finding a specific peak point in a table

Hello,

attached is a table of data where we are intrested in locating the peak point according to a certain criteria,

1. it should be located within a range of 200-700 for the variable GDALT
2. It is the first peak that is found in that interval

to do that I am applying the folling lines

profileTable.INDEX = (1:height(profileTable))';
subTable = profileTable(((profileTable.GDALT>=200) & (profileTable.GDALT<=700)), :);

[pks,locs] = findpeaks(subTable.NE8); 
result = subTable(locs,:);

x = profileTable.NE8 ; 
y = profileTable.GDALT ; 

peak_y = profileTable.GDALT(result.INDEX(1)) ; 
peak_x = profileTable.NE8(result.INDEX(1)) ; 
idx = y <= peak_y ; 

But I am getting the following error

Index exceeds the number of array elements. Index must not exceed 0.

Error in tabular/dotParenReference (line 114)
        b = b(rowIndices);

Error in PeakConstruction (line 94)
    peak_y = profileTable.GDALT(result.INDEX(1)) ; 

I am attaching the table "profileTable" where I am getting the error, and another table 'profileTable2' where the code is running fine.

[KSSV]

result = subTable(locs,:); 

The above table is empty....check your code properly.

Answer 1

[Image Analyst]

There are some nan values in the one that doesn't work and you have to basically ignore them. One way is by interpolating those nan values. Try this:

% Demo by Image Analyst
% Initialization Steps.
clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 20;

%====================================================================================
% Read in data.
profileTable = readtable('profileTable.txt')
profileTable.INDEX = (1:height(profileTable))';
rowsToExtract = ((profileTable.GDALT>=200) & (profileTable.GDALT<=700));
subTable = profileTable(rowsToExtract, :);
fprintf('Extracted %d rows from profileTable into subTable.\n', height(subTable));

%====================================================================================
% Get data that we're interested in.
x = subTable.INDEX;
y = subTable.NE8;
% Plot x and y data as blue markers with lines between them.
% Nan locations will not have a line to them and will have no marker there.
plot(x, y, 'b.-', 'LineWidth', 2, 'MarkerSize', 15);

%====================================================================================
% There are some NANs in y that prevent it from finding peaks.  
% Replace those by interpolation
nanLocations = isnan(y); % Logical indexes at first so we can extract good values.
% Get the good (x,y) locations for use in the interpolation.
xGood = x(~nanLocations);
yGood = y(~nanLocations);
nanLocations = find(isnan(y)); % Get linear indexes this time.
for k = 1 : length(nanLocations)
    % Get the index.
    thisIndex = nanLocations(k);
    % Get the x value of this nan location.
    xNan = x(thisIndex);
    y(thisIndex) = interp1(xGood, yGood, xNan, 'linear');
    fprintf('Interpolated value for x=%d is %f.\n',xNan, y(thisIndex))
end
% Show interpolated values as a red spot.
hold on;
plot(x(nanLocations), y(nanLocations), 'r.', 'LineWidth', 2, 'MarkerSize', 20);
grid on;
title('subTable.NE8  Interpolated NAN values are in red.')

%====================================================================================
% Now find Peak Values.
[peakYValues, indexesOfPeaks] = findpeaks(y);
if isempty(peakYValues)
    warningMessage = sprintf('WARNING: there are no peaks in the signal');
    uiwait(warndlg(warningMessage));
    return;
end
% Get the x values at the peak index locations.
xPeaks = x(indexesOfPeaks);
% Plot triangle over the peaks.
plot(xPeaks, peakYValues, 'rv', 'LineWidth', 2, 'MarkerSize', 14);
% Print out all peak x and y values.
for k = 1 : length(xPeaks)
    fprintf('Peak value #%d is at x = %d is y = %f.\n', k, xPeaks(k), peakYValues(k))
end

Extracted 33 rows from profileTable into subTable.

Interpolated value for x=22 is 1898000000000.000000.

Interpolated value for x=25 is 1335000000000.000000.

Peak value #1 is at x = 21 is y = 2365000000000.000000.

[Salma fathi]

Thank you, I have not thought of using the interpoloating approach. it worked quite well with me. Appreciate all the effort thank you so much