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How to extract rows based on column values in a matrix?

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qualiaMachine
qualiaMachine on 22 Jun 2015
Answered: Steven Lord on 26 Nov 2020
I am working with data that is in a 152867x2 matrix. The first column contains one of three values ranging from 1-3. The second column, however, has a unique value for each row (see example data below). I would like to know how I can write a program that can extract 3 matrices according to the value of the first column (see example output). I'm new to MATLAB, so explicit instructions would be fantastic. Thanks.
Example Data
3, 0.1234
1, 0.1345
1, 0.1456
2, 0.1567
1, 0.1678
1, 0.1789
Desired Output - Three Matrices
1) 3, 0.1234
2) 2, 0.1567
3) 1, 0.1345
1, 0.1456
1, 0.1678
1, 0.1789

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Accepted Answer

Mukul Rao
Mukul Rao on 23 Jun 2015
Edited: Mukul Rao on 24 Jun 2015
Here are links to a few concepts that would be useful for performing such operations:
1. Matrix Indexing : Using Logicals in Array Indexing
2. Find array elements based on a specified condition
Finally, here is an example on how you can implement these concepts. There are many ways to do this, here is one such method.
%Example matrix
A = [3, 0.1234
1, 0.1345
1, 0.1456
2, 0.1567
1, 0.1678
1, 0.1789];
%Find indices to elements in first column of A that satisfy the equality
ind1 = A(:,1) == 1;
ind2 = A(:,1) == 2;
ind3 = A(:,1) == 3;
%Use the logical indices to index into A to return required sub-matrices
A1 = A(ind1,:);
A2 = A(ind2,:);
A3 = A(ind3,:);

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Suchet Nanda
Suchet Nanda on 24 Feb 2020
Hi, What would be the simplest way to automate it, e.g, for ind(i), with i=n?

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More Answers (1)

Steven Lord
Steven Lord on 26 Nov 2020
This wasn't an option when the question was asked originally but depending what you're trying to do with those "3 matrices", using some of the functions in the 'Grouping and Binning Data' section on this documentation page may be useful. For example, to compute the mean of values in the second column that have the same value in the first column:
A = [3, 0.1234;
1, 0.1345;
1, 0.1456;
2, 0.1567;
1, 0.1678;
1, 0.1789];
G = groupsummary(A, A(:, 1), @mean)
G = 3×2
1.0000 0.1567 2.0000 0.1567 3.0000 0.1234

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