs two data sets along with a set of labels for each. Each row of trainData and testData is an signal. Each element of trainLabels and testLabels contains the class label for the corresponding row of the data matrices.
percent_train = 70;
[trainData,testData,trainLabels,testLabels] = ...
helperRandomSplit(percent_train,Data);
make sure to have the proper toolbox to use it.
  1 件のコメント
Lucrezia Cester
Lucrezia Cester 2021 年 2 月 7 日
could you please send a link to this function?

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sidra ashiq
sidra ashiq 2018 年 11 月 23 日
Training = A(idx(1:round(P*m)),:) ;
what is the A function??
  2 件のコメント
Mohamed Marei
Mohamed Marei 2018 年 12 月 17 日
A is the vector or array indexed by the elements inside the bracket. It is not a function.
madhan ravi
madhan ravi 2018 年 12 月 17 日
A is a matrix

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Mehernaz Savai
Mehernaz Savai 2022 年 5 月 26 日
編集済み: Mehernaz Savai 2022 年 5 月 26 日
You can partition data in a number of ways:
Let X be your input matrix. You can also use similar workflow for Tables.
If you have the Statistics and Machine Learning Toolbox, you can use cvpartition as follows:
% Partiion with 40% data as testing
hpartition = cvpartition(size(X,1),'Holdout',0.4);
% Extract indices for training and test
trainId = training(hpartition);
testId = test(hpartition);
% Use Indices to parition the matrix
trainData = X(trainId,:);
testData = X(testId,:);
If you have the Deep Learning Toolbox, you can use dividerand as follows:
% Partiion with 60:20:20 ratio for training,validation and testing
% respectively
[trainId,valId,testId] = dividerand(size(X,1),0.6,0.2,0.2);
% Use Indices to parition the matrix
trainData = X(trainId,:);
valData = X(valInd,:);
testData = X(testId,:);

Pramod Hullole
Pramod Hullole 2019 年 3 月 5 日
hello sir,
iI'm new to the neuralnetworks..now i am working on my projects which is leaf disease detections using image processing. i am done with feature extraction and now not getting what is the next step..i know that i should apply nn and divide it in training and testing data set.. but in practically how to procced that's what i am not getting .please help me through this... please send steps..each steps in details. .
  1 件のコメント
Savas Yaguzluk
Savas Yaguzluk 2019 年 3 月 8 日
Dear Pramod,
Open a new topic and ask your question there. So, people can see your topic title and help you.

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Hossein Amini
Hossein Amini 2019 年 7 月 15 日
Hi there, it worked for me but I have problem in rest of the code. In newrb doc, it has been witten how to write the code but the more tried that I did, I got error like below.error.JPG

Hossein Amini
Hossein Amini 2019 年 7 月 15 日
[z,r] = size(X);
idx = randperm(z);
TrainX = (X(idx(1:round(Ptrain.*z)),:))';
TrainY = (Y(idx(1:round(Ptrain.*z)),:))';
TestX = (X(idx(round(Ptrain.*z)+1:end),:))';
TestY = (Y(idx(round(Ptrain.*z)+1:end),:))';
If I'm not mistaken, in newrb doc, the size of input data and output data should be same like (4x266 and 1x266), that's why I transposed that matrixes. But the error which I got is specifying zeros matrix. I don't know how to prepare that.

ranjana roy chowdhury
ranjana roy chowdhury 2019 年 7 月 15 日
the dataset is WS Dream dataset with 339*5825.The entries have values between 0 and 0.1,few entries are -1.I want to make 96% of this dataset 0 excluding the entries having -1 in dataset.

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