How can I rewrite this to get peak locations?

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HC98 2023-2-10

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1910445-how-can-i-rewrite-this-to-get-peak-locations

评论： Star Strider 2023-2-14

I have this

amps = arrayfun(@(h2) max((findpeaks(udata(h2, :), x) < 0.99)...

.* findpeaks(udata(h2, :), x)), 1:size(udata, 1));

And I want to rewrite it to get peak locations. When I call x(amps) it gives a logic error...

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Answer 1

Star Strider 2023-2-10

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https://ww2.mathworks.cn/matlabcentral/answers/1910445-how-can-i-rewrite-this-to-get-peak-locations#answer_1168700

编辑：Star Strider 2023-2-10

在 MATLAB Online 中打开

The locations are the second output from findpeaks, so I doubt that arrayfun will do what you want.

I would do something like this —

udata = randn(10);                                          % Create 'udata' Array
for h2 = 1:size(udata,2)
    [pks,locs] = findpeaks(udata(h2, :));
    pksc{h2,:} = pks;                                       % Cell Array Of Peak Values
    locsc{h2,:} = locs;                                     % Cell Array Of Location Values
end
Lv = cellfun(@(x)x<0.99, pksc, 'Unif',0)                    % Cell Array Of Logical Vectors
Lv = 10×1 cell array
    {[  1 1 1]}
    {[  0 1 1]}
    {[  1 1 0]}
    {[0 1 1 1]}
    {[    0 1]}
    {[  1 1 1]}
    {[  1 1 1]}
    {[  1 1 1]}
    {[    0 1]}
    {[  0 1 1]}
pksm = cellfun(@(x,y)x(y), pksc, Lv, 'Unif',0)              % Peak Values Result
pksm = 10×1 cell array
    {[  0.8875 0.9108 0.2496]}
    {[       -0.3314 -0.1289]}
    {[         0.3672 0.2208]}
    {[  0.6036 0.8317 0.5501]}
    {[                0.0153]}
    {[-0.1968 -0.5469 0.4821]}
    {[  0.5430 0.8575 0.9401]}
    {[ 0.5961 0.3895 -0.1598]}
    {[                0.9421]}
    {[         0.5154 0.5672]}
locsm = cellfun(@(x,y)x(y), locsc, Lv, 'Unif',0)            % Location Values Result
locsm = 10×1 cell array
    {[3 5 8]}
    {[  7 9]}
    {[  4 6]}
    {[5 7 9]}
    {[    4]}
    {[4 6 8]}
    {[2 4 8]}
    {[4 6 8]}
    {[    7]}
    {[  7 9]}

This will store the ‘pks’ and ‘locs’ vectors in their respective cell arrays. Process them appropriately later.

The ‘Lv’ (logical vector cell array) can have more than one condition, so that you can use a range of values such as:

@(x)(x<0.99) & (x>0.5)

if that is what you want to do. The rest of the code is unchanged.

EDIT — Corrected typographical errors.

.

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Star Strider 2023-2-13

在 MATLAB Online 中打开

Try this —

Lv = cellfun(@(x) (x>1) & (x<2), pksc, 'Unif',0) % Cell Array Of Logical Vectors

Testing it with a random matrix —

udata = rand(500,15)*3;                                     % Create 'udata' Array
for h2 = 1:size(udata,2)
    [pks,locs] = findpeaks(udata(h2, :));
    pksc{h2,:} = pks;                                       % Cell Array Of Peak Values
    locsc{h2,:} = locs;                                     % Cell Array Of Location Values
end
Lv = cellfun(@(x)(x>1) & (x<2), pksc, 'Unif',0)             % Cell Array Of Logical Vectors
Lv = 15×1 cell array
    {[0 0 0 0 0]}
    {[0 0 0 0 0]}
    {[  1 0 0 0]}
    {[  0 0 1 0]}
    {[0 1 1 1 0]}
    {[0 0 1 0 1]}
    {[  1 0 0 0]}
    {[  0 0 0 1]}
    {[    0 1 0]}
    {[1 0 0 0 0]}
    {[  1 1 0 0]}
    {[  0 1 0 0]}
    {[  0 0 0 0]}
    {[    0 0 0]}
    {[  0 0 1 0]}
pksm = cellfun(@(x,y)x(y), pksc, Lv, 'Unif',0)              % Peak Values Result
pksm = 15×1 cell array
    {1×0 double            }
    {1×0 double            }
    {[              1.3583]}
    {[              1.4681]}
    {[1.6804 1.5822 1.5507]}
    {[       1.6386 1.5791]}
    {[              1.8833]}
    {[              1.5120]}
    {[              1.9282]}
    {[              1.8557]}
    {[       1.7491 1.1003]}
    {[              1.5079]}
    {1×0 double            }
    {1×0 double            }
    {[              1.8701]}
locsm = cellfun(@(x,y)x(y), locsc, Lv, 'Unif',0)            % Location Values Result
locsm = 15×1 cell array
    {1×0 double}
    {1×0 double}
    {[       4]}
    {[       9]}
    {[  5 7 11]}
    {[    9 14]}
    {[       4]}
    {[      14]}
    {[       9]}
    {[       2]}
    {[     3 6]}
    {[       4]}
    {1×0 double}
    {1×0 double}
    {[      11]}
x = linspace(0, (size(udata,1)-1)/100, size(udata,1));          % Create Indepdent Variable Vector
for k = 1:numel(locsm)
    if any(locsm{k})
        xv{k,:} = x(locsm{k});
    else
        xv{k,:} = NaN;
    end
end
xv
xv = 15×1 cell array
    {[                 NaN]}
    {[                 NaN]}
    {[              0.0300]}
    {[              0.0800]}
    {[0.0400 0.0600 0.1000]}
    {[       0.0800 0.1300]}
    {[              0.0300]}
    {[              0.1300]}
    {[              0.0800]}
    {[              0.0100]}
    {[       0.0200 0.0500]}
    {[              0.0300]}
    {[                 NaN]}
    {[                 NaN]}
    {[              0.1000]}

There are peaks in every vector, however only some of them meet the criteria (the ‘1’ values in ‘Lv’). The corresponding peak values are returned in ‘pksm’ and their location indices in the peak array are returned in ‘locsm’.

These results will need to be processed as individual cells (not as matrices), since it is unlikely that there will be the same number of columns in every cell.

Get the actual values of the independent variable vector (here ‘x’) at the peak locations as illustrated in the last loop (added here to my original code to demonstrate that). It includes a test to be certain that only ‘locsm’ cells with valid indices are addressed, returning NaN for the others. You can use that approach for other processing steps, as well.

.

HC98 2023-2-14

Yes so sorry for not being very clear there. What I'd like to do is find peaks between 0.96 and 1.96 but when I do that it just returns an array of zeros for locsm :(

Star Strider 2023-2-14

That means that there are no peaks in that column that meet the criteria. You can verify that by looking at the ‘pksc’ cell for that particular column:

pksv = pksc{column_number}

You can see that also occurs in my demonstration in my previous Comment. There may not be any peaks in a particular column that meet the criteria, and in that instance, my code assigns the cell as NaN (all detected peaks are false).

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