How do I find peaks in a matrix and extract those values?

I have a matrix that is 3 columns wide. How do I find all the peaks in column 2 and pull them out with their associated values in columns 1 and 3? I would like to save these values in a new matrix.

Edit: which column I want, I didn't desribe the correct situation the first time.

Also Included image of what happens when I try to used findpeaks()

Answer 1

[Star Strider]

I would do something like this:

YourMatrix = rand(20, 3);                               % Create Matrix
[pks,locs] = findpeaks(YourMatrix(:,1));                % Column #1 Peaks & Indices
for k = 1:2
    PkVals(:,k) = YourMatrix(locs, k+1);                % Corresponding Values For Columns #2 & #3
end

Experiment to get the results you want.

[Zach Rohde]

Would this also apply if I wanted to find the peaks of column 2?

Lets say column 1 is my x values, column 2 is my y values, and column 3 is a different variable that is associated with these two columns. Could I use your code but use "findpeaks(YourMatrix(:,2));" instead?

[Star Strider]

Yes, although the code changes slightly:

YourMatrix = rand(20, 3);                               % Create Matrix
[pks,locs] = findpeaks(YourMatrix(:,2));                % Column #2 Peaks & Indices
Cols = [1 3];
for k = 1:numel(Cols)
    PkVals(:,k) = YourMatrix(locs, Cols(k));            % Corresponding Values For Columns #1 & #3
end

Note that in my code, findpeaks is always presented a vector.

[Gabriel G]

Hi, Can somebody help me out, I'm trying to use the above piece of code as:

[pks,locs] = findpeaks(T.B(T.B>6800),T.time(T.B>6800));

Cols = [T.A];

for k = 1:numel(T.time)

PkVals(:,k) = T.B(locs, T.A(k));

end

Table T:

In essence, What I'm trying to do is to find the peaks in column B above a certain threshold (in this case 6800) and have also the corresponding value of column A.

The error(s) I'm getting:

Index in position 1 is invalid. Array indices must be positive integers or logical values.

Error in tabular/dotParenReference (line 114)

b = b(rowIndices,colIndices);

Thanks.

[Star Strider]

@Gabriel G —

This should actually be a new Question.

That aside:

time = (0:0.5:10).';
B = 1E4*exp(-(time-5.5).^2);
T = table(time,B)

T = 21×2 table
    time        B     
    ____    __________

      0     7.2877e-10
    0.5     1.3888e-07
      1     1.6052e-05
    1.5      0.0011254
      2       0.047851
    2.5         1.2341
      3         19.305
    3.5         183.16
      4           1054
    4.5         3678.8
      5           7788
    5.5          10000
      6           7788
    6.5         3678.8
      7           1054
    7.5         183.16

[pks,locs] = findpeaks(T.B, 'MinPeakHeight',6800);  
times = T.time(locs,:)

times = 5.5000

A Vote would be appreciated!

.

[Gabriel G]

Thanks, it works well for any other column in the table now.

I'll keep in mind for the next time to create a new question.

Answer 2

[Image Analyst]

Have you tried findpeaks() in the Signal Processing Toolbox?

How exactly do you define a peak? It's not so clear cut as you might think.

You forgot to attach your data or any screenshots. Make it easy for people to help you and you'll get more useful answers.

[Zach Rohde]

I edited my question to clairfy better and inlcuded an image. Sorry about that.

[Image Analyst]

Yes but no data is attached yet, and there's not even a screenshot of your signal plotted.

Please read this link