Find min/max values for multiple vectors

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Kathy on 29 Jun 2012

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I have data that consists of 29000 data points, the data points represent knee flexion and extension while walking. I want to identify the range of knee flexion/extension for each step so would like to set up a code that will identify the max (peak knee flexion) and the min (essentially, the extreme of knee extension) and place them in columns so that I am able to mathematically determine the full range of motion. I have attached a small section of my data. Any help would be greatly appreciated.

Thank you,

KTATC

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Walter Roberson on 29 Jun 2012

I think it means the peaks of the vector represent the extreme in one case, and the troughs represent the extreme in the other case.

Kathy on 29 Jun 2012

Hi all, thank you for the comments and answers. This is for my research and I have used the findpeaks command, the findvalleys command was kind of a pain to use so I will try the suggestions provided below. Yes, the peak value does represent the extreme in this case, knee flexion, and the troughs represent the extreme in knee extension. In my 29000 data points there are approximately 295 strides (consisting of R and L leg knee flexion/extension) and I want to find the extreme values for each of the strides..... in answer to Holei's question, I do have multiple 29000 data point files (14 subjects in all) and 2 columns (R leg, L leg) for each subject.

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Answer 1

Star Strider on 29 Jun 2012

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I had some fun with this, since I used a similar strategy for my own research not long ago. See 'findpeaks' (I think it's in the Signal Processing Toolbox) to check out its options. You might be able to make it do exactly what you want.

The code:

        [YOUR KTATC VECTOR GOES HERE]
        xvct = 0:length(KTATC)-1;       % Create a reference x-vector
        [PeaksKTATC, PeaksIdx] = findpeaks(KTATC);  % Find peaks of KTATC
        FlipKTATC = 1.001*max(KTATC)-KTATC;     % Flip it upside down to turn Valleys into Peaks
        [ValleyKTATC1,ValleyIdx] = findpeaks(FlipKTATC);    % Find Valleys
        ValleyKTATC = 1.001*max(KTATC)-ValleyKTATC1;    % Reconstitute Valley values
        figure(205)
        plot(xvct, KTATC)
        hold on
        plot(xvct(PeaksIdx), KTATC(PeaksIdx), '+r')
        plot(xvct(ValleyIdx), KTATC(ValleyIdx), '+g')
        hold off
        legend('Knee Goniometry Data', 'Peaks', 'Valleys')
        grid

Putting your data in columns is likely to be a bit more of a challenge, since they're not all going to be equal. I didn't address that because after plotting your data, I don't know what you want to define as where your data vectors start and stop.

There are several ways to go about this, creating a cell array is likely the easiest. That wasn't the way I opted to work with my data (for the time being at least), preferring instead to create a 'zeros' matrix with the column length defined by the longest data column. I keep a separate matrix of the original data set start times and indices, stop times and indices, and lengths of the various data vectors. My routines are set up to load and work with those data in their matrix form.

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Kathy on 29 Jun 2012

Thank you. I have used the findpeaks function and it works very nicely with my data although I do have to establish a 'minpeakdistance' so that each peak isn't picked up. There is an option to provide peak values and locations to check your data... was hoping I might be able to accomplish something like that with this. I am going to try it now.

Thank you again.

Star Strider on 29 Jun 2012

Edited: Star Strider on 29 Jun 2012

My pleasure! I plotted your data because I was curious to see what it looked like. The answer was fairly straightforward after that. The relatively high-frequency ringing at full extension was interesting.

Thank you for accepting my answer.

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