Find two maximum points

Question

I have a two arrays that I plot: A a 1x101 vector and B the same

A = [0.140673450903833  0.143148937279028   0.145430952171596   0.147474938627147   0.149581060870114   0.151187105347571   0.152646348246015   0.153892222566265   0.154913060187075   0.155701930397674   0.156253328260122   0.156562551841967   0.156625533585493   0.156438787610539   0.155999394209637   0.155304997895017   0.154353810555534   0.153144616301392   0.151676776486360   0.149950234280632   0.147965519042205   0.145723755995511   0.143229676241241   0.140476287800831   0.137475884805212   0.134228713435530   0.130738812449387   0.127010778531129   0.123049766057659   0.118861478234099   0.114452155847321   0.109828564345449   0.104997979409803   0.0999681710919947  0.0947473865690700  0.0893443315667412  0.0837681505026921  0.0780284054055798  0.0721350536699391  0.0660984247128685  0.0599291956061627  0.0536383657701255  0.0472372308396036  0.0407373558685518  0.0341505481893574  0.0274888307202605  0.0207644183921863  0.0139897104812690  0.00717740846258673 0.000358034181698980    0.00651349557333709 0.0133637955715171  0.0202018404602034  0.0270147225489001  0.0337898204971252  0.0405146415132138  0.0471768260462406  0.0537641715916784  0.0602646603043279  0.0666664873507057  0.0729580891146200  0.0791281709097673  0.0851657340195109  0.0910601019446384  0.0968009457657087  0.102378308539557   0.107782628657363   0.113004762097380   0.118036003510261   0.122868106079509   0.127493300104313   0.131904310257409   0.136094371477732   0.140057243469708   0.143787223810476   0.147279159770258   0.150528459504324   0.153531108836772   0.156280444813554   0.158783035106175   0.161027296288627   0.163014562505352   0.164743731117677   0.166214276765471   0.167426257040343   0.168380310331524   0.169077651806683   0.169520068571722   0.169709914896378   0.169650109087113   0.169344135453180   0.168796059816963   0.168010582212876   0.166993205517562   0.165750858213848   0.164295206012858   0.162692813100379   0.160590402150861   0.158550181408264   0.156271984944015   0.153800366335689]

B = [-2 -1.96000000000000   -1.92000000000000   -1.88000000000000   -1.84000000000000   -1.80000000000000   -1.76000000000000   -1.72000000000000   -1.68000000000000   -1.64000000000000   -1.60000000000000   -1.56000000000000   -1.52000000000000   -1.48000000000000   -1.44000000000000   -1.40000000000000   -1.36000000000000   -1.32000000000000   -1.28000000000000   -1.24000000000000   -1.20000000000000   -1.16000000000000   -1.12000000000000   -1.08000000000000   -1.04000000000000   -1  -0.960000000000000  -0.920000000000000  -0.880000000000000  -0.840000000000000  -0.800000000000000  -0.760000000000000  -0.720000000000000  -0.680000000000000  -0.640000000000000  -0.600000000000000  -0.560000000000000  -0.520000000000000  -0.480000000000000  -0.440000000000000  -0.400000000000000  -0.360000000000000  -0.320000000000000  -0.280000000000000  -0.240000000000000  -0.200000000000000  -0.160000000000000  -0.120000000000000  -0.0800000000000001 -0.0400000000000000 0   0.0400000000000000  0.0800000000000001  0.120000000000000   0.160000000000000   0.200000000000000   0.240000000000000   0.280000000000000   0.320000000000000   0.360000000000000   0.400000000000000   0.440000000000000   0.480000000000000   0.520000000000000   0.560000000000000   0.600000000000000   0.640000000000000   0.680000000000000   0.720000000000000   0.760000000000000   0.800000000000000   0.840000000000000   0.880000000000000   0.920000000000000   0.960000000000000   1   1.04000000000000    1.08000000000000    1.12000000000000    1.16000000000000    1.20000000000000    1.24000000000000    1.28000000000000    1.32000000000000    1.36000000000000    1.40000000000000    1.44000000000000    1.48000000000000    1.52000000000000    1.56000000000000    1.60000000000000    1.64000000000000    1.68000000000000    1.72000000000000    1.76000000000000    1.80000000000000    1.84000000000000    1.88000000000000    1.92000000000000    1.96000000000000    2];

Plotting these two plot(B,A) I get this with two maximum points at B = -1.52 and B = +1.52

I want to add automatically a point as marker in the two maximum values, a horizontal line above the highest point and a two way row pointing from the line to the second peak like this

I tried to sort A and find the position of the two maximum

[val ind] = sort(A,'descend');
max_values = val(1:2)
index = ind(1:2)
r_max = A(ind(1:2))

but the second peak is not the the second position of val because I get this sort:

  Columns 1 through 13

    0.1697    0.1697    0.1695    0.1693    0.1691    0.1688    0.1684    0.1680    0.1674    0.1670    0.1662    0.1658    0.1647

  Columns 14 through 26

    0.1643    0.1630    0.1627    0.1610    0.1606    0.1588    0.1586    0.1566    0.1566    0.1564    0.1563    0.1563    0.1563

The first value 0.1697 (in this case) is the correct one, but the second peak is not in the second position but at the 22nd position.

Looking at the plot, how can I get easily the two maximum points? Once I know the two coordinates I can easily add all the objects that I need.

Answer 1

Using findpeaks (requires Signal Processing Toolbox), yline (introduced in R2018b) and annotation :

A = [0.140673450903833  0.143148937279028   0.145430952171596   0.147474938627147   0.149581060870114   0.151187105347571   0.152646348246015   0.153892222566265   0.154913060187075   0.155701930397674   0.156253328260122   0.156562551841967   0.156625533585493   0.156438787610539   0.155999394209637   0.155304997895017   0.154353810555534   0.153144616301392   0.151676776486360   0.149950234280632   0.147965519042205   0.145723755995511   0.143229676241241   0.140476287800831   0.137475884805212   0.134228713435530   0.130738812449387   0.127010778531129   0.123049766057659   0.118861478234099   0.114452155847321   0.109828564345449   0.104997979409803   0.0999681710919947  0.0947473865690700  0.0893443315667412  0.0837681505026921  0.0780284054055798  0.0721350536699391  0.0660984247128685  0.0599291956061627  0.0536383657701255  0.0472372308396036  0.0407373558685518  0.0341505481893574  0.0274888307202605  0.0207644183921863  0.0139897104812690  0.00717740846258673 0.000358034181698980    0.00651349557333709 0.0133637955715171  0.0202018404602034  0.0270147225489001  0.0337898204971252  0.0405146415132138  0.0471768260462406  0.0537641715916784  0.0602646603043279  0.0666664873507057  0.0729580891146200  0.0791281709097673  0.0851657340195109  0.0910601019446384  0.0968009457657087  0.102378308539557   0.107782628657363   0.113004762097380   0.118036003510261   0.122868106079509   0.127493300104313   0.131904310257409   0.136094371477732   0.140057243469708   0.143787223810476   0.147279159770258   0.150528459504324   0.153531108836772   0.156280444813554   0.158783035106175   0.161027296288627   0.163014562505352   0.164743731117677   0.166214276765471   0.167426257040343   0.168380310331524   0.169077651806683   0.169520068571722   0.169709914896378   0.169650109087113   0.169344135453180   0.168796059816963   0.168010582212876   0.166993205517562   0.165750858213848   0.164295206012858   0.162692813100379   0.160590402150861   0.158550181408264   0.156271984944015   0.153800366335689];    
B = [-2 -1.96000000000000   -1.92000000000000   -1.88000000000000   -1.84000000000000   -1.80000000000000   -1.76000000000000   -1.72000000000000   -1.68000000000000   -1.64000000000000   -1.60000000000000   -1.56000000000000   -1.52000000000000   -1.48000000000000   -1.44000000000000   -1.40000000000000   -1.36000000000000   -1.32000000000000   -1.28000000000000   -1.24000000000000   -1.20000000000000   -1.16000000000000   -1.12000000000000   -1.08000000000000   -1.04000000000000   -1  -0.960000000000000  -0.920000000000000  -0.880000000000000  -0.840000000000000  -0.800000000000000  -0.760000000000000  -0.720000000000000  -0.680000000000000  -0.640000000000000  -0.600000000000000  -0.560000000000000  -0.520000000000000  -0.480000000000000  -0.440000000000000  -0.400000000000000  -0.360000000000000  -0.320000000000000  -0.280000000000000  -0.240000000000000  -0.200000000000000  -0.160000000000000  -0.120000000000000  -0.0800000000000001 -0.0400000000000000 0   0.0400000000000000  0.0800000000000001  0.120000000000000   0.160000000000000   0.200000000000000   0.240000000000000   0.280000000000000   0.320000000000000   0.360000000000000   0.400000000000000   0.440000000000000   0.480000000000000   0.520000000000000   0.560000000000000   0.600000000000000   0.640000000000000   0.680000000000000   0.720000000000000   0.760000000000000   0.800000000000000   0.840000000000000   0.880000000000000   0.920000000000000   0.960000000000000   1   1.04000000000000    1.08000000000000    1.12000000000000    1.16000000000000    1.20000000000000    1.24000000000000    1.28000000000000    1.32000000000000    1.36000000000000    1.40000000000000    1.44000000000000    1.48000000000000    1.52000000000000    1.56000000000000    1.60000000000000    1.64000000000000    1.68000000000000    1.72000000000000    1.76000000000000    1.80000000000000    1.84000000000000    1.88000000000000    1.92000000000000    1.96000000000000    2];
plot(B,A)

% Find peaks.
[maxValuesY,isMaxY]=findpeaks(A);
maxValuesX = B(isMaxY);

% Plot horizontal line.
yline(maxValuesY(2));

% Create arrow.
ar = annotation('arrow');
ar.Parent = gca;
ar.X = [maxValuesX(1), maxValuesX(1)];
ar.Y = [maxValuesY(2), maxValuesY(1)];
ar.Color = 'black';
ar.HeadLength = 3;

Thanks to marsei for tip on the position of annotation.

Answer 2

If you specifically have such plots, you can go with the following solution, which just excludes n neighbours around the first found maximum.

% Input (copy from above...)
A = [ .. ];
B = [ .. ];

% Index of max value.
[max_val, max_idx] = max(A);

% Find second max value by excluding n neighbourhood.
n = 10;
AA = A;
AA(max_idx - n : max_idx + n) = [];
sec_max_val = max(AA);
sec_max_idx = find(A == sec_max_val);

% Output.
figure(1);
hold on;

% Graph.
plot(B, A);

% Black line.
plot([B(1) B(end)], [max_val max_val], 'k');

% Black arrow.
p1 = [B(sec_max_idx) B(sec_max_idx)];
p2 = [max_val sec_max_val];
dp = p2 - p1;
quiver(p1(1), p2(1), p1(2) - p1(1), p2(2) - p2(1), 0, 'k'); 

hold off;

You'll get such an output: